# Proof that the square root of $5$ is not a natural number

I am a student and I want to know if this proof is correct.

Proof. If $\sqrt5$ is a natural number, it should be even or odd. If it would be even, we could express it as $2k$ for some integer $k$. If it would be odd, we could express it as $2j+1$ for some integer $j$. But those integers do not exist what is a contradiction. Then, $\sqrt5$ is not a natural number. Q.E.D.

Using Joofan's idea, I have a second proof. Please give me your opinion!

Proof. If $\sqrt5$ is a natural number, then there exists a natural number $n^2=5$, but $2^2<5<3^2$, and there is no natural number between $2$ and $3$. Therefore, we can conclude that there does not exist a natural number $n^2=5$, and that is a contradiction. Q.E.D.

• There are just 4 natural numbers less than 5. If the square root of 5 were a natural number, it must be from among them. You can check if any of them is. – P Vanchinathan Feb 4 '15 at 0:31
• This is not a correct proof. For one thing, why would your proof not also work for $9$? – quid Feb 4 '15 at 0:33
• @AdamHughes this is not the same question. We are doing natural roots here not rational roots. – quid Feb 4 '15 at 0:37
• @AdamHughes my question is not about how to proof that square root of 5 is irrational. I want to know if my proof is correct. – Beginner Feb 4 '15 at 0:38
• @AdamHughes obviously this question here is a consequence of that one, however this does not mean it is a duplicate. – quid Feb 4 '15 at 0:43

OK, let's think about the proof as presented:

If square root of 5 is a natural number, it should be even or odd.
If it would be even, we could express it as 2k for some integer k.
If it would be odd, we could express it as 2j+1 for some integer j.
But those integers do not exist what is a contradiction.
Then, square root of 5 is not a natural number.


So the question is, why have you specified $k$ and $j$ and what did you do or check to ensure that they do not exist? You separated into even and odd cases: what was the purpose of this separation?

An alternative:

For $n,k>0$, observe that $(n+k)^2 = n^2+2kn+k^2 > n^2$.

So $m>n \implies m^2>n^2$

Note that $3^2 = 9 > 5$, so for all $n>3, n^2>3^2>5$

Therefore the square root of $5$ is less than $3$. Only $1$ and $2$ are natural numbers less than $3$, and $1^2 = 1 \neq 5$ and also $2^2 = 4 \neq 5$

Therefore there is no natural number whose square is $5$.

• Those who wish to add zero to the natural numbers will have to write their own demonstration for that – Joffan Feb 4 '15 at 0:40
• @Beginner - sure, but what did you do to show that the odd case was not possible? And similarly the even case? – Joffan Feb 4 '15 at 1:01
• @Beginner Saying that "we know that those integers do not exist" is called begging the question, which means assuming what you're trying to prove. You must explain why they do not exist! – Théophile Feb 4 '15 at 1:10
• @Beginner I think it is an improvement. However it seems like you are in a little bit of a hurry to get to the end. You could mention the (obvious) fact that there is no natural number between 2 and 3. The amount of proof needed depends on the context; it might be that you have done enough now. – Joffan Feb 4 '15 at 4:35

If the square root of 5 is a natural number, call it $n,$ then $n^2=5,$ by definition of the square root function. Let $n=p_1^{e_1} p_2^{e_2} \cdots p_k^{e_k}$ be the unique factorization of $n$ into prime powers then $5=n^2=p_1^{2 e_1} p_2^{2 e_2} \cdots p_k^{2 e_k}$ must hold, i.e., all the prime factors of $5$ must appear in its factorization an even number of times. This is a contradiction since $5=5^1$.

• But can you assume that 5 is prime for this problem? – Paul Feb 4 '15 at 0:51
• @Paul Assuming $5$ is prime is nothing compared to the invocation of the high-power Fundamental Theorem of Arithmetic. The problem can be solved by much simpler methods. – Bill Dubuque Feb 4 '15 at 0:58

Let's assume, to the contrary, that $\sqrt{5}$ is a natural number, $n \in \mathbb{N}.$ Then, since $f(x)=\sqrt{x}$ is an increasing function, we know that $\sqrt{4}<\sqrt{5}<\sqrt{9}.$ Thus, $$\sqrt{4}<n<\sqrt{9}\implies 2<n < 3$$ Producing a contradiction. So, it must be the case that $n \notin \mathbb{N}$.

• Shouldnt it be $2 < n^2 < 3$? – Amad27 Feb 25 '15 at 13:00
• Nope, I'm not squaring my inequalities, I just took the square root of 4 and 9 – Paddling Ghost Feb 25 '15 at 22:00
• didnt you set $n = sqrt{5}$? – Amad27 Feb 26 '15 at 13:01
• I sure did. In doing so I showed that $sqrt(5)$ is between 2 and 3, and thus it can't be an integer. – Paddling Ghost Feb 26 '15 at 13:24
• IF $\sqrt{4} < n < \sqrt{9}$ then how is it possible: $2 < n < 3$ without squaring EVERYTHING? – Amad27 Feb 26 '15 at 13:30

You can prove that 5 is not rational. Suppose, without lost of generality that there exists natural numbers $p, q$ such that:

$$\sqrt5 =p/q$$ $$5q^2 = p^2$$

Then, $p$ has to be multiple of 5, so $p=5 k$. Therefore:

$$q^2=5k^2$$

So $p, q$ cannot be relatively prime. $\sqrt 5$ cannot be expressed as a ratio of integers, so it is irrational.