4.6: Exercise- Applying the Impulse Response Function to a Series of Impulses
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For finite impulse response function filters, the output of the filter for a complex waveform is simply the sum of the filter’s impulse response function for the voltages at each time point, scaled by the input amplitude at each time point. That’s a pretty complicated sentence, so in this exercise we’ll look at a couple simple examples.
You can start by loading and plotting the ERPset file names impulse1.erp . This ERPset contains one bin, and each channel has a different impulse in it. Channel 1 has an impulse of 1 µV at 100 ms. Channels 2, 3, and 4 have impulses at 200, 300, and 400 ms with amplitudes of 0.5, 1, and 0.5 µV, respectively. Channel 5 has all four of the impulses in it (see Screenshot 4.9.A).
Filter this ERPset using a half-amplitude cutoff at 30 Hz and a slope of 12 dB/octave (following the same steps you used in the previous exercises) and plot the result (which should look like Screenshot 4.9.B). Each impulse has been replaced by the impulse response function of the filter, shifted so that it is centered at the latency of the impulse, and scaled (multiplied) by the size of the impulse. Importantly, filtering the waveform with four impulses (Channel 5) gives you a waveform that is equivalent to the sum of the four filtered waveforms for the individual impulses (Channels 1-4). This shows you what I meant when I said that “the output of the filter for a complex waveform is simply the sum of the filter’s impulse response function for the voltages at each time point, scaled by the input amplitude at each time point.” That is, the output of the filter for the waveform with four impulses (Channel 5) is equivalent to replacing each individual impulse with a copy of the impulse response function that has been shifted to be centered on a given impulse and scaled by the amplitude of that impulse.
To make this even clearer, we’re going to take the four impulses and make them consecutive sample points (just as an ERP waveform typically consists of a sequence of consecutive nonzero voltage values). Load the ERPset named impulse2.erp and plot it. You’ll see that now our four impulses are at 100, 105, 110, and 115 ms, which are consecutive because we have a sampling rate of 200 Hz and therefore a voltage value every 5 ms. Filter this ERPset using a half-amplitude cutoff at 30 Hz and a slope of 12 dB/octave (just as before) and plot the result. Just as in the previous example, filtering the set of four consecutive impulses is equivalent to filtering each impulse separately and then summing them together. In other words, the filtered waveform is equivalent to replacing each impulse in the unfiltered waveform with a copy of the impulse response function, centered on each impulse and scaled by the height of each impulse.
In this exercise, we used impulses to create 4 time points in an ERP waveform. Figure 4.1 extends this idea to a more realistic ERP waveform. Panel A is the same artificial waveform shown at the beginning of the chapter in Screenshot 4.1, but blown up. We have a voltage value every 5 ms, and these voltage values are connected by lines to create a waveform. Panel B is the same set of voltage values, but with the voltage at each time point shown as an impulse. This is conceptually identical to the set of four impulses shown in Screenshot 4.9.A, except now we have an impulse at each time point. To filter this waveform, we just replace each of these impulses with a copy of the impulse response function, centered at each impulse and scaled (multiplied) by the amplitude of the impulse. We then sum together these scaled copies of the impulse response function to obtain our filtered ERP waveform.
The process of replacing each point in one waveform with a scaled copy of another waveform is called convolving the two waveforms. So, filtering an ERP waveform is achieved by convolving the waveform with the impulse response function. It turns out that convolution is not as complicated (or convoluted) as it sounds!
I hope you can now see that filtering in the time domain is conceptually very simple as long as you know the impulse response function of the filter. That’s why we designed the filtering tool in ERPLAB to show you this function. Many EEG/ERP analysis systems don’t show you the impulse response function, but you can always figure it out by making a waveform that consists of a single impulse (like the one shown in Screenshot 4.7.A) and passing it through the filter.
By knowing the impulse response function, you can make a pretty good guess about how the filter might distort your data. For example, do you remember the artificial negative peak produced by the filter with the 10 Hz cutoff and 48 dB/octave roll-off (Screenshot 4.6)? That artificial peak makes perfect sense once you see the impulse response function of the filter. To see the impulse response function for this filter, load the impulse1.erp ERPset (or make it the active ERPset if it’s already loaded) and filter it using a 10 Hz half-amplitude cutoff and a roll-off of 48 dB/octave. If you plot the filtered data, you’ll see something like Screenshot 4.10.
Channel 1 shows you what the impulse response function looks like (but centered at 100 ms rather than 0 ms because the impulse was at 100 ms). It has a negative dip on each side of the peak. And when we filter the set of four impulses in Channel 5, we can see this dip just before the first positive peak. Now imagine what happens when you apply this filter to the more realistic waveform shown in Figure 4.1.A. This would involve replacing each point in the waveform with a scaled copy of the impulse response function. When we replace the positive impulses that start around 50 ms with this function, the negative part of the impulse response function generates the negative dip prior to 50 ms.
This sort of distortion is easy to understand when you think about filtering in the time domain using the impulse response function. However, the distortion is not at all obvious when you think about filtering using the frequency response function. That’s why I prefer to think about filtering in the time domain. However, it’s still useful to know the frequency response function, especially if you know something about the frequency content of the noise in your data. This is why we designed the ERPLAB filtering tool to provide you with both functions. Also, the frequency response function and impulse response function are very closely related: The frequency response function is simply the Fourier transform of the impulse response function, and the impulse response function is simply the inverse Fourier transform of the frequency response function.


1.3: Varying solubility of ionic compounds
Le Chatelier's principle
When ionic compounds dissolve in water, they dissociate into ions. An equilibrium is established between the ions in water and the undissolved compound. Changes in the concentration of the ions can stress the equilibrium and vary the solubility of the ionic compounds.
Le Chatelier's principle states, “When a system at equilibrium is subjected to a change in concentration, temperature, volume, or pressure, the system will change to a new equilibrium, such that the applied change is partially counteracted.”
If the ions in the solubility equilibrium are increased or decreased by another reaction going on in parallel, the equilibrium will counteract by decreasing or increasing the compound's solubility. The following explains how to use Le Chatelier's principle to vary the solubility of sparingly soluble ionic compounds with examples.
Common ion effect
Consider dissolution of a sparingly soluble ionic compound \(\ce{CaF2}\) in water:
\[\ce{CaF2(s) <=> Ca^{2+}(aq) + 2F^{-}(aq)},\quad K_{sp} = \ce{[Ca^{2+}][F^{-}]^2} = 1.5\times 10^{-10}\nonumber\]
The solubility (S) can be expressed in the units of mol/L or molarity (M). Similarly, the concentration of any species in square brackets, as [\(\ce{Ca^{2+}}\)] in the above-mentioned \(\ce{K_{sp}}\) expression, is also in the units of mol/L or M.
\(\ce{NaF}\) is a water-soluble ionic compound that has \(\ce{F^-}\) in common with the above equilibrium. The addition of \(\ce{NaF}\) into the mixture will increase the concentration of \(\ce{F^-}\), causing a decrease in the solubility of \(\ce{CaF2}\) because the solubility equilibrium will move in the reverse direction to counteract the rise in the concentration of the common ion. This is called the common ion effect .
The common ion effect refers to the decrease in the solubility of a sparingly soluble ionic compound by adding a soluble ionic compound with an ion in common with the sparingly soluble one.
A quantitative estimate of this common ion effect is given with the help of the following calculations. If solubility of \(\ce{CaF2}\) in pure water is S mol/L, then [ \(\ce{Ca^{2+}}\)] = S, and [ \(\ce{F^-}\)] = 2S. Plugging in these values in the \(\ce{K_{sp}}\) expression and rearranging shows that the solubility of \(\ce{CaF2}\) in pure water is 3.3 x 10 -4 M:
\[K_{sp} = \ce{[Ca^{2+}][F^{-}]^2}\nonumber\]
\[1.5 \times 10^{-10} = S(2S)^2\nonumber\]
\[S=\sqrt[3]{1.5 \times 10^{-10} / 4}=3.310^{-4} \mathrm{~M}\nonumber\]
If \(\ce{F^-}\) concentration is raised to 0.1M by dissolving \(\ce{NaF}\) in the solution, then the molar solubility of \(\ce{CaF2}\) changes to a new value S i , [ \(\ce{Ca^{2+}}\)] = S i , and [ \(\ce{F^-}\)] = (0.1 + S i ) = 0.1 (S i cancels because it is negligible compared to 0.1). Plugging in these values in the \(\ce{K_{sp}}\) expression and rearranging shows that the new solubility (S i ) of \(\ce{CaF2}\) is 1.5 x 10 -8 M:
\[K_{sp} = \ce{[Ca^{2+}][F^{-}]^2}\nonumber\]
\[1.5 \times 10^{-10} = S_{i}(0.1)^2\nonumber\]
\[S_{i} = \frac{1.5 \times 10^{-10}}{(0.1)^2} = 1.5\times10^{-8} \mathrm{~M}\nonumber\]
It means the common ion effect described above decreases the solubility of \(\ce{CaF2}\) by more than twenty thousand times.
Generally, the solubility of sparingly soluble ionic compounds decreases when a common ion is added to the equilibrium mixture.
An example of a common ion effect is in the separation of \(\ce{PbCl2}\) from \(\ce{AgCl}\) and \(\ce{Hg2Cl2}\) precipitates. \(\ce{PbCl}\) is the most soluble in hot water among these three sparingly soluble compounds. So, \(\ce{PbCl2}\) is selectively dissolved in hot water and separated. The solution is then cooled to room temperate and \(\ce{HCl}\) is added to it as a source of common ion \(\ce{Cl^-}\) to enforce re-precipitation of \(\ce{PbCl2}\):
\[\ce{Pb^{2+}(aq) + 2Cl^{-}(aq) <=> PbCl2(s)}\nonumber\]
Effect of pH
The pH is related to the concentration of \(\ce{H3O^+}\) and \(\ce{OH^-}\) in the solution. Increasing pH increases \(\ce{OH^-}\) and decreases \(\ce{H3O^+}\) concentration in the solution, and decreasing pH has the opposite effect. If one of the ions in the solubility equilibrium of a sparingly soluble ionic compound is an acid or a base, its concentration will change with changes in the pH. It is because acids will neutralize with \(\ce{OH^-}\) at high pH, and bases will neutralize with \(\ce{H3O^+}\) at low pH. For example, consider the dissolution of \(\ce{Mg(OH)2}\) in pure water.
\[\ce{Mg(OH)2(s) <=> Mg^{2+}(aq) + 2OH^{-}(aq)},\quad K_{sp} = \ce{[Mg^{2+}][OH^{-}]^2} = 2.1\times 10^{-13}\nonumber\]
Making the solution acidic, i.e., a decrease in pH adds more \(\ce{H3O^+}\) ion that removes \(\ce{OH^-}\) by the following neutralization reaction.\[\ce{H3O^{+}(aq) + OH^{-}(aq) <=> 2H2O(l)}\nonumber\]
According to Le Chatelier's principle, the system moves forward to compensate for the loss of \(\ce{OH^-}\). In other words, \(\ce{Mg(OH)2}\) is insoluble in neutral or alkaline water and becomes soluble in acidic water.
