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'Prove by contradiction that $\sqrt5$ is irrational.'

Proof: Assume that $\sqrt5$ is rational i.e. $\sqrt5 = p/q,$ where $p,q \in \Bbb Z$.

Then $\sqrt5q = p$

Now for p to be an integer, $\sqrt5q$ must be an integer, i.e $q=\sqrt5$, $2\sqrt5$, $3\sqrt5$, ...

$\implies$ q must be irrational for p to be an integer.

$\implies$ $p,q \notin \Bbb Z$.

Contradiction. Therefore $\sqrt5$ is irrational. #

Sonjov
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    '(therefore) $q$ must be irrational for $p$ to be an integer'. How do you know this? That's effectively the step that you have to prove - as it stands your logic is circular. How do I know that, say, $305\sqrt{5}$ isn't an integer? – Steven Stadnicki Aug 25 '17 at 06:12
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    what was the contradiction here? – MAN-MADE Aug 25 '17 at 06:12
  • "$\implies$ q must be irrational for $p$ to be an integer." That's mere assertion. I mean, that's really what you are trying to prove, no? – Thomas Andrews Aug 25 '17 at 06:13
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    Why does the fact $\sqrt{5}q \in \mathbb{Z}$ imply that $q$ is an integer multiple of $\sqrt{5}$? – M. Van Aug 25 '17 at 06:14
  • True, $q$ could be $\frac{9}{\sqrt{5}}$. @M.Van – Thomas Andrews Aug 25 '17 at 06:15
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    Your proof is not correct but +1) from me for showing your thoughts and attempt – Marios Gretsas Aug 25 '17 at 06:28
  • Possible duplicate of $\sqrt a$ is either an integer or an irrational number. – Davood Aug 25 '17 at 06:29
  • @Shansss ; See each of the following: https://math.stackexchange.com/questions/4467/sqrt-a-is-either-an-integer-or-an-irrational-number https://math.stackexchange.com/questions/189130/prove-that-if-n-is-not-the-square-of-a-natural-number-then-sqrtn-is-irra https://math.stackexchange.com/questions/2405026/if-p-1-p-n-are-positive-prime-numbers-then-sqrtp-1-cdots-p-n-notin-b – Davood Aug 25 '17 at 06:31
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    @Famke This is not a duplicate. OP is proposing an alternative proof and is asking if it is a valid proof. It's not, but that doesn't make the question a duplicate. – 5xum Aug 25 '17 at 06:34

4 Answers4

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Your proof is not correct because you have to prove that $q= k\sqrt{5}$ is irrational for every integer $k$ by proving that $\sqrt{5}$ is irrational.

Note that if we multiply a non-zero integer and an irrational the product will be an irrational number.(You can prove this for practise)

So you assume what you want to prove.

Here it is a valid proof which i hope it will help you:

Assume that $\sqrt{5}=\frac{m}{n}$ where $(m,n)=1$.

We assume that $(m,n)=1$ because if its not then we can cancel a priori every common factor of the numerator and denominator until we remain with a fraction $\frac{s}{l}$ in its lowest terms namely $(s,l)=1$.

Thus $$5n^2=m^2 \Rightarrow 5|m^2 \Rightarrow 5|m$$ because $5$ is a prime number.

Also because $5|m$ we have that $m^2=25s^2$

Thus $$5n^2=25m^2 \Rightarrow n^2=5m^2 \Rightarrow 5|n$$

Now we have a contradiction because $5|n$ and $5|m$ and we assumed that $(n,m)=1$

So $\sqrt{5}$ is an irrational.

Community
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Marios Gretsas
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This is wrong.

Your statement that "$\sqrt{5}q$ must be an integer" does not imply that "$q$ is irrational".

marty cohen
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Note that "$\Longrightarrow$ q must be irrational for p to be an integer." is not a correct statement. It includes the assumption that $\sqrt5$ is not a rational number which you want to prove. (See more in the end)

Here is an alternative way:

Let $\gcd(p,q)=1$,

$\sqrt5p=q\Rightarrow 5p^2=q^2\Rightarrow 6p^2=p^2+q^2$.

Then $3\mid p^2+q^2$, but $p^2+q^2\equiv 1,2(\mod 3)$ since $\gcd(p,q)=1$ and $p^2\equiv 0,1 (\mod 3)$ and $q^2\equiv 0,1 (\mod 3)$, which leads to contradiction.

Added:

"Now for p to be an integer, $\sqrt5 q$ must be an integer, i.e $q=\sqrt5,2\sqrt5,3\sqrt5\dots$" also a wrong statement in a way.

For example, let $2=\dfrac{p}{q}\Rightarrow q=\dfrac{p}{2}$. Now you you would write...

"Now for q to be an integer, $\dfrac{p}{2}$ must be an integer, i.e $p=\dfrac{1}{2},\dfrac{2}{2},\dfrac{3}{2},\dots$"

But your assumption was $p$ is an integer. But here isn't for each value of $q$. Does it imply $2$ is not a rational number!!!

MAN-MADE
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No, your proof is not valid. The mistake happens in two lines:

First mistake:

Now for p to be an integer, $\sqrt5q$ must be an integer, i.e $q=\sqrt5$, $2\sqrt5$, $3\sqrt5$, ...

You didn't list all the possibilities here. For example, $\sqrt{5}q$ is an integer if $q=\frac{9}{\sqrt{5}}$, but you didn't list that possibility.

Second, much bigger mistake:

$\implies$ q must be irrational for p to be an integer.

How do you know this? This is not true for $\sqrt{4}$, so why should it be true for $\sqrt{5}$?

(yes, it is true for $\sqrt 5$, but you have to prove it).

5xum
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