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We will attempt a proof by contradiction.
Assume there exists two integers $p$ and $q$ such that $\left(\frac{p}{q}\right)^2 =64$.
Hence, $p^2 = 6 \cdot q^2$.
Thus, $6 | p$ and $3 | p$.
Let $3 \cdot r=p$, where $r$ is an integer. Then, $9 \cdot r^2=6 \cdot q^2$.
Thus, $q^2=\frac{3}{2}\cdot r^2$.
Suppose $r =5$. Then, $q^2=37.5$.

Therefore, $q$ cannot be an integer, which is in contradiction with the original statement.

Note: very new to proofs (2nd one) so apologies for bad formatting etc.

Marco Ripà
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philip
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1 Answers1

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Good $2$nd attempt at a proof but unfortunately you are missing the mark slightly. Let me rewrite your proof with corrections:

We will attempt a proof by contradiction. Assume there exists integers $p,q$ such that $(p/q)^2 =6$ and $\text{gcd}(p,q)=1$.

Note that any fraction can be written in most simplified form and in fact this fact is critical for the contradiction you are trying to arrive at

Therefore $p^2 = 6q^2$. Therefore p divisible by $6$ and $p$ divisible by $3$. Let $3r=p$,where $r$ is an integer: $9r^2=6q^2$.

This section does not lead to a contradiction, even if everything except the last statement is true

Therefore $q^2=3/2(r^2)$. Suppose $r =5$: therefore $q^2=37.5$. Therefore Q cannot be an integer, which is in contradiction with original statement.

Instead, write

Divide both sides by $3$ to get $3r^2=2q^2$. Since $3$ does not divide $2$, $3$ must divide $q$.

Do you see why this is the case?

But this is a contradiction since $\text{gcd}(q,p)\geq 3$.

QC_QAOA
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