Generally, the solubility of an ionic compound containing basic anion increases by decreasing pH, i.e., in an acidic medium.
In a qualitative analysis of cations, dissociation of \(\ce{H2S}\) is used as a source of \(\ce{S^{2-}}\) ions:
\[\ce{H2S(g) + 2H2O(l) <=> 2H3O^+(aq) + S^{2-}(aq)}\nonumber\]
The reaction is pH-dependent, i.e., the extent of dissociation of \(\ce{H2S}\) can be decreased by adding \(\ce{HCl}\) as a source of common ion \(\ce{H3O^+}\) or increased by adding a base as a source of \(\ce{OH^-}\) that removes \(\ce{H3O^+}\) from the products:
\[\ce{OH^{-}(aq) + H3O^{+}(aq) <=> 2H2O(l)}\nonumber\]
Generally, the solubility of weak acids can be increased by increasing the pH and decreased by decreasing the pH. The opposite is true for the weak bases.
Complex ion equilibria
Transition metal ions, like \(\ce{Ag^+}\), \(\ce{Cu^{2+}}\), \(\ce{Ni^{2+}}\), etc. tend to be strong Lewis acids, i.e., they can accept a lone pair of electrons from Lewis bases. Neutral or anionic species with a lone pair of electrons, like \(\ce{H2{\!\overset{\Large{\cdot\cdot}}{O}}\!:}\), \(\ce{:\!{NH3}}\), \(\ce{:\!\overset{-}{C}N\!:}\), \(\ce{:\!\overset{\Large{\cdot\cdot}}{\underset{\Large{\cdot\cdot}}{Cl}}\!:^{-}}\), etc. can act as Lewis bases in these reactions. The bond formed by donating a lone pair of electrons of a Lewis base to a Lewis acid is called a coordinate covalent bond . The neutral compound or ion that results from the Lewis acid-base reaction is called a coordination complex or a complex ion . For example, silver ion dissolved in water is often written as Ag + (aq) , but, in reality, it exists as complex ion \(\ce{Ag(H2O)2^+}\) in which \(\ce{Ag^+}\) accepts lone pair of electrons from oxygen atoms in water molecules. Transition metal ion in a coordination complex or complex ion, e.g., \(\ce{Ag^+}\) in \(\ce{Ag(H2O)2^+}\) is called central metal ion and the Lewis base like \(\ce{H2{\!\overset{\Large{\cdot\cdot}}{O}}\!:}\), in \(\ce{Ag(H2O)2^+}\), is called a ligand . A ligand's strength is its ability to donate its lone pair of electrons to a central metal ion. If a stronger ligand is added to the solution, it displaces a weaker ligand. For example, if \(\ce{:\!{NH3}}\) is dissolved in the solution containing \(\ce{Ag(H2O)2^+}\), the \(\ce{:\!{NH3}}\) displaces \(\ce{H2{\!\overset{\Large{\cdot\cdot}}{O}}\!:}\) from the complex ion:
\[\ce{Ag(H2O)2^{+}(aq) + 2NH3(aq) <=> Ag(NH3)2^{+}(aq) + 2H2O(aq)}\nonumber\]
The lone pair on the ligand is omitted from the equation above and the following equations. Water is usually omitted from the equation for simplicity, that reduces the above reaction to the following:
\[\ce{Ag^{+}(aq) + 2NH3(aq) <=> Ag(NH3)2^{+}(aq)}\quad K_f = 1.7\times10^7\nonumber\]
The equilibrium constant for the formation of complex ion is called formation constant (\(\ce{K_{f}}\)), e.g., in the case of the above reaction:
\[K_f = \frac{\ce{[Ag(NH3)2^{+}]}}{\ce{[Ag^+]\times[NH3]^2}} = 1.7\times10^7\nonumber\]
A large value of \(\ce{K_{f}}\) in the above reaction shows that the reaction is highly favored in the forward direction. If ammonia is present in water, it increases the solubility of \(\ce{AgCl}\) by removing the \(\ce{Ag^+}\) ion from the products, just like acid (\(\ce{H3O^+}\)) increases the solubility of \(\ce{Mg(OH)2}\) by removing \(\ce{OH^-}\) from the products:
\[\ce{AgCl(s) <<=> Ag^{+}(aq) + Cl^{-}(aq)}\quad K_f = 1.8\times10^{-10}\nonumber\]
\[\ce{Ag^{+}(aq) + 2NH3(aq) <=>> Ag(NH3)2^{+}(aq)}\quad K_f = 1.7\times10^7\nonumber\]
\[\text{Adding above reactions:}~\ce{AgCl(aq) + 2NH3(aq) <=> Ag(NH3)2^{+}(aq) + Cl^{-}(aq)}\quad K = 3.0\times10^{-3}\nonumber\]
The equilibrium constant for the dissolution of \(\ce{AgCl}(s)\) changes from 1.8 x 10 -10 in pure water to 3.0 x 10 -3 in the water containing dissolved ammonia, i.e., a 17 million times increase. It makes insoluble \(\ce{AgCl}(s)\) quite soluble. This reaction is used to separate silver ions from mercury ions in a mixture of \(\ce{AgCl}\) and \(\ce{Hg2Cl2}\) mixture precipitates.
Generally, adding a strong ligand to the solution increases the solubility of metal compounds containing metals capable of coordinate complex formation.
The chemical equations can be manipulated like algebraic equations, i.e., multiplied or divided by a constant, added, and subtracted, as demonstrated in the example of the silver ammonia complex formation reactions above. Note that the species on the right side of the equation cancels the same species on the left side of any other equation like algebraic equations, e.g., \(Ag^{+}\) is canceled in the final equation.
When two equilibrium reactions are added, their equilibrium constants are multiplied to get the equilibrium constant of the overall reaction, i.e., \(K = K_{sp}\times{K_f}\) in the above reactions.
Redox reactions
There are three major types of chemical reactions: precipitation reactions, acid-base reactions, and redox reactions.
Precipitation reactions of ionic compounds are double replacement reactions where the cation of one compound combines with the anion of another and vice versa, such that one of the new combinations is an insoluble salt.
For example, when silver nitrate (\(\ce{AgNO3}\)) solution is mixed with sodium chloride (\(\ce{NaCl}\)) solution, an insoluble compound silver chloride (\(\ce{AgCl}\)) precipitates out of the solution:
\[\ce{AgNO3(aq) + HCl(aq) -> AgCl(s)v + NaNO3(aq)}\nonumber\]
Acid -base reactions are the reactions involving the transfer of a proton.
For example, \(\ce{H2S}\) dissociates in water by donating its proton to water molecules:
\[\ce{H2S(g) + 2H2O(l) <=> 2H3O^{+}(aq) + S^{2-}(aq)}\nonumber\]
Redox reactions are reactions involving the transfer of electrons.
For example, when sodium metal (\(\ce{Na}\)) reacts with chlorine gas (\(\ce{Cl2}\)), sodium loses electrons and becomes \(\ce{Na^+}\) cation and chlorine gains electrons and becomes \(\ce{Cl^-}\) anion that combine to form \(\ce{NaCl}\) salt:
\[\ce{2Na(s) + 2Cl2(g) -> NaCl(s)}\nonumber\]
An example of a redox reaction in qualitative analysis of cations is the dissolution of \(\ce{NiS}\) precipitate by adding an oxidizing acid \(\ce{HNO3}\). The \(\ce{S^{2-}}\) is a weak base that can be removed from the product by adding a strong acid like \(\ce{HCl}\):
\[\ce{S^{2-}(aq) + 2H3O^{+}(aq) <=>> H2S(aq) + 2H2O(l)}\nonumber\]
Therefore, addition of \(\ce{HCl}\) is sufficient to dissolve \(\ce{FS}\) precipitate by removal of \(\ce{S^{2-}}\) from the products:
\[\ce{FeS(s) + 2H3O^{+}(aq) <=>> Fe^{2+}(aq) + H2S(aq) + 2H2O(l)}\nonumber\]
However, the addition of \(\ce{HCl}\) does not remove \(\ce{S^{2-}}\) sufficient enough to dissolve a relatively less soluble \(\ce{NiS}\) precipitate. Nitric acid (\(\ce{HNO3}\)) that is a source of an oxidizing agent \(\ce{NO3^{-}}\) is needed to remove \(\ce{S^{2-}}\) to a higher extent for dissolving \(\ce{NiS}\):
\[\ce{3S^{2-}(aq) + 2NO3^{-}(aq) + 8H3O^{+}(aq) -> 3S(s, yellow)(v) + 2NO(g)(^) + 12H2O(l)}\nonumber\]
In this reaction, sulfur is oxidized from an oxidation state of -2 in \(\ce{S^{2-}}\) to an oxidation state of zero in \(\ce{S}\), and nitrogen is reduced from an oxidation state of +5 in \(\ce{NO3^{-}}\) to an oxidation state of +2 in \(\ce{NO}\).


14.2: Central Processing
Learning Objectives
- Describe the pathways that sensory systems follow into the central nervous system
- Differentiate between the two major ascending pathways in the spinal cord
- Describe the pathway of somatosensory input from the face and compare it to the ascending pathways in the spinal cord
- Explain topographical representations of sensory information in at least two systems
- Describe two pathways of visual processing and the functions associated with each
Sensory Pathways
Specific regions of the CNS coordinate different somatic processes using sensory inputs and motor outputs of peripheral nerves. A simple case is a reflex caused by a synapse between a dorsal sensory neuron axon and a motor neuron in the ventral horn. More complex arrangements are possible to integrate peripheral sensory information with higher processes. The important regions of the CNS that play a role in somatic processes can be separated into the spinal cord brain stem, diencephalon, cerebral cortex, and subcortical structures.
