Real Analysis

1. add Prove that $\sqrt{3}$ is irrational. Does a similar argument work to show $\sqrt{6}$ is irrational?

   Solution: Assume, for contradiction, that there exist integers $p$ and $q$ such that $gcd(p,q)=1$ and: $$\begin{array}{}{\left(\frac{p}{q}\right)}^2=3& \text{(1)}\end{array}$$ Next, equation (1) implies: $$\begin{array}{}{p}^2=3q^2& \text{(2)}\end{array}$$ This shows that $p^2$ is a multiple of 3. The only way for a perfect square to be a multiple of 3 is if its factors also include 3. Therefore, $p$ must also be a multiple of 3. Let $p=3k$, where $k$ is an integer. Substituting this into (2) gives: $$(3k{)}^2=3q^2$$ $$9k^2=3q^2$$ $$\begin{array}{}3k^2=q^2& \text{(3)}\end{array}$$ This shows that $q^2$ is a multiple of 3, and hence, $q$ is a multiple of 3. This contradicts our hypothesis that $gcd(p,q)=1$. From this reasoning, we can only assume that equation (1) cannot hold for any integers $p$ and $q$. Thus, $\sqrt{3}$ is irrational.
   

   Q.E.D.

   Yes, a similar argument works to show that $\sqrt{6}$ is irrational. We can replace equation (2) with: $$\begin{array}{}{p}^2=6q^2& \text{(4)}\end{array}$$ This shows that $p$ is a multiple of 6. Let $p=6k$ and substitute into (4): $$(6k{)}^2=6q^2$$ $$36k^2=6q^2$$ $$6k^2=q^2$$ This tells us that $q$ is a multiple of 6. By the same reasoning as above, $\sqrt{6}$ is irrational.

2. add Show that there is no rational number $r$ satisfying $2^r=3$

   Solution: Suppose that, for contradiction, there exists a rational number such that $2^r=3$. Let $r=p/q$ where $gcd(p,q)=1$ for $p,q\in Z$. We will consider three cases:

   1. $p/q>0$
   2. $p/q<0$
   3. $p/q=0$

   Case 1:

   Suppose that $p/q>0$. Then $$2^{p/q}=3⟹2^p=3^q$$ Since $2^p$ is even and $3^q$ is odd for all $p,q\in N$, this is a contradiction and no such $r>0$ exists.

   Case 2:

   Suppose that $p/q<0$. Without loss of generality, assume that $p<0$ and $q>0$. Then $$2^{-p/q}=\frac{1}{2^{p/q}}=3⟹\frac{1}{2^p}=3^q$$ Then $\frac{1}{2^p}<1$ while $3^q>1$. This is a contradiction and no such $r<0$ exists.

   Case 3:

   Suppose that $p/q=0$. Then $$2^0=3⟹1=3$$ This is a contradiction and no such $r=0$ exists.

   Hence, there is no rational number $r$ satisfying $2^r=3$.

   Q.E.D.