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

• Jul 20th 2010, 06:47 AM
mfetch22
What is the basic method for proving the irrationality of a number?
I know to prove the irrationality of $\displaystyle \sqrt{2}$ we simply minipulate the following:

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

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 $\displaystyle 2n$ or $\displaystyle 2n+1$ for $\displaystyle n=1,2,3...$; and the fact that if $\displaystyle x^2$ is an even number then $\displaystyle x$ 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 $\displaystyle n=1, 2, 3...$ that $\displaystyle \sqrt{n}$ is always irrational?

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

• Jul 20th 2010, 06:51 AM
Ackbeet
[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.
• Jul 20th 2010, 07:03 AM
mfetch22
Opps. That was a big mistake. For [1] I meant this:

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

My apologies, that was a very idiotic error in clearity. Hopefully that makes sense now.
• Jul 20th 2010, 07:05 AM
Ackbeet
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.
• Jul 20th 2010, 07:17 AM
Quote:

Originally Posted by mfetch22
I know to prove the irrationality of $\displaystyle \sqrt{2}$ we simply minipulate the following:

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

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 $\displaystyle 2n$ or $\displaystyle 2n+1$ for $\displaystyle n=1,2,3...$; and the fact that if $\displaystyle x^2$ is an even number then $\displaystyle x$ 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 $\displaystyle n=1, 2, 3...$ that $\displaystyle \sqrt{n}$ is always irrational?

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

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

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

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

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

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

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

$\displaystyle n/3 \in Z$

We see that both m and n have a common factor of 3 which contradicts our assumption.
• Jul 20th 2010, 07:28 AM
Also sprach Zarathustra
It is a nice exercise to prove following:

Every square root of a prime number is irrational.
• Jul 20th 2010, 07:36 AM
mfetch22
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 $\displaystyle \sqrt{n}$ is irrational, as long as no numbers $\displaystyle x = 1, 2, 3....$ satisfy $\displaystyle x^2 = n$.

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

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

[ii] This leads to the following equation:

$\displaystyle a^2 = nb^2$

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

[Case 1]

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

$\displaystyle a^2 = 2p = nb^2$

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

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

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

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

This proves $\displaystyle nb^2$ to be even.

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

$\displaystyle b^2 = 2t$

thus, $\displaystyle b = 2g$

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

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

Is this correct? For this one case where $\displaystyle a = 2k$ for some $\displaystyle k = 1, 2, 3...$ and $\displaystyle n = 2g$ for some $\displaystyle g=1, 2, 3...$? I don't know exactly how to go about proving it with the odd numbers, I tried on pen and paper using the form $\displaystyle 2k+1$ for $\displaystyle k = 1, 2, 3...$ but the "common factor" never seemed to show up. Can somebody give me a hint or show me the right direction?
• Jul 20th 2010, 07:40 AM
melese
Quote:

[1] How would you prove that if that is always irrational?
When you use the Unique Factorization Theorem, you get a simple proof.

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

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

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

This way you can even prove a more general result: For any positive integers $\displaystyle n, m$; if $\displaystyle \sqrt[m]{n}$ is rational, then $\displaystyle n$ must be a perfect $\displaystyle m$th power.
• Jul 21st 2010, 05:57 AM
melese
An alternative solution.
There is a nice proof I saw once. It can be generalized to any positive integer that is not a perfect sqaure.

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

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

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