Spinal Cord and Brain Stem
A sensory pathway that carries peripheral sensations to the brain is referred to as an ascending pathway , or ascending tract. The various sensory modalities each follow specific pathways through the CNS. Tactile and other somatosensory stimuli activate receptors in the skin, muscles, tendons, and joints throughout the entire body. However, the somatosensory pathways are divided into two separate systems on the basis of the location of the receptor neurons. Somatosensory stimuli from below the neck pass along the sensory pathways of the spinal cord, whereas somatosensory stimuli from the head and neck travel through the cranial nerves—specifically, the trigeminal system.
The dorsal column system (sometimes referred to as the dorsal column–medial lemniscus) and the spinothalamic tract are two major pathways that bring sensory information to the brain (Figure \(\PageIndex{1}\)). The sensory pathways in each of these systems are composed of three successive neurons.
The dorsal column system begins with the axon of a dorsal root ganglion neuron entering the dorsal root and joining the dorsal column white matter in the spinal cord. As axons of this pathway enter the dorsal column, they take on a positional arrangement so that axons from lower levels of the body position themselves medially, whereas axons from upper levels of the body position themselves laterally. The dorsal column is separated into two component tracts, the fasciculus gracilis that contains axons from the legs and lower body, and the fasciculus cuneatus that contains axons from the upper body and arms.
The axons in the dorsal column terminate in the nuclei of the medulla, where each synapses with the second neuron in their respective pathway. The nucleus gracilis is the target of fibers in the fasciculus gracilis, whereas the nucleus cuneatus is the target of fibers in the fasciculus cuneatus. The second neuron in the system projects from one of the two nuclei and then decussates , or crosses the midline of the medulla. These axons then continue to ascend the brain stem as a bundle called the medial lemniscus . These axons terminate in the thalamus, where each synapses with the third neuron in their respective pathway. The third neuron in the system projects its axons to the postcentral gyrus of the cerebral cortex, where somatosensory stimuli are initially processed and the conscious perception of the stimulus occurs.
The spinothalamic tract also begins with neurons in a dorsal root ganglion. These neurons extend their axons to the dorsal horn, where they synapse with the second neuron in their respective pathway. The name “spinothalamic” comes from this second neuron, which has its cell body in the spinal cord gray matter and connects to the thalamus. Axons from these second neurons then decussate within the spinal cord and ascend to the brain and enter the thalamus, where each synapses with the third neuron in its respective pathway. The neurons in the thalamus then project their axons to the spinothalamic tract, which synapses in the postcentral gyrus of the cerebral cortex.
These two systems are similar in that they both begin with dorsal root ganglion cells, as with most general sensory information. The dorsal column system is primarily responsible for touch sensations and proprioception, whereas the spinothalamic tract pathway is primarily responsible for pain and temperature sensations. Another similarity is that the second neurons in both of these pathways are contralateral, because they project across the midline to the other side of the brain or spinal cord. In the dorsal column system, this decussation takes place in the brain stem; in the spinothalamic pathway, it takes place in the spinal cord at the same spinal cord level at which the information entered. The third neurons in the two pathways are essentially the same. In both, the second neuron synapses in the thalamus, and the thalamic neuron projects to the somatosensory cortex.
The trigeminal pathway carries somatosensory information from the face, head, mouth, and nasal cavity. As with the previously discussed nerve tracts, the sensory pathways of the trigeminal pathway each involve three successive neurons. First, axons from the trigeminal ganglion enter the brain stem at the level of the pons. These axons project to one of three locations. The spinal trigeminal nucleus of the medulla receives information similar to that carried by spinothalamic tract, such as pain and temperature sensations. Other axons go to either the chief sensory nucleus in the pons or the mesencephalic nuclei in the midbrain. These nuclei receive information like that carried by the dorsal column system, such as touch, pressure, vibration, and proprioception. Axons from the second neuron decussate and ascend to the thalamus along the trigeminothalamic tract. In the thalamus, each axon synapses with the third neuron in its respective pathway. Axons from the third neuron then project from the thalamus to the primary somatosensory cortex of the cerebrum.
The sensory pathway for gustation travels along the facial and glossopharyngeal cranial nerves, which synapse with neurons of the solitary nucleus in the brain stem. Axons from the solitary nucleus then project to the ventral posterior nucleus of the thalamus. Finally, axons from the ventral posterior nucleus project to the gustatory cortex of the cerebral cortex, where taste is processed and consciously perceived.
The sensory pathway for audition travels along the vestibulocochlear nerve, which synapses with neurons in the cochlear nuclei of the superior medulla. Within the brain stem, input from either ear is combined to extract location information from the auditory stimuli. Whereas the initial auditory stimuli received at the cochlea strictly represent the frequency—or pitch—of the stimuli, the locations of sounds can be determined by comparing information arriving at both ears.
Sound localization is a feature of central processing in the auditory nuclei of the brain stem. Sound localization is achieved by the brain calculating the interaural time difference and the interaural intensity difference . A sound originating from a specific location will arrive at each ear at different times, unless the sound is directly in front of the listener. If the sound source is slightly to the left of the listener, the sound will arrive at the left ear microseconds before it arrives at the right ear (Figure \(\PageIndex{2}\)). This time difference is an example of an interaural time difference. Also, the sound will be slightly louder in the left ear than in the right ear because some of the sound waves reaching the opposite ear are blocked by the head. This is an example of an interaural intensity difference.
Auditory processing continues on to a nucleus in the midbrain called the inferior colliculus . Axons from the inferior colliculus project to two locations, the thalamus and the superior colliculus . The medial geniculate nucleus of the thalamus receives the auditory information and then projects that information to the auditory cortex in the temporal lobe of the cerebral cortex. The superior colliculus receives input from the visual and somatosensory systems, as well as the ears, to initiate stimulation of the muscles that turn the head and neck toward the auditory stimulus.
Balance is coordinated through the vestibular system, the nerves of which are composed of axons from the vestibular ganglion that carries information from the utricle, saccule, and semicircular canals. The system contributes to controlling head and neck movements in response to vestibular signals. An important function of the vestibular system is coordinating eye and head movements to maintain visual attention. Most of the axons terminate in the vestibular nuclei of the medulla. Some axons project from the vestibular ganglion directly to the cerebellum, with no intervening synapse in the vestibular nuclei. The cerebellum is primarily responsible for initiating movements on the basis of equilibrium information.
Neurons in the vestibular nuclei project their axons to targets in the brain stem. One target is the reticular formation, which influences respiratory and cardiovascular functions in relation to body movements. A second target of the axons of neurons in the vestibular nuclei is the spinal cord, which initiates the spinal reflexes involved with posture and balance. To assist the visual system, fibers of the vestibular nuclei project to the oculomotor, trochlear, and abducens nuclei to influence signals sent along the cranial nerves. These connections constitute the pathway of the vestibulo-ocular reflex (VOR) , which compensates for head and body movement by stabilizing images on the retina (Figure \(\PageIndex{3}\)). Finally, the vestibular nuclei project to the thalamus to join the proprioceptive pathway of the dorsal column system, allowing conscious perception of equilibrium.
The connections of the optic nerve are more complicated than those of other cranial nerves. Instead of the connections being between each eye and the brain, visual information is segregated between the left and right sides of the visual field. In addition, some of the information from one side of the visual field projects to the opposite side of the brain. Within each eye, the axons projecting from the medial side of the retina decussate at the optic chiasm . For example, the axons from the medial retina of the left eye cross over to the right side of the brain at the optic chiasm. However, within each eye, the axons projecting from the lateral side of the retina do not decussate. For example, the axons from the lateral retina of the right eye project back to the right side of the brain. Therefore the left field of view of each eye is processed on the right side of the brain, whereas the right field of view of each eye is processed on the left side of the brain (Figure \(\PageIndex{4}\)).
A unique clinical presentation that relates to this anatomic arrangement is the loss of lateral peripheral vision, known as bilateral hemianopia. This is different from “tunnel vision” because the superior and inferior peripheral fields are not lost. Visual field deficits can be disturbing for a patient, but in this case, the cause is not within the visual system itself. A growth of the pituitary gland presses against the optic chiasm and interferes with signal transmission. However, the axons projecting to the same side of the brain are unaffected. Therefore, the patient loses the outermost areas of their field of vision and cannot see objects to their right and left.
Extending from the optic chiasm, the axons of the visual system are referred to as the optic tract instead of the optic nerve. The optic tract has three major targets, two in the diencephalon and one in the midbrain. The connection between the eyes and diencephalon is demonstrated during development, in which the neural tissue of the retina differentiates from that of the diencephalon by the growth of the secondary vesicles. The connections of the retina into the CNS are a holdover from this developmental association. The majority of the connections of the optic tract are to the thalamus—specifically, the lateral geniculate nucleus . Axons from this nucleus then project to the visual cortex of the cerebrum, located in the occipital lobe. Another target of the optic tract is the superior colliculus.
In addition, a very small number of RGC axons project from the optic chiasm to the suprachiasmatic nucleus of the hypothalamus. These RGCs are photosensitive, in that they respond to the presence or absence of light. Unlike the photoreceptors, however, these photosensitive RGCs cannot be used to perceive images. By simply responding to the absence or presence of light, these RGCs can send information about day length. The perceived proportion of sunlight to darkness establishes the circadian rhythm of our bodies, allowing certain physiological events to occur at approximately the same time every day.
Diencephalon
The diencephalon is beneath the cerebrum and includes the thalamus and hypothalamus. In the somatic nervous system, the thalamus is an important relay for communication between the cerebrum and the rest of the nervous system. The hypothalamus has both somatic and autonomic functions. In addition, the hypothalamus communicates with the limbic system, which controls emotions and memory functions.
Sensory input to the thalamus comes from most of the special senses and ascending somatosensory tracts. Each sensory system is relayed through a particular nucleus in the thalamus. The thalamus is a required transfer point for most sensory tracts that reach the cerebral cortex, where conscious sensory perception begins. The one exception to this rule is the olfactory system. The olfactory tract axons from the olfactory bulb project directly to the cerebral cortex, along with the limbic system and hypothalamus.
The thalamus is a collection of several nuclei that can be categorized into three anatomical groups. White matter running through the thalamus defines the three major regions of the thalamus, which are an anterior nucleus, a medial nucleus, and a lateral group of nuclei. The anterior nucleus serves as a relay between the hypothalamus and the emotion and memory-producing limbic system. The medial nuclei serve as a relay for information from the limbic system and basal ganglia to the cerebral cortex. This allows memory creation during learning, but also determines alertness. The special and somatic senses connect to the lateral nuclei, where their information is relayed to the appropriate sensory cortex of the cerebrum.
