<s>[INST] Show that $0 \le a^{\frac{1}{2}} \le b^{\frac{1}{2}}$ 
Show that if $0 \le a \le b$, then  $0 \le a^{\frac{1}{2}} \le b^{\frac{1}{2}}$ in a $C^*$-algebra $A$. 

Assume that $A$ is unital. 
Suppose that $a$ and $b$ are invertible. Then $$0 \le a \le b \implies 0 \le a^{\frac{-1}{2}}aa^{\frac{-1}{2}} \le a^{\frac{-1}{2}}ba^{\frac{-1}{2}} \implies 0 \le e \le a^{\frac{-1}{2}}ba^{\frac{-1}{2}} \implies 0 \le a^{\frac{1}{2}}b^{-1}a^{\frac{1}{2}} \le e$$
Thus we have $$0 \le a^{\frac{1}{2}}b^{\frac{-1}{2}}b^{\frac{-1}{2}}a^{\frac{1}{2}} \le e \implies 0 \le \left(a^{\frac{1}{2}}b^{\frac{-1}{2}}\right)\left(a^{\frac{1}{2}}b^{\frac{-1}{2}}\right)^* \le e$$
Hence, $\|a^{\frac{1}{2}}b^{\frac{-1}{2}}\|\le 1$ which implies that $r(a^{\frac{1}{2}}b^{\frac{-1}{2}}) \le 1$
But $$\sigma(a^{\frac{1}{2}}b^{\frac{-1}{2}}) \cup \{0\}=\sigma(b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}}) \cup \{0\}$$ gives that $r(b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}}) \le 1$. Let $n=b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}}$. Then $n$ is normal. Let $B=C^*(1,n)$. Then $B$ is an abelian $C^*$-algebra. Moreover $$\sigma_A(1-n)=\sigma_B(1-n)=\{\tau(1-n):\tau \in \Omega(B)\}=\{\tau(1)-\tau(n): \tau \in \Omega(B)\}=\{1-\tau(n): \tau \in \Omega(B)\} \subset \mathbb{R}^+$$
Hence $0 \le n \le 1$ which is same as $$b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}} \le 1$$ implying $a^{\frac{1}{2}} \le b^{\frac{1}{2}}$. 
How do I adapt this proof for any general $a$ and $b$? Since $1-b \le 1-a$, dividing through  out by $2||1-a||$ we will have invertible elements. Then I can do as above. 
Thanks for the help!
 [/INST] You need to note that if $0 \leq x \leq y$ then for $\epsilon > 0$ that 
$0 \leq x+ \epsilon 1 \leq y + \epsilon 1$. Now, for $\epsilon >0$, $-\epsilon \notin \sigma(x),\sigma(y)$ so we have $x+ \epsilon 1$ and $y + \epsilon 1$ are both invertible. Hence, from your work above we have that $(x+\epsilon 1)^{1/2} \leq (y+\epsilon 1)^{1/2}$.
It is not hard to see that 
$(x + \epsilon 1)^{1/2} \to_{|| \cdot ||} x$ and $(y+ \epsilon 1)^{1/2} \to_{|| \cdot ||} y$. 
Hence, the case you have already proved generalizes the result after noting these facts. 
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