Prove that $\sqrt{8}$ is an irrational number.
As know that $2\lt \sqrt{8} \lt 3$, so if assume (to attempt proof by contradiction) that $\sqrt{8} = \frac pq$, where $p,q$ are coprime integers; then $2\lt \frac pq \lt 3$.
Attempt 1:
Subtracting $2$ from all terms, get : $0\lt \frac {p-2q}q \lt 1$.
This means $p-2q$ is an integer with no common terms with $q$. Also $q$ has no common terms with $p$.
But, $p-2q$ is a linear combination of $p,q$.
Am unable to use any property of linear combination of co-prime integers to directly prove by contradiction. Request help by this approach.
Attempt 2:
$2\lt \frac pq \lt 3 \implies 2q \lt p \lt 3q \implies 0\lt p-2q \lt q$. So, $\frac{p-2q}{q}$ is not an integer value & must be $\lt 1$.
As $\sqrt{8}=\frac pq$ is assumed to be a rational, so its product with $p-2q$ is also rational. But, this product cannot be an integer, as $\frac pq (p-2q)= p.\frac {p-2q}q$.
But $\frac pq(p-2q) = \frac {p^2}{q} -2p = 8q -2p$ which is a linear combination of integers. This contradicts the earlier statement.
Answer
Suppose $\sqrt8 = \frac{p}q$ where $gcd(p,q)=1$.
then we can find integer $x,y$, such that $$px+qy =1\tag{1}$$
$$8q^2=p^2$$
Hence $p$ is an even number, $p=2k$, $$2q^2=k^2$$
Hence $k$ must also be an even number. $k=2l$.
$$2q^2=(2l)^2$$
$$q^2=2l^2$$
Hence $q$ must be an even number.
Since $p$ and $q$ are both even. $px+qy$ must be even. They cannot be equal to $1$.
Remark:
Once you lose track of the property of $\sqrt8$. Your proof shouldn't work. After all, we know that there are rational numbers between $2$ and $3$.
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