Cortical Processing
As described earlier, many of the sensory axons are positioned in the same way as their corresponding receptor cells in the body. This allows identification of the position of a stimulus on the basis of which receptor cells are sending information. The cerebral cortex also maintains this sensory topography in the particular areas of the cortex that correspond to the position of the receptor cells. The somatosensory cortex provides an example in which, in essence, the locations of the somatosensory receptors in the body are mapped onto the somatosensory cortex. This mapping is often depicted using a sensory homunculus (Figure \(\PageIndex{5}\)).
The term homunculus comes from the Latin word for “little man” and refers to a map of the human body that is laid across a portion of the cerebral cortex. In the somatosensory cortex, the external genitals, feet, and lower legs are represented on the medial face of the gyrus within the longitudinal fissure. As the gyrus curves out of the fissure and along the surface of the parietal lobe, the body map continues through the thighs, hips, trunk, shoulders, arms, and hands.  The representation of the body in this topographical map is medial to lateral from the lower to upper body. It is a continuation of the topographical arrangement seen in the dorsal column system, where axons from the lower body are carried in the fasciculus gracilis, whereas axons from the upper body are carried in the fasciculus cuneatus. As the dorsal column system continues into the medial lemniscus, these relationships are maintained. Also, the head and neck axons running from the trigeminal nuclei to the thalamus run adjacent to the upper body fibers. The connections through the thalamus maintain topography such that the anatomic information is preserved. Note that this correspondence does not result in a perfectly miniature scale version of the body, but rather exaggerates the more sensitive areas of the body, such as the fingers and lower face. Less sensitive areas of the body, such as the shoulders and back, are mapped to smaller areas on the cortex.
Likewise, the topographic relationship between the retina and the visual cortex is maintained throughout the visual pathway. The visual field is projected onto the two retinae, as described above, with sorting at the optic chiasm. The right peripheral visual field falls on the medial portion of the right retina and the lateral portion of the left retina. The right medial retina then projects across the midline through the optic chiasm. This results in the right visual field being processed in the left visual cortex. Likewise, the left visual field is processed in the right visual cortex (see Figure \(\PageIndex{4}\)). Though the chiasm is helping to sort right and left visual information, superior and inferior visual information is maintained topographically in the visual pathway. Light from the superior visual field falls on the inferior retina, and light from the inferior visual field falls on the superior retina. This topography is maintained such that the superior region of the visual cortex processes the inferior visual field and vice versa. Therefore, the visual field information is inverted and reversed as it enters the visual cortex—up is down, and left is right. However, the cortex processes the visual information such that the final conscious perception of the visual field is correct. The topographic relationship is evident in that information from the foveal region of the retina is processed in the center of the primary visual cortex. Information from the peripheral regions of the retina are correspondingly processed toward the edges of the visual cortex. Similar to the exaggerations in the sensory homunculus of the somatosensory cortex, the foveal-processing area of the visual cortex is disproportionately larger than the areas processing peripheral vision.
In an experiment performed in the 1960s, subjects wore prism glasses so that the visual field was inverted before reaching the eye. On the first day of the experiment, subjects would duck when walking up to a table, thinking it was suspended from the ceiling. However, after a few days of acclimation, the subjects behaved as if everything were represented correctly. Therefore, the visual cortex is somewhat flexible in adapting to the information it receives from our eyes (Figure \(\PageIndex{6}\)).
The cortex has been described as having specific regions that are responsible for processing specific information; there is the visual cortex, somatosensory cortex, gustatory cortex, etc. However, our experience of these senses is not divided. Instead, we experience what can be referred to as a seamless percept. Our perceptions of the various sensory modalities—though distinct in their content—are integrated by the brain so that we experience the world as a continuous whole.
In the cerebral cortex, sensory processing begins at the primary sensory cortex , then proceeds to an association area , and finally, into a multimodal integration area . For example, the visual pathway projects from the retinae through the thalamus to the primary visual cortex in the occipital lobe. This area is primarily in the medial wall within the longitudinal fissure. Here, visual stimuli begin to be recognized as basic shapes. Edges of objects are recognized and built into more complex shapes. Also, inputs from both eyes are compared to extract depth information. Because of the overlapping field of view between the two eyes, the brain can begin to estimate the distance of stimuli based on binocular depth cues .
Watch this video to learn more about how the brain perceives 3-D motion. Similar to how retinal disparity offers 3-D moviegoers a way to extract 3-D information from the two-dimensional visual field projected onto the retina, the brain can extract information about movement in space by comparing what the two eyes see. If movement of a visual stimulus is leftward in one eye and rightward in the opposite eye, the brain interprets this as movement toward (or away) from the face along the midline. If both eyes see an object moving in the same direction, but at different rates, what would that mean for spatial movement?
Depth Perception, 3-D Movies, and Optical Illusions
The visual field is projected onto the retinal surface, where photoreceptors transduce light energy into neural signals for the brain to interpret. The retina is a two-dimensional surface, so it does not encode three-dimensional information. However, we can perceive depth. How is that accomplished?
Two ways in which we can extract depth information from the two-dimensional retinal signal are based on monocular cues and binocular cues, respectively. Monocular depth cues are those that are the result of information within the two-dimensional visual field. One object that overlaps another object has to be in front. Relative size differences are also a cue. For example, if a basketball appears larger than the basket, then the basket must be further away. On the basis of experience, we can estimate how far away the basket is. Binocular depth cues compare information represented in the two retinae because they do not see the visual field exactly the same.
The centers of the two eyes are separated by a small distance, which is approximately 6 to 6.5 cm in most people. Because of this offset, visual stimuli do not fall on exactly the same spot on both retinae unless we are fixated directly on them and they fall on the fovea of each retina. All other objects in the visual field, either closer or farther away than the fixated object, will fall on different spots on the retina. When vision is fixed on an object in space, closer objects will fall on the lateral retina of each eye, and more distant objects will fall on the medial retina of either eye (Figure \(\PageIndex{7}\)).  You will see two images of your finger that represent the two disparate images that are falling on either retina.
These depth cues, both monocular and binocular, can be exploited to make the brain think there are three dimensions in two-dimensional information. This is the basis of 3-D movies. The projected image on the screen is two dimensional, but it has disparate information embedded in it. The 3-D glasses that are available at the theater filter the information so that only one eye sees one version of what is on the screen, and the other eye sees the other version. If you take the glasses off, the image on the screen will have varying amounts of blur because both eyes are seeing both layers of information, and the third dimension will not be evident. Some optical illusions can take advantage of depth cues as well, though those are more often using monocular cues to fool the brain into seeing different parts of the scene as being at different depths.
There are two main regions that surround the primary cortex that are usually referred to as areas V2 and V3 (the primary visual cortex is area V1). These surrounding areas are the visual association cortex. The visual association regions develop more complex visual perceptions by adding color and motion information. The information processed in these areas is then sent to regions of the temporal and parietal lobes. Visual processing has two separate streams of processing: one into the temporal lobe and one into the parietal lobe. These are the ventral and dorsal streams, respectively (Figure \(\PageIndex{8}\)). The ventral stream identifies visual stimuli and their significance. Because the ventral stream uses temporal lobe structures, it begins to interact with the non-visual cortex and may be important in visual stimuli becoming part of memories. The dorsal stream locates objects in space and helps in guiding movements of the body in response to visual inputs. The dorsal stream enters the parietal lobe, where it interacts with somatosensory cortical areas that are important for our perception of the body and its movements. The dorsal stream can then influence frontal lobe activity where motor functions originate.
DISORDERS OF THE...
Brain: Prosopagnosia
The failures of sensory perception can be unusual and debilitating. A particular sensory deficit that inhibits an important social function of humans is prosopagnosia, or face blindness. The word comes from the Greek words prosopa, that means “faces,” and agnosia, that means “not knowing.” Some people may feel that they cannot recognize people easily by their faces. However, a person with prosopagnosia cannot recognize the most recognizable people in their respective cultures. They would not recognize the face of a celebrity, an important historical figure, or even a family member like their mother. They may not even recognize their own face.
Prosopagnosia can be caused by trauma to the brain, or it can be present from birth. The exact cause of proposagnosia and the reason that it happens to some people is unclear. A study of the brains of people born with the deficit found that a specific region of the brain, the anterior fusiform gyrus of the temporal lobe, is often underdeveloped. This region of the brain is concerned with the recognition of visual stimuli and its possible association with memories. Though the evidence is not yet definitive, this region is likely to be where facial recognition occurs.
Though this can be a devastating condition, people who suffer from it can get by—often by using other cues to recognize the people they see. Often, the sound of a person’s voice, or the presence of unique cues such as distinct facial features (a mole, for example) or hair color can help the sufferer recognize a familiar person. In the video on prosopagnosia provided in this section, a woman is shown having trouble recognizing celebrities, family members, and herself. In some situations, she can use other cues to help her recognize faces.
The inability to recognize people by their faces is a troublesome problem. It can be caused by trauma, or it may be inborn. Watch this video to learn more about a person who lost the ability to recognize faces as the result of an injury. She cannot recognize the faces of close family members or herself. What other information can a person suffering from prosopagnosia use to figure out whom they are seeing?
Chapter Review
Sensory input to the brain enters through pathways that travel through either the spinal cord (for somatosensory input from the body) or the brain stem (for everything else, except the visual and olfactory systems) to reach the diencephalon. In the diencephalon, sensory pathways reach the thalamus. This is necessary for all sensory systems to reach the cerebral cortex, except for the olfactory system that is directly connected to the frontal and temporal lobes.
The two major tracts in the spinal cord, originating from sensory neurons in the dorsal root ganglia, are the dorsal column system and the spinothalamic tract. The major differences between the two are in the type of information that is relayed to the brain and where the tracts decussate. The dorsal column system primarily carries information about touch and proprioception and crosses the midline in the medulla. The spinothalamic tract is primarily responsible for pain and temperature sensation and crosses the midline in the spinal cord at the level at which it enters. The trigeminal nerve adds similar sensation information from the head to these pathways.
The auditory pathway passes through multiple nuclei in the brain stem in which additional information is extracted from the basic frequency stimuli processed by the cochlea. Sound localization is made possible through the activity of these brain stem structures. The vestibular system enters the brain stem and influences activity in the cerebellum, spinal cord, and cerebral cortex.
The visual pathway segregates information from the two eyes so that one half of the visual field projects to the other side of the brain. Within visual cortical areas, the perception of the stimuli and their location is passed along two streams, one ventral and one dorsal. The ventral visual stream connects to structures in the temporal lobe that are important for long-term memory formation. The dorsal visual stream interacts with the somatosensory cortex in the parietal lobe, and together they can influence the activity in the frontal lobe to generate movements of the body in relation to visual information.
