Request help on proof of irrationality of $\sqrt {8}$. 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.
 A: Why don't you just use $2\sqrt{2} = \sqrt{8}$?
Once you do this just prove by contradiction that if $\sqrt{8}$ is rational, so must $\sqrt{2}$ be.
A: 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 $3$.
A: There are faster methods, but this is essentially the same as showing square-roots of prime numbers are irrational 


*

*Suppose $\sqrt{8}$ can be written as $\frac pq$ in lowest terms, 

*So $p^2=8q^2$, so $p^2$ is a multiple of $8$, which requires $p$ to be a multiple of $2$, so write $p=2a$ for some integer $a$

*Then $(2a)^2=8q^2$ and $4a^2=8q^2$ and $a^2=2q^2$, so $a^2$ is a multiple of $2$, which requires $a$ to be a multiple of $2$, so write $a=2b$ for some integer $b$

*Then $(2b)^2=2q^2$ and $4b^2=2q^2$ and $2b^2=q^2$, so $q^2$ is a multiple of $2$, which requires $q$ to be a multiple of $2$

*But then $p$ and $q$ are both divisible by $2$, contrary to the supposition that  $\frac pq$ is in lowest terms
A: At first glance, I think the proof will not work as the following statement hold for all real numbers, rational or irrational, between $2$ and $3$. I can't see how to use the linear combination of $p$, $q$.

$2\lt \frac pq \lt 3 \implies 2q \lt p \lt 3q \implies 0\lt p-2q \lt q$. So, $\frac{p-2q}{q}$ is not an integer value & must be $\lt 1$.

Then we have

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 $\frac pq (p-2q)= p\cdot\frac {p-2q}q$.

I can't see why $\frac pq (p-2q)$ cannot be an integer just because it is equal to $ p\cdot\frac {p-2q}q$. But I think the conclusion is correct, as $\gcd(p,q)=1$ implies that
$$\gcd(p(p-2q),q)=\gcd(p(p-2q)+2p(q),q)=\gcd(p^2,q)=1$$
So,  $\frac pq (p-2q)$ is not an integer. This holds for any rational number between $2$ and $3$ when it is written in the lowest term.
Then jiten makes a beautiful argument and complete the proof.

But $\frac pq(p-2q) = \frac {p^2}{q} -2p = 8q -2p$ which is a linear combination of integers. This contradicts the earlier statement.

Here note that $\frac{p^2}{q}=\left(\frac{p}{q}\right)^2q=8q$.
The second attempt is a nice proof, although there is one argument I can't really follow (but the conclusion is correct.)
