# prove irrational

• Feb 5th 2008, 02:31 PM
mandy123
prove irrational
Prove that sqrt(1 + sqrt(3)) is irrational!
• Feb 5th 2008, 02:59 PM
Jhevon
Quote:

Originally Posted by mandy123
Prove that sqrt(1 + sqrt(3)) is irrational!

you can do a proof by contradiction

assume it is rational. then we can write $\sqrt{1 + \sqrt {3}} = \frac ab$ for $a,b \in \mathbb{Z}$, $b \ne 0$.

we square both sides to obtain: $1 + \sqrt{3} = \frac {a^2}{b^2}$

so $\sqrt{3} = \frac {a^2}{b^2} - 1$

now the right side of the equation is a rational number (by our assumption). thus we must have that $\sqrt{3}$ is rational. therefore, we can prove the claim by showing that this is false.

now proving $\sqrt{3}$ is irrational is also done by contradiction, and usually done the same way $\sqrt{2}$ is proven to be irrational (see here)

but there is a shorter way.

consider the equation $x^2 - 3 = 0$

by the rational roots theorem, if rational roots exist, they will be one of $\pm 1, \mbox{ or } \pm 3$. but none of these solve the equation. however, $\sqrt{3}$ solves the equation and is therefore not a rational number.

so seeing that $\sqrt{3}$ is not rational, we have our contradiction
• Feb 6th 2008, 04:47 AM
mr fantastic
Quote:

Originally Posted by Jhevon
[snip]

now proving $\sqrt{3}$ is irrational is also done by contradiction, and usually done the same way $\sqrt{2}$ is proven to be irrational (see here)

[snip]

I'll just toss in my 2 cents worth here and say that proving $\sqrt{3}$ is irrational is subtlely more difficulty than proving $\sqrt{2}$ is irrational ......

If you're assuming the usual thing, $\sqrt{3} = \frac{m}{n}$ where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).
• Feb 6th 2008, 05:11 AM
CaptainBlack
Quote:

Originally Posted by mr fantastic
I'll just toss in my 2 cents worth here and say that proving $\sqrt{3}$ is irrational is subtlely more difficulty than proving $\sqrt{2}$ is irrational ......

If you're assuming the usual thing, $\sqrt{3} = \frac{m}{n}$ where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).

I think this may be overkill, they may already know that the square root of
any integer that is not a perfect square is irrational (or even n-th root of
an integer not an n-th power).

In this form of question I woukd assume they know that $\sqrt{3}$ is irrational.

RonL
• Feb 6th 2008, 09:01 AM
ThePerfectHacker
Quote:

Originally Posted by mr fantastic
I'll just toss in my 2 cents worth here and say that proving $\sqrt{3}$ is irrational is subtlely more difficulty than proving $\sqrt{2}$ is irrational ......

If you're assuming the usual thing, $\sqrt{3} = \frac{m}{n}$ where m,n are relatively prime integers, then at some stage you must prove that if m^2 is divisible by 3, then m is also divisible by 3.

This is slightly more difficult than proving that if m^2 is divisible by 2 then m is divisible by 2 .... (but can nevertheless be easily done by using the fact that every positive integer has a unique prime factorisation).

It is extremely simple.

Definition: Given a positive integer $a$ and a prime $p$ we define $\mbox{ord}_p a = n$ ( $n\geq 0$)as the number of times that $p$ divides $a$, i.e. $p^n | a$ but $p^{n+1} \not | a$. (Note, this definition is a good definition because eventually powers of $p$ will exceeded $a$ and therefore will not divide it any further thus such an $n$ has to exist).

Lemma 1: Given two positive integers $a \mbox{ and }b$ with $p$ prime we have that $\mbox{ord}_p (ab) = \mbox{ord}_p a + \mbox{ord}_p b$.

Proof: If $\mbox{ord}_p a= n$ then $p^n|a$ but $p^n\not | a$. If $\mbox{ord}_p b = m$ then $p^m | b$ but $p^m \not | b$. This means that $p^{n+m}| ab$ but if $p^{n+m+1}|ab$ then $p^{n+1}$ (or $p^{m+1})$ divides $a$ (or $b$ respectively) which is a contradiction.

Lemma 2: Given a positive integer $a$ the integer is a power of $n$ if and only if $\mbox{ord}_p a$ is divisible by $n$ for all primes $p$.

Proof: If $a=b^n$ then $\mbox{ord}_p a = \mbox{ord}_p b^n = n\mbox{ord}_p b$. Conversely, if $\mbox{ord}_p a$ is divisible by $n$ then by unique factorization we know that $a= \prod_{p \text{ prime}} p^{\text{ord}_p a}$ and thereofore each exponent is prime and divisible by $n$ hence $a$ is a power of $n$.

Theorem: If $a\geq 1$ is not a power of $n$ then $\sqrt[n]{a}$ is irrational.

Proof: Proof by contrapositive. Suppose that $\sqrt[n]{a} = r/s$ for positive integers $r,s$. Then, $(r/s)^n = a$ thus $r^n = as^n$. Now for any prime $p$ we have $\mbox{ord}_p r^n = \mbox{ord}_p as^n$ thus $n \mbox{ord}_p r = n\mbox{ord}_p s + \mbox{ord}_p a$. This shows us that $\mbox{ord}_p a$ is divisible by $n$ for any prime $p$. Thus, by Lemma 2 we have that $a$ is a power of $n$.