Interactive Link Questions
Watch this video to learn more about how the brain perceives 3-D motion. Similar to how retinal disparity offers 3-D moviegoers a way to extract 3-D information from the two-dimensional visual field projected onto the retina, the brain can extract information about movement in space by comparing what the two eyes see. If movement of a visual stimulus is leftward in one eye and rightward in the opposite eye, the brain interprets this as movement toward (or away) from the face along the midline. If both eyes see an object moving in the same direction, but at different rates, what would that mean for spatial movement?
Answer: Whereas the video shows opposite movement information in each eye for an object moving toward the face on the midline, movement past one side of the head will result in movement in the same direction on both retinae, but it will be slower in the eye on the side nearer to the object.
The inability to recognize people by their faces is a troublesome problem. It can be caused by trauma, or it may be inborn. Watch this video to learn more about a person who lost the ability to recognize faces as the result of an injury. She cannot recognize the faces of close family members or herself. What other information can a person suffering from prosopagnosia use to figure out whom they are seeing?
Answer: Even if a person cannot recognize a person’s face, other cues such as clothing, hairstyle, or a particular feature such as a prominent nose or facial hair, can help make an identification.
Review Questions
Q. Which of these sensory modalities does not pass through the ventral posterior thalamus?
A. gustatory
B. proprioception
C. audition
D. nociception
Answer: C
Q. Which nucleus in the medulla is connected to the inferior colliculus?
A. solitary nucleus
B. vestibular nucleus
C. chief sensory nucleus
D. cochlear nucleus
Answer: D
Q. Visual stimuli in the upper-left visual field will be processed in what region of the primary visual cortex?
A. inferior right
B. inferior left
C. superior right
D. superior left
Answer: A
Q. Which location on the body has the largest region of somatosensory cortex representing it, according to the sensory homunculus?
A. lips
B. thigh
C. elbow
D. neck
Answer: A
Q. Which of the following is a direct target of the vestibular ganglion?
A. superior colliculus
B. cerebellum
C. thalamus
D. optic chiasm
Answer: B
Critical Thinking Questions
Q. Following a motorcycle accident, the victim loses the ability to move the right leg but has normal control over the left one, suggesting a hemisection somewhere in the thoracic region of the spinal cord. What sensory deficits would be expected in terms of touch versus pain? Explain your answer.
A. The right leg would feel painful stimuli, but not touch, because the spinothalamic tract decussates at the level of entry, which would be below the injury, whereas the dorsal column system does not decussate until reaching the brain stem, which would be above the injury and thus those fibers would be damaged.
Q. A pituitary tumor can cause perceptual losses in the lateral visual field. The pituitary gland is located directly inferior to the hypothalamus. Why would this happen?
A. As the tumor enlarges, it would press against the optic chiasm, and fibers from the medial retina would be disrupted. These fibers carry information about the lateral visual field because the visual scene is reversed as the light passes through the pupil and lens.
Glossary
- ascending pathway
- fiber structure that relays sensory information from the periphery through the spinal cord and brain stem to other structures of the brain
- association area
- region of cortex connected to a primary sensory cortical area that further processes the information to generate more complex sensory perceptions
- binocular depth cues
- indications of the distance of visual stimuli on the basis of slight differences in the images projected onto either retina
- chief sensory nucleus
- component of the trigeminal nuclei that is found in the pons
- circadian rhythm
- internal perception of the daily cycle of light and dark based on retinal activity related to sunlight
- decussate
- to cross the midline, as in fibers that project from one side of the body to the other
- dorsal column system
- ascending tract of the spinal cord associated with fine touch and proprioceptive sensations
- dorsal stream
- connections between cortical areas from the occipital to parietal lobes that are responsible for the perception of visual motion and guiding movement of the body in relation to that motion
- fasciculus cuneatus
- lateral division of the dorsal column system composed of fibers from sensory neurons in the upper body
- fasciculus gracilis
- medial division of the dorsal column system composed of fibers from sensory neurons in the lower body
- inferior colliculus
- last structure in the auditory brainstem pathway that projects to the thalamus and superior colliculus
- interaural intensity difference
- cue used to aid sound localization in the horizontal plane that compares the relative loudness of sounds at the two ears, because the ear closer to the sound source will hear a slightly more intense sound
- interaural time difference
- cue used to help with sound localization in the horizontal plane that compares the relative time of arrival of sounds at the two ears, because the ear closer to the sound source will receive the stimulus microseconds before the other ear
- lateral geniculate nucleus
- thalamic target of the RGCs that projects to the visual cortex
- medial geniculate nucleus
- thalamic target of the auditory brain stem that projects to the auditory cortex
- medial lemniscus
- fiber tract of the dorsal column system that extends from the nuclei gracilis and cuneatus to the thalamus, and decussates
- mesencephalic nucleus
- component of the trigeminal nuclei that is found in the midbrain
- multimodal integration area
- region of the cerebral cortex in which information from more than one sensory modality is processed to arrive at higher level cortical functions such as memory, learning, or cognition
- nucleus cuneatus
- medullary nucleus at which first-order neurons of the dorsal column system synapse specifically from the upper body and arms
- nucleus gracilis
- medullary nucleus at which first-order neurons of the dorsal column system synapse specifically from the lower body and legs
- optic chiasm
- decussation point in the visual system at which medial retina fibers cross to the other side of the brain
- optic tract
- name for the fiber structure containing axons from the retina posterior to the optic chiasm representing their CNS location
- primary sensory cortex
- region of the cerebral cortex that initially receives sensory input from an ascending pathway from the thalamus and begins the processing that will result in conscious perception of that modality
- sensory homunculus
- topographic representation of the body within the somatosensory cortex demonstrating the correspondence between neurons processing stimuli and sensitivity
- solitary nucleus
- medullar nucleus that receives taste information from the facial and glossopharyngeal nerves
- spinal trigeminal nucleus
- component of the trigeminal nuclei that is found in the medulla
- spinothalamic tract
- ascending tract of the spinal cord associated with pain and temperature sensations
- superior colliculus
- structure in the midbrain that combines visual, auditory, and somatosensory input to coordinate spatial and topographic representations of the three sensory systems
- suprachiasmatic nucleus
- hypothalamic target of the retina that helps to establish the circadian rhythm of the body on the basis of the presence or absence of daylight
- ventral posterior nucleus
- nucleus in the thalamus that is the target of gustatory sensations and projects to the cerebral cortex
- ventral stream
- connections between cortical areas from the occipital lobe to the temporal lobe that are responsible for identification of visual stimuli
- vestibular nuclei
- targets of the vestibular component of the eighth cranial nerve
- vestibulo-ocular reflex (VOR)
- reflex based on connections between the vestibular system and the cranial nerves of eye movements that ensures images are stabilized on the retina as the head and body move
Contributors and Attributions
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OpenStax Anatomy & Physiology (CC BY 4.0). Access for free at https://openstax.org/books/anatomy-and-physiology


7.23B: Applications of Genetic Engineering
- Describe the major applications of genetic engineering
Genetic engineering, also called genetic modification, is the direct manipulation of an organism’s genome using biotechnology.
New DNA may be inserted in the host genome by first isolating and copying the genetic material of interest, using molecular-cloning methods to generate a DNA sequence; or by synthesizing the DNA, and then inserting this construct into the host organism. Genes may be removed, or “knocked out”, using a nuclease.
Gene targeting is a different technique that uses homologous recombination to change an endogenous gene, and can be used to delete a gene, remove exons, add a gene, or introduce point mutations. Genetic engineering has applications in medicine, research, industry and agriculture and can be used on a wide range of plants, animals and microorganisms.
Genetic engineering has produced a variety of drugs and hormones for medical use. For example, one of its earliest uses in pharmaceuticals was gene splicing to manufacture large amounts of insulin, made using cells of E. coli bacteria. Interferon, which is used to eliminate certain viruses and kill cancer cells, also is a product of genetic engineering, as are tissue plasminogen activator and urokinase, which are used to dissolve blood clots.
Another byproduct is a type of human growth hormone; it’s used to treat dwarfism and is produced through genetically-engineered bacteria and yeasts. The evolving field of gene therapy involves manipulating human genes to treat or cure genetic diseases and disorders. Modified plasmids or viruses often are the messengers to deliver genetic material to the body’s cells, resulting in the production of substances that should correct the illness. Sometimes cells are genetically altered inside the body; other times scientists modify them in the laboratory and return them to the patient’s body.
Since the 1990s, gene therapy has been used in clinical trials to treat diseases and conditions such as AIDS, cystic fibrosis, cancer, and high cholesterol. Drawbacks of gene therapy are that sometimes the person’s immune system destroys the cells that have been genetically altered, and also that it is hard to get the genetic material into enough cells to have the desired effect.
Key Points
- Genetic engineering has applications in medicine, research, industry and agriculture and can be used on a wide range of plants, animals and microorganisms.
- In medicine, genetic engineering has been used to mass-produce insulin, human growth hormones, follistim (for treating infertility), human albumin, monoclonal antibodies, antihemophilic factors, vaccines, and many other drugs.
- In research, organisms are genetically engineered to discover the functions of certain genes.
- Industrial applications include transforming microorganisms such as bacteria or yeast, or insect mammalian cells with a gene coding for a useful protein. Mass quantities of the protein can be produced by growing the transformed organism in bioreactors using fermentation, then purifying the protein.
- Genetic engineering is also used in agriculture to create genetically-modified crops or genetically-modified organisms.
Key Terms
- biotechnology : The use of living organisms (especially microorganisms) in industrial, agricultural, medical, and other technological applications.
- cloning : The production of a cloned embryo by transplanting the nucleus of a somatic cell into an ovum.


1.10: Properties of Real Numbers
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Learning Objectives
By the end of this section, you will be able to:
- Use the commutative and associative properties
- Use the identity and inverse properties of addition and multiplication
- Use the properties of zero
- Simplify expressions using the distributive property
Note
A more thorough introduction to the topics covered in this section can be found in the Prealgebra chapter, The Properties of Real Numbers .
Use the Commutative and Associative Properties
Think about adding two numbers, say 5 and 3. The order we add them doesn’t affect the result, does it?
\[\begin{array} { cc } { 5 + 3 } & { 3 + 5 } \\ { 8 } & { 8 } \\ { 5 + 3 = } & { 3 + 5 } \end{array} \nonumber\]
The results are the same.
As we can see, the order in which we add does not matter!
