Prove that $\sqrt{8}$ is irrational in different method

Question

I tried to prove that $\sqrt{8}$ is irrational. I said let $\sqrt{8}$ be rational then $\sqrt{8}$ = $a/b$ where $a$ and $b$ are relatively prime.

Then $2\sqrt{2}=a/b$ , and $\sqrt{2} =a/(2b)$. it is obvious that $RHS$ is rational and $LHS$ is irrational (assumed that $\sqrt{2}$ is proved). So there is a contradiction and proof done.

My question is that is there other ways to prove that $\sqrt{8}$ is irrational?

Answer 1

let $\sqrt{8}$ is equal to $\frac {a}{b}$ where $(a,b)=1$

Then $8 = \frac {(a^2)}{(b^2)}$ , so $8(b^2)= a^2$

Because of the fact that $(a,b)$ are relatively prime $b$ cannot divide $a$.Hence $8|a^2$.

Then ,let's say $a=2k$ where $k$ is a positive integer.

Then $8(b^2)=(2k)^2$. Then,by simplification $2(b^2)=k^2$ .It is obvious that $k$ is an even integer, so k=2m where m is positive integer.

so $2(b^2)=(k^2)=(2m)^2$.then $b=8m b^2 = 2(m^2)$.As a result $2|b$ .

we concluded that a and b have common divisor which is $2$. it is a contradiction.

Answer 2

Another way would be to write $\sqrt{8}$ as a continued fraction and then apply Lagrange´s theorem to conclude that it is irrational.

Answer 3

You could use the equation $x^2-8=0$ and apply the Rational Root Theorem.

Answer 4

Use the fact (that works for any non perfect square (i'e. is a square of a whole number) ) that :

If $$\sqrt{n} =a/b$$ then $$a^2 n=b^2$$

Then because the exponents of prime decomposition are all even at the right side of the expression and somes are odd at the left (because $n$ isn't a perfect square)

$\sqrt{n} $ cannot be rational as $n$ isn't a perfect square.