# Irrational numbers

• Jun 13th 2011, 08:09 AM
TeaWithoutMussolini
Irrational numbers
Say whether the following are true or false, backing up with examples, if $\displaystyle \alpha \in R$ is rational, and $\displaystyle \beta \in R$ is irrational:

$\displaystyle \alpha + \beta$ is irrational
$\displaystyle \alpha \beta$ is irrational
• Jun 13th 2011, 08:18 AM
Plato
With having fifty postings, you should understand that this is not a homework service nor is it a tutorial service. Please either post some of your own work on these problems or explain what you do not understand about the question.
• Jun 13th 2011, 08:49 AM
TeaWithoutMussolini
Sorry dude but I am posting the aspects of the course I am having trouble with before the exam (tomorrow).

These are exam questions from the Further discrete maths exam past papers.

The question is I dont know how to work out the answer?

You have to show whether or not they are irrational.

i know rational means it can be expressed as m/n, both integers. I think both are true that both are irrational? but I dont know how to prove it?

I had a little think about the first one using my brain and I think its a proof by contradiction, by writing m/n + b = p/q, so b= p/q - m/n = pn - mq/qn which is rational so if a + b is rational b is rational too. but the product i am not too sure?
• Jun 13th 2011, 11:27 AM
Also sprach Zarathustra
Re: Irrational numbers
Quote:

Originally Posted by TeaWithoutMussolini
Say whether the following are true or false, backing up with examples, if $\displaystyle \alpha \in R$ is rational, and $\displaystyle \beta \in R$ is irrational:

$\displaystyle \alpha + \beta$ is irrational
$\displaystyle \alpha \beta$ is irrational

Plato, he called you a dude... (Giggle)(Surprised)(Lipssealed)

$\displaystyle \alpha + \beta$

$\displaystyle \alpha =\frac{a}{b}$

Now assume that $\displaystyle \frac{a}{b}+\beta=\frac{c}{d}$, why this is a contradiction?

Try the second one alone?
• Jun 13th 2011, 12:03 PM
emakarov
Re: Irrational numbers
Quote:

Originally Posted by TeaWithoutMussolini
I had a little think about the first one using my brain and I think its a proof by contradiction, by writing m/n + b = p/q, so b= p/q - m/n = (pn - mq)/qn which is rational so if a + b is rational b is rational too.

Correct.

Quote:

but the product i am not too sure?
Similarly, assume that $\displaystyle \alpha=m/n$ and $\displaystyle \alpha\beta=p/q$ for some integer m, n, p, q and isolate $\displaystyle \beta$.
• Jun 15th 2011, 12:09 PM
Amer
Re: Irrational numbers
suppose that a+b rational

thats mean a+b = n/m
so b = n/m - a and you that the rational numbers is closed under addition so b is rational contradiction
the other is same