What about multiplying 5 and 3?
\[\begin{array} { c c } { 5 \cdot 3 } & { 3 \cdot 5 } \\ { 15 } & { 15 } \\ { 5 \cdot 3=} &{3 \cdot 5 } \end{array} \nonumber\]
Again, the results are the same!
The order in which we multiply does not matter!
These examples illustrate the commutative property . When adding or multiplying, changing the order gives the same result.
COMMUTATIVE PROPERTY
\[\begin{array} { l l } { \textbf { of Addition } } & { \text { If } a , b \text { are real numbers, then } \quad a + b = b + a } \\ { \textbf { of Multiplication } } & { \text { If } a , b \text { are real numbers, then } \quad a \cdot b = b \cdot a } \end{array} \nonumber\]
When adding or multiplying, changing the order gives the same result.
The commutative property has to do with order. If you change the order of the numbers when adding or multiplying, the result is the same.
What about subtraction? Does order matter when we subtract numbers? Does 7−3 give the same result as 3−7?
\[\begin{array} { c c } { 7 - 3 } & { 3 - 7 } \\ { 4 } & { - 4 } \end{array} \nonumber\]
\[\begin{aligned} 4 & \neq - 4 \\ 7 - 3 & \neq 3 - 7 \end{aligned}\]
The results are not the same.
Since changing the order of the subtraction did not give the same result, we know that subtraction is not commutative .
Let’s see what happens when we divide two numbers. Is division commutative?
\[\begin{array} { cc} { 12 \div 4 } & { 4 \div 12 } \\ { \frac { 12 } { 4 } } & { \frac { 4 } { 12 } } \\ { 3 } & { \frac { 1 } { 3 } } \end{array} \nonumber\]
\[\begin{aligned} 3 \neq & \frac { 1 } { 3 } \\ 12 \div 4 & \neq 4 \div 12 \end{aligned}\]
The results are not the same.
Since changing the order of the division did not give the same result, division is not commutative . The commutative properties only apply to addition and multiplication!
- Addition and multiplication are commutative.
- Subtraction and Division are not commutative.
If you were asked to simplify this expression, how would you do it and what would your answer be?
\[7 + 8 + 2 \nonumber\]
Some people would think \(7+8\) is 15 and then \(15+2\) is 17. Others might start with \(8+2\) makes 10 and then \(7+10\) makes 17.
Either way gives the same result. Remember, we use parentheses as grouping symbols to indicate which operation should be done first.
\[\begin{array} { ll } { \text{ Add } 7 + 8 . } & { ( 7 + 8 ) + 2 } \\ { \text { Add. } } & { 15 + 2 } \\ { \text { Add. } } & { 17 } \\ \\ { } & { 7 + ( 8 + 2 ) } \\ { \text { Add } 8 + 2 . } & { 7 + 10 } \\ { \text { Add. } } & { 77 } \\\\ { ( 7 + 8 ) + 2 = 7 + ( 8 + 2 ) } \end{array} \nonumber\]
When adding three numbers, changing the grouping of the numbers gives the same result.
This is true for multiplication, too.
\[\begin{array} { ll } { } & { (5\cdot \frac{1}{3})\cdot 3 } \\ { \text { Multiply. } 5\cdot \frac{1}{3} } & { \frac{5}{3}\cdot 3 } \\ { \text { Multiply. } } & { 5 } \\ \\ { } & { 5\cdot (\frac{1}{3}\cdot 3) } \\ { \text { Multiply. } \frac{1}{3}\cdot 3 } & { 5\cdot 1 } \\ { \text { Multiply. } } & { 5 } \\ \\ { (5\cdot \frac{1}{3})\cdot 3 = 5\cdot (\frac{1}{3}\cdot 3) } \end{array} \nonumber\]
When multiplying three numbers, changing the grouping of the numbers gives the same result.
You probably know this, but the terminology may be new to you. These examples illustrate the associative property .
ASSOCIATIVE PROPERTY
\[\begin{array} { l l } { \textbf { of Addition } } & { \text { If } a , b , c \text { are real numbers, then } ( a + b ) + c = a + ( b + c ) } \\ { \textbf { of Multiplication } } & { \text { If } a , b , c \text { are real numbers, then } ( a \cdot b ) \cdot c = a \cdot ( b \cdot c ) } \end{array} \nonumber\]
When adding or multiplying, changing the grouping gives the same result.
Let’s think again about multiplying \(5\cdot \frac{1}{3}\cdot 3\). We got the same result both ways, but which way was easier? Multiplying \(\frac{1}{3}\) and 3 first, as shown above on the right side, eliminates the fraction in the first step. Using the associative property can make the math easier!
The associative property has to do with grouping. If we change how the numbers are grouped, the result will be the same. Notice it is the same three numbers in the same order—the only difference is the grouping.
We saw that subtraction and division were not commutative. They are not associative either.
When simplifying an expression, it is always a good idea to plan what the steps will be. In order to combine like terms in the next example, we will use the commutative property of addition to write the like terms together.
Example \(\PageIndex{1}\)
Simplify: \(18p+6q+15p+5q\).
Solution
\[\begin{array} { l l} {} &{18p+6q+15p+5q}\\ \\{ \text { Use the commutative property of addition } } &{} \\ { \text {to re-order so that like terms are together.} } &{18p+15p+ 6q+5q} \\ \\ {\text{Add like terms.}} &{33p + 11q} \end{array} \nonumber\]
Try It \(\PageIndex{2}\)
Simplify: \(23r+14s+9r+15s\).
- Answer
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\(32r+29s\)
Try It \(\PageIndex{3}\)
Simplify: \(37m+21n+4m−15n\).
- Answer
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\(41m+6n\)
When we have to simplify algebraic expression s, we can often make the work easier by applying the commutative or associative property first, instead of automatically following the order of operations. When adding or subtracting fractions, combine those with a common denominator first.
Example \(\PageIndex{4}\)
Simplify: \((\frac{5}{13} + \frac{3}{4}) + \frac{1}{4}\)
Solution
\[\begin{array} { l l } {} &{(\frac{5}{13} + \frac{3}{4}) + \frac{1}{4}} \\{ \text { Notice that the last } 2 \text { terms have a } } \\ { \text { common denominator, so change the } } &{\frac { 5 } { 13 } + \left( \frac { 3 } { 4 } + \frac { 1 } { 4 } \right)}\\ { \text { grouping. } } &{}\\ \\ {\text{Add in parentheses first.}} &{\frac{5}{13} + (\frac{4}{4})} \\ \\ {\text{Simplify the fraction.}} &{\frac{5}{13} + 1} \\ \\ {\text{Add.}} &{1\frac{5}{13}} \\ \\ {\text{Convert to an improper fraction.}} &{\frac{18}{13}} \end{array} \nonumber\]
Try It \(\PageIndex{5}\)
Simplify: \((\frac{7}{15} + \frac{5}{8}) + \frac{3}{8}\)
- Answer
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\(1\frac{7}{15}\)
Try It \(\PageIndex{6}\)
Simplify: \((\frac{2}{9} + \frac{7}{12}) + \frac{5}{12}\)
- Answer
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\(1\frac{2}{9}\)
Example \(\PageIndex{7}\)
Use the associative property to simplify \(6(3x)\).
Solution
Use the associative property of multiplication, \((a\cdot b)\cdot c=a\cdot (b\cdot c)\), to change the grouping.
\[\begin{array} { ll } {} &{ 6 ( 3 x ) } \\ { \text { Change the grouping. } } &{(6\cdot 3)x} \\ { \text { Multiply in the parentheses. } } &{18} \end{array} \nonumber\]
Notice that we can multiply \(6\cdot 3\) but we could not multiply \(3x\) without having a value for \(x\).
Try It \(\PageIndex{8}\)
Use the associative property to simplify \(8(4x)\).
- Answer
-
\(32x\)
Try It \(\PageIndex{9}\)
Use the associative property to simplify \(-9(7y)\).
- Answer
-
\(-63y\)
Use the Identity and Inverse Properties of Addition and Multiplication
What happens when we add 0 to any number? Adding 0 doesn’t change the value. For this reason, we call 0 the additive identity .
For example,
\[\begin{array} { c c c } { 13 + 0 } & { - 14 + 0 } & { 0 + ( - 8 ) } \\ { 13 } & { - 14 } & { - 8 } \end{array} \nonumber\]
These examples illustrate the Identity Property of Addition that states that for any real number \(a\), \(a+0=a\) and \(0+a=a\).
What happens when we multiply any number by one? Multiplying by 1 doesn’t change the value. So we call 1 the multiplicative identity .
For example, \[\begin{array} { r r r } { 43 \cdot 1 } & { - 27 \cdot 1 } & { 1 \cdot \frac { 3 } { 5 } } \\ { 43 } & { - 27 } & { \frac { 3 } { 5 } } \end{array} \nonumber\]
These examples illustrate the Identity Property of Multiplication that states that for any real number \(a\), \(a\cdot 1=a\) and \(1\cdot a=a\).
We summarize the Identity Properties below.
IDENTITY PROPERTY
\[\begin{array} { l l} { \textbf {of addition}\text{ For any real number } a : } &{ a + 0 = a \quad 0 + a = a } \\ { \textbf{0} \text { is the}\textbf{ additive identity } } \\ {\textbf {of multiplication}\text{ For any real number } a : } &{ a \cdot 1 = a \quad 1 \cdot a = a } \\ { \textbf{1}\text{ is the}\textbf{ multiplicative identity } } \end{array} \nonumber\]
Notice that in each case, the missing number was the opposite of the number!
We call \(−a\). the additive inverse of a . The opposite of a number is its additive inverse. A number and its opposite add to zero, which is the additive identity. This leads to the Inverse Property of Addition that states for any real number \(a, a+(−a)=0\). Remember, a number and its opposite add to zero.
What number multiplied by \(\frac{2}{3}\) gives the multiplicative identity, 1? In other words, \(\frac{2}{3}\) times what results in 1?
What number multiplied by 2 gives the multiplicative identity, 1? In other words 2 times what results in 1?
Notice that in each case, the missing number was the reciprocal of the number!
We call \(\frac{1}{a}\) the multiplicative inverse of a . The reciprocal of aa number is its multiplicative inverse. A number and its reciprocal multiply to one, which is the multiplicative identity. This leads to the Inverse Property of Multiplication that states that for any real number \(a, a\neq 0, a\cdot \frac{1}{a}=1\).
