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

Question

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 04 '15 at 00:31
This is not a correct proof. For one thing, why would your proof not also work for $9$? — quid, Feb 04 '15 at 00:33
possible duplicate of use contradiction to prove that the square root of $p$ is irrational — Adam Hughes, Feb 04 '15 at 00:35
My question is what specific part of your proof does not work for $9$ instead of $5$? If there is not step that does not work you proof is no good as $9$ has a natural squareroot, namely $3$. — quid, Feb 04 '15 at 00:36
@AdamHughes this is not the same question. We are doing natural roots here not rational roots. — quid, Feb 04 '15 at 00:37
@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 04 '15 at 00:38
@AdamHughes obviously this question here is a consequence of that one, however this does not mean it is a duplicate. — quid, Feb 04 '15 at 00:43
@Beginner It seems that you may be attempting to derive a contradiction from the inference that $,5=n^2,\Rightarrow,n,$ is odd. If so, you need to say more explicitly how you derive a contradiction from that. — Bill Dubuque, Feb 04 '15 at 00:48
possible duplicate of http://math.stackexchange.com/questions/451700/prove-that-sqrt-5-is-irrational. — lhf, Feb 04 '15 at 00:58
@Beginner: Yes, it is called proving by cases; it is worth pursuing when the number of cases is small. Actually for your question the proof rests on the fact that positive integers can be written as a product of powers of prime numbers in a unique manner.(The Fundamental Theorem of Arithmetic) — P Vanchinathan, Feb 04 '15 at 03:16
@Beginner A proof by FTA already exists in kodlu's answer. Using FTA is overkill. If you don't know how to finish the proof you started then say so and we can help you. — Bill Dubuque, Feb 04 '15 at 03:32
@Beginner It's one simple way. Another: $,n^2 =5,\Rightarrow,n^2\equiv 2\pmod 3,,$ but modulo $,3,$ the only squares are ${0^2,1^2,2^2}\equiv {0,1}.\ $ — Bill Dubuque, Feb 04 '15 at 03:57
@Beginner That's a less detailed form of the proof in Joffan's answer. — Bill Dubuque, Feb 04 '15 at 04:25
@Beginner Why did you delete all but one of your comments on this page? Doing that makes it more difficult for other readers to learn from the discussions (as you did). — Bill Dubuque, Feb 04 '15 at 22:42
@BillDubuque I am so sorry, I thought that was a good idea. I will not do it again. — Beginner, Feb 04 '15 at 23:00

score 4 · Answer 1 · answered Feb 04 '15 at 00:40

4

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 $5=5^1$.

answered Feb 04 '15 at 00:40

kodlu

9,338

But can you assume that 5 is prime for this problem? – Paul Feb 04 '15 at 00:51
2

@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 04 '15 at 00:58

score 2 · Answer 2 · answered Feb 04 '15 at 05:15

2

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 $n \notin \mathbb{N}$.

answered Feb 04 '15 at 05:15

Paddling Ghost

1,999

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
What do you mean?? $\sqrt 4 = 2$, right? $\sqrt 9 = 3$, right? So... if $\sqrt 4 < n < \sqrt 9$, then $2<n<3$. Again, I'm not squaring anything, all I'm doing is recognizing that $\sqrt 4 = 2$ and $ \sqrt 9 = 3$. – Paddling Ghost Feb 26 '15 at 17:38
If I were to have squared my inequality, I would have ended up with $4<n^2<9$ which doesn't really help me. – Paddling Ghost Feb 26 '15 at 17:46
Sorry, I seemed to have gone blank!! (+1) – Amad27 Feb 26 '15 at 18:29

Joffan · Accepted Answer · 2015-02-04T00:51:35.520

1

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 $5$.

edited Feb 04 '15 at 00:51

answered Feb 04 '15 at 00:39

Joffan

39,627

Those who wish to add zero to the natural numbers will have to write their own demonstration for that – Joffan Feb 04 '15 at 00:40
@Beginner - sure, but what did you do to show that the odd case was not possible? And similarly the even case? – Joffan Feb 04 '15 at 01:01
1

@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 04 '15 at 01:10
@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 04 '15 at 04:35

score 0 · Answer 4 · answered Aug 16 '15 at 09:56

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.

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

4 Answers4