What is the basic method for proving the irrationality of a number?

**mfetch22** · July 20th 2010, 06:47 AM

I know to prove the irrationality of $\sqrt{2}$ we simply minipulate the following:

$(\frac{p}{q})^2 = 2$

until we hit the contradiction of not being in reduced terms. I also know that we must take into account that all number can be written in the form $2n$ or $2n+1$ for $n=1,2,3...$ ; and the fact that if $x^2$ is an even number then $x$ is an even number. Heres my question, how do you prove the irrationality of a certain opertation on numbers in general? And for odd square roots?

Two examples of what I mean:

[1] How would you prove that if $n=1, 2, 3...$ that $\sqrt{n}$ is always irrational?

[2] How would you prove $\sqrt{3}$ is irrational? Would you find a way to define numbers in terms of 3's ? Just like with the $\sqrt{2}$ ? So all numbers would be of the form $3n$ , $3n+1$ , and $3n+2$ ; and would you some how use that fact to prove the irrationality of $\sqrt{3}$ ?

Thanks in advance

**Ackbeet** · July 20th 2010, 06:51 AM

[1] You wouldn't. It's not true: the square root of 4 is rational.
[2] I think this proof follows pretty much the same lines as for the square root of 2.

**mfetch22** · July 20th 2010, 07:03 AM

Opps. That was a big mistake. For [1] I meant this:

[1] How would one prove that $\sqrt{x}$ is irrational for all $x = 1, 2, 3...$ such that $x \neq n^2$ for any natural number $n$ such that $n = 1, 2, 3...$

My apologies, that was a very idiotic error in clearity. Hopefully that makes sense now.

**Ackbeet** · July 20th 2010, 07:05 AM

I think generally that that proof would go along the same lines as for the square root of 2. Try it and see what happens.

**mathaddict** · July 20th 2010, 07:17 AM

Originally Posted by mfetch22

I know to prove the irrationality of $\sqrt{2}$ we simply minipulate the following:

$(\frac{p}{q})^2 = 2$

until we hit the contradiction of not being in reduced terms. I also know that we must take into account that all number can be written in the form $2n$ or $2n+1$ for $n=1,2,3...$ ; and the fact that if $x^2$ is an even number then $x$ is an even number. Heres my question, how do you prove the irrationality of a certain opertation on numbers in general? And for odd square roots?

Two examples of what I mean:

[1] How would you prove that if $n=1, 2, 3...$ that $\sqrt{n}$ is always irrational?

[2] How would you prove $\sqrt{3}$ is irrational? Would you find a way to define numbers in terms of 3's ? Just like with the $\sqrt{2}$ ? So all numbers would be of the form $3n$ , $3n+1$ , and $3n+2$ ; and would you some how use that fact to prove the irrationality of $\sqrt{3}$ ?

Thanks in advance

For (2), it's the same way how you would prove the irrationality of root(2).

Assuming that $\sqrt{3}$ is rational, then it takes the form $\sqrt{3}=\frac{m}{n}$ , where $m,n\in Z$ and m and n do not have common factors.

$3=\frac{m^2}{n^2}\Rightarrow n^2=\frac{m}{3}\cdot m$

Since $n^2$ is an integer, it follows that m/3 is an integer too. Thus, 3 is a factor of m.

Let $m=3p$ where $p\in Z$ , $n^2=3p^2$

$p^2=\frac{n}{3}\cdot n$

$n/3 \in Z$

We see that both m and n have a common factor of 3 which contradicts our assumption.

**Also sprach Zarathustra** · July 20th 2010, 07:28 AM

It is a nice exercise to prove following:

Every square root of a prime number is irrational.

**mfetch22** · July 20th 2010, 07:36 AM

Okay, your right, I think I've figured out the proof for all even numbers, but I'm alittle lost for the odd numbers. Here is what I have, please point out any mistakes I made.

-> We want to prove that $\sqrt{n}$ is irrational, as long as no numbers $x = 1, 2, 3....$ satisfy $x^2 = n$ .

[i] If $\sqrt{n}$ is rational, then it can be written in the following form:

$(\frac{a}{b}) = \sqrt{n}$ thus $(\frac{a}{b})^2 = n$

[ii] This leads to the following equation:

$a^2 = nb^2$

$a^2$ is either even or odd. Lets start with even.

[Case 1]

Say $a^2$ is even, then for some $p = 1, 2, 3...$ we have

$a^2 = 2p = nb^2$

Thus, if $a^2$ is even, [of the form $2p$ ]

then $a$ is even, and of the form $2k$
for some natural number $k = 1, 2, 3....$ .

Thus, we have $a = 2k$ and :

$a^2 = 4k^2 = 2p = nb^2$

This proves $nb^2$ to be even.

If $n$ is even, then we'd arrive at the form (for some $t=1,2,3...$ )

$b^2 = 2t$

thus, $b = 2g$

for some $g =1,2,3....$

Thus $a$ and $b$ have a common divisor, and this is a contradiction.

Is this correct? For this one case where $a = 2k$ for some $k = 1, 2, 3...$ and $n = 2g$ for some $g=1, 2, 3...$ ? I don't know exactly how to go about proving it with the odd numbers, I tried on pen and paper using the form $2k+1$ for $k = 1, 2, 3...$ but the "common factor" never seemed to show up. Can somebody give me a hint or show me the right direction?

**melese** · July 20th 2010, 07:40 AM

[1] How would you prove that if that is always irrational?

When you use the Unique Factorization Theorem, you get a simple proof.

Let $n$ be a positive integer. We can show that if $\sqrt{n}$ is rational, then $n$ must be a perfect square. From here it follows that if $n$ is not a perfect square, then $\sqrt{n}$ is irrational.

Suppose that $\sqrt{n}=a/b$ for $a, b$ positive. Then $a^2n=b^2$ . For any prime $q$ , let $q^A, q^N,$ and $q^B$ be the highest powers of $q$ that divide $a, n,$ and $b$ , respectively.

Now, because $a^2n=b^2$ and using the Unique Factorication Theorem, we must have $2A+N=2B$ and then $2|N$ . This means that $n$ is a perfect square.

This way you can even prove a more general result: For any positive integers $n, m$ ; if $\sqrt[m]{n}$ is rational, then $n$ must be a perfect $m$ th power.

**melese** · July 21st 2010, 05:57 AM

There is a nice proof I saw once. It can be generalized to any positive integer that is not a perfect sqaure.

Soppose that $\sqrt{2}=a/b$ , where $a, b$ are postive integers that are relatively prime. Then $\sqrt{2}=\sqrt{2}\cdot1=\sqrt{2}\cdot(ar+bs)$ for some integers $r, s$ . (Bézout's Identity)

Because $\sqrt{2}=a/b$ , we have $a=\sqrt{2}b$ and $b=a/\sqrt{2}$ . Then $\sqrt{2}=\sqrt{2}ar+\sqrt{2}bs=2br+as$ , and $\sqrt{2}ar+\sqrt{2}bs=2br+as$ is clearly an integer.

This means that $\sqrt{2}$ is an integer, but $1<\sqrt{2}<2$ ; contradiction.

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