We’ll formally state the inverse properties here:
INVERSE PROPERTY
\[\begin{array} { l l l } { \textbf { of addition } } &{ \text { For any real number } a,} &{a + (-a) = 0}\\{} &{-a \text{. is the}\textbf{ additive inverse} \text{ of }a} &{}\\ {} &{ \text { A number and its opposite add to zero. } }&{}\\ \\{ \textbf { of multiplication } } &{ \text { For any real number } a, a\neq 0} &{a\cdot \frac{1}{a} = 1}\\{} &{\frac{1}{a} \text{. is the}\textbf{ multiplicative inverse} \text{ of }a} &{}\\ {} &{ \text { A number and its reciprocal multiply to one. } }&{} \end{array} \nonumber\]
Example \(\PageIndex{10}\)
Find the additive inverse of
- \(\frac{5}{8}\)
- \(0.6\)
- \(-8\)
- \(-\frac{4}{3}\)
Solution
To find the additive inverse, we find the opposite.
- The additive inverse of \(\frac{5}{8}\) is the opposite of \(\frac{5}{8}\). The additive inverse of \(\frac{5}{8}\) is \(-\frac{5}{8}\)
- The additive inverse of \(0.6\) is the opposite of \(0.6\). The additive inverse of \(0.6\) is \(-0.6\).
- The additive inverse of \(-8\) is the opposite of \(-8\). We write the opposite of \(-8\) as \(-(-8)\), and then simplify it to \(8\). Therefore, the additive inverse of \(-8\) is \(8\).
- The additive inverse of \(-\frac{4}{3}\) is the opposite of \(-\frac{4}{3}\). We write this as \(-(-\frac{4}{3})\), and then simplify to \(\frac{4}{3}\). Thus, the additive inverse of \(-\frac{4}{3}\) is \(\frac{4}{3}\).
Try It \(\PageIndex{11}\)
Find the additive inverse of
- \(\frac{7}{9}\)
- \(1.2\)
- \(-14\)
- \(-\frac{9}{4}\)
- Answer
-
- \(-\frac{7}{9}\)
- \(-1.2\)
- \(14\)
- \(\frac{9}{4}\)
Try It \(\PageIndex{12}\)
Find the additive inverse of
- \(\frac{7}{13}\)
- \(8.4\)
- \(-46\)
- \(-\frac{5}{2}\)
- Answer
-
- \(-\frac{7}{13}\)
- \(-8.4\)
- \(46\)
- \(\frac{5}{2}\)
Example \(\PageIndex{13}\)
Find the multiplicative inverse of
- \(9\)
- \(-\frac{1}{9}\)
- \(0.9\)
Solution
To find the multiplicative inverse, we find the reciprocal.
- The multiplicative inverse of \(9\) is the reciprocal of \(9\), which is \(\frac{1}{9}\). Therefore, the multiplicative inverse of \(9\) is \(\frac{1}{9}\).
- The multiplicative inverse of \(-\frac{1}{9}\) is the reciprocal of \(-\frac{1}{9}\), which is \(−9\). Thus, the multiplicative inverse of \(-\frac{1}{9}\) is \(-9\).
- To find the multiplicative inverse of \(0.9\), we first convert \(0.9\) to a fraction, \(\frac{9}{10}\). Then we find the reciprocal of the fraction. The reciprocal of \(\frac{9}{10}\) is \(\frac{10}{9}\). So the multiplicative inverse of \(0.9\) is \(\frac{10}{9}\).
Try It \(\PageIndex{14}\)
Find the multiplicative inverse of
- \(4\)
- \(-\frac{1}{7}\)
- \(0.3\)
- Answer
-
- \(\frac{1}{4}\)
- \(-7\)
- \(\frac{10}{3}\)
Try It \(\PageIndex{15}\)
Find the multiplicative inverse of
- \(18\)
- \(-\frac{4}{5}\)
- \(0.6\)
- Answer
-
- \(\frac{1}{18}\)
- \(-\frac{5}{4}\)
- \(\frac{5}{3}\)
Use the Properties of Zero
The identity property of addition says that when we add 0 to any number, the result is that same number. What happens when we multiply a number by 0? Multiplying by 0 makes the product equal zero.
MULTIPLICATION BY ZERO
For any real number \(a\).
\[a \cdot 0 = 0 \quad 0 \cdot a = 0 \nonumber\]
The product of any real number and 0 is 0.
What about division involving zero? What is \(0\div 3\)? Think about a real example: If there are no cookies in the cookie jar and 3 people are to share them, how many cookies does each person get? There are no cookies to share, so each person gets 0 cookies. So,
\[0\div 3 = 0 \nonumber\]
We can check division with the related multiplication fact.
\[12 \div 6 = 2 \text { because } 2 \cdot 6 = 12 \nonumber\]
So we know \(0\div 3=0\) because \(0\cdot 3=0\).
DIVISION OF ZERO
For any real number \(a\), except \(0, \frac{0}{a}=0\) and \(0\div a=0\).
Zero divided by any real number except zero is zero.
Now think about dividing by zero. What is the result of dividing 4 by 0? Think about the related multiplication fact: \(4\div 0=?\) means \(?\cdot 0=4\). Is there a number that multiplied by 0 gives 4? Since any real number multiplied by 0 gives 0, there is no real number that can be multiplied by 0 to obtain 4.
We conclude that there is no answer to \(4\div 0\) and so we say that division by 0 is undefined.
DIVISION BY ZERO
For any real number \(a\), \( \frac{a}{0}\) and \(a\div 0\) are undefined.
Division by zero is undefined.
We summarize the properties of zero below.
PROPERTIES OF ZERO
Multiplication by Zero: For any real number \(a\),
\[a \cdot 0 = 0 \quad 0 \cdot a = 0 \quad \text { The product of any number and } 0 \text { is } 0 \nonumber\]
Division of Zero, Division by Zero: For any real number \(a, a\neq 0\)
\[\begin{array} { l l } { \frac { 0 } { a } = 0 } & { \text { Zero divided by any real number, except itself is zero. } } \\ { \frac { a } { 0 } \text { is undefined } } & { \text { Division by zero is undefined. } } \end{array} \nonumber\]
Example \(\PageIndex{16}\)
Simplify:
- \(-8\cdot 0\)
- \(\frac{0}{-2}\)
- \(\frac{-32}{0}\)
Solution
- \[\begin{array} { cc } { } &{-8\cdot 0}\\{\text{The product of any real number and 0 is 0}} &{0}\end{array} \nonumber\]
- \[\begin{array} { ll } { } &{\frac{0}{-2}}\\{\text{Zero divided by any real number, except}} &{} \\ {\text{itself, is 0}} &{0}\end{array} \nonumber\]
- \[\begin{array} { ll } { } &{\frac{-32}{0}}\\ {\text{Division by 0 is undefined.}} &{\text{undefined}} \end{array} \nonumber\]
Try It \(\PageIndex{17}\)
Simplify:
- \(-14\cdot 0\)
- \(\frac{0}{-6}\)
- \(\frac{-2}{0}\)
- Answer
-
- \(0\)
- \(0\)
- undefined
Try It \(\PageIndex{18}\)
Simplify:
- \(0(-17)\)
- \(\frac{0}{-10}\)
- \(\frac{-5}{0}\)
- Answer
-
- \(0\)
- \(0\)
- undefined
We will now practice using the properties of identities, inverses, and zero to simplify expressions.
Example \(\PageIndex{19}\)
Simplify:
- \(\frac{0}{n + 5}\), where \(n\neq −5\)
- \(\frac{10 - 3p}{0}\) where \(10 - 3p \neq 0\)
Solution
- \[\begin{array} { ll } { } &{\frac{0}{n + 5}}\\ {\text { Zero divided by any real number except }} &{0} \\ { \text { itself is } 0.} &{} \end{array} \nonumber\]
- \[\begin{array} { ll } { } &{\frac{10 - 3p}{0}}\\ {\text { Division by 0 is undefined }} &{\text{undefined}} \end{array} \nonumber\]
Try It \(\PageIndex{26}\)
Simplify:
- \(\frac{0}{m + 7}\), where \(m \neq -7\)
- \(\frac{18 - 6c}{0}\), where \(18 - 6c \neq 0\)
- Answer
-
- 0
- undefined
Try It \(\PageIndex{27}\)
Simplify:
- \(\frac{0}{d - 4}\), where \(d \neq 4\)
- \(\frac{15 - 4q}{0}\), where \(15 - 4q \neq 0\)
- Answer
-
- 0
- undefined
Example \(\PageIndex{20}\)
Simplify: \(−84n+(−73n)+84n\).
Solution
\[\begin{array} { l l } { } &{−84n+(−73n)+84n} \\ { \text { Notice that the first and third terms are } } &{}\\ { \text { opposites; use the commutative property of } } &{- 84 n + 84 n + ( - 73 n ) } \\ { \text { addition to re-order the terms. } } &{} \\ \\ { \text { Add left to right. } } &{0 + (-73)}\\ \\{ \text { Add. } } &{-73n} \end{array} \nonumber\]
Try It \(\PageIndex{21}\)
Simplify: \(−27a+(−48a)+27a\).
- Answer
-
\(−48a\)
Try It \(\PageIndex{22}\)
Simplify: \(39x+(−92x)+(−39x)\).
- Answer
-
\(−92x\)
Now we will see how recognizing reciprocals is helpful. Before multiplying left to right, look for reciprocals—their product is 1.
Example \(\PageIndex{23}\)
Simplify: \(\frac{7}{15}\cdot\frac{8}{23}\cdot\frac{15}{7}\)
Solution
\[\begin{array} { l l } { } &{\frac{7}{15}\cdot\frac{8}{23}\cdot\frac{15}{7}} \\ { \text { Notice that the first and third terms are } } &{}\\ { \text { reciprocals, so use the commutative } } &{\frac{7}{15}\cdot\frac{15}{7}\cdot\frac{8}{23}} \\ { \text { property of multiplication to re-order the } } &{} \\ { \text { factors. } } &{}\\ \\{ \text { Multiply left to right. } } &{1\cdot\frac{8}{23}} \\\\{\text{Multiply.}} &{\frac{8}{23}}\end{array} \nonumber\]
Exercise \(\PageIndex{24}\)
Simplify: \(\frac{9}{16}\cdot\frac{5}{49}\cdot\frac{16}{9}\)
- Answer
-
\(\frac{5}{49}\)
Try It \(\PageIndex{25}\)
Simplify: \(\frac{6}{17}\cdot\frac{11}{25}\cdot\frac{17}{6}\)
- Answer
-
\(\frac{11}{25}\)
Example \(\PageIndex{28}\)
Simplify: \(\frac{3}{4}\cdot\frac{4}{3}(6x + 12)\)
Solution
\[\begin{array} { l l } { } &{\frac{3}{4}\cdot\frac{4}{3}(6x + 12)} \\ { \text { There is nothing to do in the parentheses, } } &{}\\ { \text { so multiply the two fractions first—notice, } } &{1(6x + 12)} \\ { \text { they are reciprocals. } } &{} \\ \\{ \text { Simplify by recognizing the multiplicative } } &{} \\{\text{ identity.}} &{6x + 12} \end{array} \nonumber\]
Try It \(\PageIndex{29}\)
Simplify: \(\frac{2}{5}\cdot\frac{5}{2}(20y + 50)\)
- Answer
-
\(20y + 50\)
Try It \(\PageIndex{30}\)
Simplify: \(\frac{3}{8}\cdot\frac{8}{3}(12z + 16)\)
- Answer
-
\(12z + 16\)
Simplify Expressions Using the Distributive Property
Suppose that three friends are going to the movies. They each need $9.25—that’s 9 dollars and 1 quarter—to pay for their tickets. How much money do they need all together?
