Proof irrational

**statman101** · May 16th 2010, 07:06 PM

Can anyone give a complete proof step by step that the:

sqrt(41)+5=irrational
given that the sqrt(41) is irrational..

Show that it is irrational by the fact that the
sqrt(41) is irrational.

**roninpro** · May 16th 2010, 07:09 PM

How about by contradiction?

What happens if $\sqrt{41}+5$ is irrational?

**statman101** · May 16th 2010, 07:13 PM

by contradiction:
I have the sqrt(41)+5=p/q
but how does it help knowing the sqrt(41)=irrat.

**roninpro** · May 16th 2010, 07:14 PM

What if you subtract 5 from both sides?

$\sqrt{41}=\frac{p}{q}-5$

Anything wrong with this?

**statman101** · May 16th 2010, 07:20 PM

still cant see it

**roninpro** · May 16th 2010, 07:24 PM

We have a rational $\frac{p}{q}$ subtract another rational $5$ . What kind of number results?

**statman101** · May 16th 2010, 07:44 PM

Rat - rat = rat
I tried the prob before I submited it..
I appreciate u trying to me to get to think but
I posted it to get a complete proof..
So anyone ELSE please prove

**roninpro** · May 16th 2010, 07:49 PM

You have it right there. First, you are given that $\sqrt{41}$ is irrational. Then, you said that $\sqrt{41}=\frac{p}{q}-5$ is rational, since a rational subtract a rational is rational. Hence, $\sqrt{41}$ is rational and irrational.

Isn't this a problem?

**statman101** · May 17th 2010, 08:06 AM

ok partner, i can see it..thanks for sticking it out

**undefined** · May 17th 2010, 08:24 AM

If it's any use to you, it's also possible to write

$\sqrt{41}=\frac{p}{q}-5=\frac{p-5q}{q}=\frac{p'}{q} \in \mathbb{Q}$ .

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