Irrationality and Division

**TitaniumX** · December 17th 2008, 10:11 AM

I have two questions in my packet that I'm not completely show how to do. Any help is appreciated.

Q1) prove that the sum of square root of 2 and square root of 3 is irrational. [hint: prove square root of six is irrational and then use this as a lemma]

Q2) Suppose a,b,c,d is a set of integers such that a/c and b/c. Show that ab/cd.

**o_O** · December 17th 2008, 11:38 AM

#1: Let: $n = \sqrt{2} + \sqrt{3} \qquad {\color{red}\star _1}$

Assume $n$ is rational.

Multiply both sides by the conjugate: $n \left(\sqrt{2} - \sqrt{3}\right) = -1 \ \Leftrightarrow \ -\frac{1}{n} = \sqrt{2} - \sqrt{3} \qquad {\color{red}\star _2}$

Add ${\color{red}\star _1}$ and ${\color{red}\star _2}$ : $n - \frac{1}{n} = 2\sqrt{2} \ \Leftrightarrow \ \frac{n}{2} - \frac{1}{2n} = \sqrt{2}$

Can you see the contradiction?

______________________________

#2: What's the question?

**Isomorphism** · December 17th 2008, 08:57 PM

Originally Posted by TitaniumX

I have two questions in my packet that I'm not completely show how to do. Any help is appreciated.

Q2) Suppose a,b,c,d is a set of integers such that a/c and b/c. Show that ab/cd.

**Grandad** · December 18th 2008, 04:42 AM

Hello -

Originally Posted by o_O

#1: Let: $n = \sqrt{2} + \sqrt{3} \qquad {\color{red}\star _1}$

Assume $n$ is rational.

Multiply both sides by the conjugate: $n \left(\sqrt{2} - \sqrt{3}\right) = -1 \ \Leftrightarrow \ -\frac{1}{n} = \sqrt{2} - \sqrt{3} \qquad {\color{red}\star _2}$

Add ${\color{red}\star _1}$ and ${\color{red}\star _2}$ : $n - \frac{1}{n} = 2\sqrt{2} \ \Leftrightarrow \ \frac{n}{2} - \frac{1}{2n} = \sqrt{2}$

Can you see the contradiction?

______________________________

#2: What's the question?

This is neat. As a matter of interest, it works for any expression of the form $\sqrt{a}+\sqrt{b}$ , where $a, b \in \mathbb{N}$ and $\sqrt{a}$ and $\sqrt{b}$ are individually irrational. Just put $\sqrt{a}+\sqrt{b}=n$ , multiply both sides by $\sqrt{a}-\sqrt{b}$ , and continue as before.

Also, if you do want to prove that $\sqrt{6}$ is irrational, do it in the same way as proving $\sqrt{2}$ is irrational, as follows:

Suppose $\sqrt{6}$ is rational, and write it as $\frac{p}{q}$ , where $p, q \in \mathbb{N}$ , and $p$ and $q$ are co-prime. (In other words the fraction $\frac{p}{q}$ is in its lowest terms.)

Then:

$\frac{p^2}{q^2} = 6$

$\implies p^2 = 6q^2$

$\implies p^2$ has a factor of $2$ (and $3$ as well, but you don't actually need that)

$\implies p$ has a factor $2$ , since $2$ is prime

So write $p=2m$ , say, where $m \in \mathbb{N}$

So $p^2=4m^2=6q^2$

$\implies 3q^2=2m^2$

$\implies q^2$ has a factor $2$

$\implies q$ has a factor $2$

$\implies p$ and $q$ have a common factor $2$ .

Contradiction.

$\implies \sqrt{6}$ is irrational.

Grandad

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