You can think about the dollars separately from the quarters. They need 3 times $9 so $27, and 3 times 1 quarter, so 75 cents. In total, they need $27.75. If you think about doing the math in this way, you are using the distributive property .
DISTRIBUTIVE PROPERTY
\[\begin{array} { rr } {\text { If } a , b , c \text { are real numbers, then }} &{a ( b + c ) = a b + a c} \\ \\{ \text { Also,} } &{( b + c ) a = b a + c a} \\ {} &{a ( b - c ) = a b - a c } &{} \\{} &{( b - c ) a = b a - c a } \end{array} \nonumber\]
Back to our friends at the movies, we could find the total amount of money they need like this:
\[\begin{array} { c } { 3 ( 9.25 ) } \\ { 3 ( 9 + 0.25 ) } \\ { 3 ( 9 ) + 3 ( 0.25 ) } \\ { 27 + 0.75 } \\ \\ { 27.75 } \end{array} \nonumber\]
In algebra, we use the distributive property to remove parentheses as we simplify expressions.
For example, if we are asked to simplify the expression \(3(x+4)\), the order of operations says to work in the parentheses first. But we cannot add x and 4, since they are not like terms. So we use the distributive property, as shown in Exercise \(\PageIndex{31}\).
Example \(\PageIndex{31}\)
Simplify: \(3(x+4)\).
Solution
\[\begin{array} { l l } { } & { 3 ( x + 4 ) } \\ { \text { Distribute. } } & { 3 \cdot x + 3 \cdot 4 } \\ { \text { Multiply. } } & { 3 x + 12 } \end{array} \nonumber\]
Try It \(\PageIndex{32}\)
Simplify: \(4(x+2)\).
- Answer
-
\(4x + 8\)
Try It \(\PageIndex{33}\)
Simplify: \(6(x+7)\).
- Answer
-
\(6x + 42\)
Some students find it helpful to draw in arrows to remind them how to use the distributive property. Then the first step in Exercise \(\PageIndex{31}\) would look like this:
Example \(\PageIndex{34}\)
Simplify: \(8(\frac{3}{8}x+\frac{1}{4})\).
Solution
| Distribute. | |
| Multiply. |
Try It \(\PageIndex{35}\)
Simplify: \(6(\frac{5}{6}y+\frac{1}{2})\).
- Answer
-
\(5y + 3\)
Try It \(\PageIndex{36}\)
Simplify: \(12(\frac{1}{3}n+\frac{3}{4})\).
- Answer
-
\(4n + 9\)
Using the distributive property as shown in Exercise \(\PageIndex{37}\) will be very useful when we solve money applications in later chapters.
Example \(\PageIndex{37}\)
Simplify: \(100(0.3+0.25q)\).
Solution
| Distribute. | |
| Multiply. |
Try It \(\PageIndex{38}\)
Simplify: \(100(0.7+0.15p)\).
- Answer
-
\(70 + 15p\)
Try It \(\PageIndex{39}\)
Simplify: \(100(0.04+0.35d)\).
- Answer
-
\(4 + 35d\)
When we distribute a negative number, we need to be extra careful to get the signs correct!
Example \(\PageIndex{40}\)
Simplify: \(−2(4y+1)\).
Solution
| Distribute. | |
| Multiply. |
Try It \(\PageIndex{41}\)
Simplify: \(−3(6m+5)\).
- Answer
-
\(−18m-15)\)
Try It \(\PageIndex{42}\)
Simplify: \(−6(8n+11)\).
- Answer
-
\(−48n- 66)\)
Example \(\PageIndex{43}\)
Simplify: \(−11(4-3a)\).
Solution
| Distribute. | |
| Multiply. | |
| Simplify. |
Notice that you could also write the result as \(33a−44\). Do you know why?
Try It \(\PageIndex{44}\)
Simplify: \(−5(2-3a)\).
- Answer
-
\(10+ 15a\)
Try It \(\PageIndex{45}\)
Simplify: \(−7(8-15y)\).
- Answer
-
\(-56 + 105y\)
Exercise \(\PageIndex{46}\) will show how to use the distributive property to find the opposite of an expression.
Example \(\PageIndex{46}\)
Simplify: \(−(y+5)\).
Solution
\[\begin{array} { ll } {} &{-(y + 5)} \\ \\{ \text {Multiplying by -1 results in the opposite.} } &{-1( y + 5 )} \\ \\ {\text{Distribute.}} &{-1\cdot y + (-1)\cdot 5}\\ \\{\text{Simplify.}} &{-y + (-5)} \\ \\ {} &{-y - 5} \end{array} \nonumber\]
Try It \(\PageIndex{47}\)
Simplify: \(−(z-11)\).
- Answer
-
\(-z + 11\)
Try It \(\PageIndex{48}\)
Simplify: \(−(x -4)\).
- Answer
-
\(-x + 4\)
There will be times when we’ll need to use the distributive property as part of the order of operations. Start by looking at the parentheses. If the expression inside the parentheses cannot be simplified, the next step would be multiply using the distributive property, which removes the parentheses. The next two examples will illustrate this.
Example \(\PageIndex{49}\)
Simplify: \(8−2(x + 3)\).
Be sure to follow the order of operations. Multiplication comes before subtraction, so we will distribute the 2 first and then subtract.
Solution
\[\begin{array} { ll } {} &{8−2(x + 3)} \\ \\{ \text {Distribute.} } &{8−2\cdot x -2\cdot 3} \\ \\ {\text{Multiply.}} &{8 - 2x - 6}\\ \\{\text{Combine like terms.}} &{-2x + 2} \end{array} \nonumber\]
Try It \(\PageIndex{50}\)
Simplify: \(9−3(x + 2)\).
- Answer
-
\(3 - 3x\)
Try It \(\PageIndex{51}\)
Simplify: \(7x−5(x + 4)\).
- Answer
-
\(2x - 20\)
Example \(\PageIndex{52}\)
Simplify: \(4(x - 8)−(x + 3)\).
Solution
\[\begin{array} { ll } {} &{4(x - 8)−(x + 3)} \\ \\{ \text {Distribute.} } &{4x - 32 - x - 3} \\ \\{\text{Combine like terms.}} &{3x - 35} \end{array} \nonumber\]
Try It \(\PageIndex{53}\)
Simplify: \(6(x - 9)−(x + 12)\).
- Answer
-
\(5x - 66\)
Try It \(\PageIndex{54}\)
Simplify: \(8(x - 1)-(x + 5)\).
- Answer
-
\(7x - 13\)
All the properties of real numbers we have used in this chapter are summarized in Table \(\PageIndex{1}\).
| Commutative Property | |
|
of addition
If a,b are real numbers, then
of multiplication If a,b are real numbers, then |
\(a+b=b+a\)
\(a\cdot b=b\cdot a\) |
| Associative Property | |
|
of addition
If a,b,c are real numbers, then
of multiplication If a,b,c are real numbers, then |
\((a+b)+c=a+(b+c)\)
\((a\cdot b)\cdot c=a\cdot (b\cdot c)\) |
| Distributive Property | |
| If a,b,c are real numbers, then | \(a(b+c)=ab+ac\) |
| Identity Property | |
|
of addition
For any real number a:
of multiplication
For any real number a:
|
\(a+0=a\)
\(0+a=a\)
\(1·a=a\) |
| Inverse Property | |
|
of addition
For any real number a,
\(−a\) is the additive inverse of a of multiplication For any real number \(a,a\neq 0\) \(\frac{1}{a}\) is the multiplicative inverse of a |
\(a+(−a)=0\)
\(a\cdot\frac{1}{a}=1\) |
| Properties of Zero | |
|
For any real number a ,
For any real number \(a,a\neq 0\)
|
\(a\cdot 0=0\) \(0\cdot a=0\)
\(\frac{0}{a} = 0\)
|
Key Concepts
-
Commutative Property of
- Addition: If a,b are real numbers, then \(a+b=b+a\).
- Multiplication: If a,b are real numbers, then \(a\cdot b=b\cdot a\). When adding or multiplying, changing the order gives the same result.
-
Associative Property of
- Addition: If a,b,c are real numbers, then \((a+b)+c=a+(b+c)\).
-
Multiplication:
If a,b,c are real numbers, then \((a\cdot b)\cdot c=a\cdot (b\cdot c)\).
When adding or multiplying, changing the grouping gives the same result.
-
Distributive Property:
If a,b,c are real numbers, then
- \(a(b+c)=ab+ac\)
- \((b+c)a=ba+ca\)
- \(a(b-c)=ab-ac\)
- \((b+c)a=ba+ca\)
-
Identity Property
-
of Addition:
For any real number a: \(a+0=a\)
0 is the additive identity -
of Multiplication:
For any real number a: \(a\cdot 1=a \quad 1·a=a\)
1 1 is the multiplicative identity
-
of Addition:
For any real number a: \(a+0=a\)
-
Inverse Property
- of Addition: For any real number \(a, a+(−a)=0\). A number and its opposite add to zero. \(−a\) is the additive inverse of a.
- of Multiplication: For any real number \(a,(a\neq 0)a\cdot\frac{1}{a}=1\). A number and its reciprocal multiply to one. \(\frac{1}{a}\) is the multiplicative inverse of a.
-
Properties of Zero
-
For any real number a,
\(a\cdot 0=0 \quad 0·a=0\) – The product of any real number and 0 is 0. - \(\frac{0}{a}=0\) for \(a\neq 0\) – Zero divided by any real number except zero is zero.
- \(\frac{a}{0}\) is undefined – Division by zero is undefined.
-
For any real number a,