Proof Strategy Practice

1.2.1

a

Theorem

$\sqrt{3}$ is irrational.

Proof
Assume for contradiction that $\sqrt{3}$ is rational, so then:
$\frac{p}{q} = \sqrt{3}$
for some $p, q \in Z$ where $q \neq 0$ and $p, q$ don't share any common factors. Squaring both sides gives that:
$p^{2} = 3 q$
so $3 | p^{2}$ , or $p^{2}$ has a factor of $3$ . Notice that because of that:
$\frac{p^{2}}{3} = \frac{p}{3} \cdot p \in Z$
so then since $p \in Z$ then clearly $3 | p$ so then set $p = 3 k$ for some $k \in Z$ so then:
$(3 k)^{2} = 3 q^{2} \Rightarrow 9 k^{2} = q^{2} \Rightarrow q^{2} = 3 p$
so then $q^{2}$ is divisible by $3$ , and by the same argument $3 | q$ which contradicts $p, q$ not sharing any factors.
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Note that the argument would follow similarly for $\sqrt{6}$ , except that when we breakup $p^{2} / 6$ notice that 6 isn't prime so then we'll get:
$\frac{p^{2}}{6} = \frac{p}{2} \cdot \frac{p}{3}$
so we show that $p$ is either divisible by 2 or by 3 and continue the argument to show that $q$ shares the same factor.

b

The proof for showing $\sqrt{4}$ would work in the same way it did for 1.2 - Set Theory Review#1.2.1#a since we'd get to:
$p^{2} = 4 q^{2} = 2 (2 q^{2})$
so then $p$ is even, so then $p = 2 k$ :
$(2 k)^{2} = 2 (2 q^{2}) \Rightarrow 4 k^{2} = 4 q^{2} \Rightarrow k^{2} = q$
but this couldn't lead to a contradiction as $k = | q |$ so then $p = 2 | q |$ so then:
$\frac{p}{q} = \frac{2 | q |}{q} = \pm 2$
which doesn't give a contradiction.

1.2.2

Theorem

There is no rational number $r$ satisfying $2^{r} = 3$ .

Proof
Assume for contradiction that there is some rational number $r \in Q$ such that $2^{r} = 3$ . Namely:

r = \frac{a}{b}; a, b \in Z; b \neq 0

So then plugging in:

\begin{aligned} 2^{r} & = 3 \\ 2^{\frac{a}{b}} & = 3 \\ 2^{a} & = 3^{b} & Raise both to b power \end{aligned}

But this can't be true. Here $2, 3$ are prime numbers, so $2^{a}$ is already in it's prime factorization and similar with $3^{b}$ . Since prime factorizations are unique and since $2^{a} = 3^{b}$ then the hidden $2^{0}$ on the RHS suggests that $a = 0$ . Doing something similar with the LHS $3^{0}$ suggests $b = 0$ but that contradicts $b \neq 0$ !

Therefore the assumption was false, so $r$ cannot exist.
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