<s>[INST] Show that the number $\sqrt{2^1}+\sqrt{2^2}+\cdots+\sqrt{2^n}$ is not rational for any integer $n>0$ I've got the first part of the proof.
So we know that the odd exponents will give us an irrational answer, and the even ones will give us a rational answer. Therefore we have an alternating sum of rationals and irrationals. I can prove that an irrational plus a rational is irrational, so now we can combine all the alternating sum into a sum of irrationals. I can't prove that for this specific case that an irrational plus an irrational equals an irrational.
 [/INST] Tricks, tricks, tricks.Here, it's the geometric summation.
$$
\sqrt{2} + \sqrt{2^2} + \ldots \sqrt{2^n} = 2^{\frac 12} + 2^\frac{2}{2} + \ldots + 2^\frac{n}{2}
$$
Use the geometric series summation formula:
$$
2^{\frac 12} + 2^\frac{2}{2} + \ldots + 2^\frac{n}{2} = (2^\frac 12)\frac{2^\frac{n+1}{2} - 1}{2^{\frac 12} - 1}
$$
Now, rationalize by multiplying top and bottom by $2^{\frac 12} + 1$:
$$
(2^\frac 12)\frac{2^\frac{n+1}{2} - 1}{2^{\frac 12} - 1} \times \frac{2^{\frac 12} + 1}{2^{\frac 12} + 1} = 2^{\frac 12} (2^{\frac 12} + 1)(2^{\frac{n+1}{2} - 1})
$$
expand this to get:
$$
{\frac 12} (2^{\frac 12} + 1)(2^{\frac{n+1}{2} - 1}) = 2^{n/2}(1 + \sqrt 2) = 2^{\frac n2} + 2^\frac{n+1}{2}
$$
Now, suppose that $n$ is even: Then $2^{n/2}$ is an integer, so the answer is of the form $a+b \sqrt 2$ where $a,b$ are integers. If this were rational, then $\sqrt 2$ would also be rational. Contradiction.
Now, suppose that $n$ is odd: Then $2^\frac{n+1}{2}$ is an integer, so again the answer is of the form $a+b \sqrt 2$ where $a,b$ are integers. If this were rational, then $\sqrt 2$ would also be rational. Contradiction yet again.
Hence, we can conclude that for all $n$, the quantity given is irrational.
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