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.

Printable View

- May 16th 2010, 07:06 PMstatman101Proof irrational
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. - May 16th 2010, 07:09 PMroninpro
How about by contradiction?

What happens if $\displaystyle \sqrt{41}+5$ is irrational? - May 16th 2010, 07:13 PMstatman101
by contradiction:

I have the sqrt(41)+5=p/q

but how does it help knowing the sqrt(41)=irrat. - May 16th 2010, 07:14 PMroninpro
What if you subtract 5 from both sides?

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

Anything wrong with this? - May 16th 2010, 07:20 PMstatman101
still cant see it

- May 16th 2010, 07:24 PMroninpro
We have a rational $\displaystyle \frac{p}{q}$ subtract another rational $\displaystyle 5$. What kind of number results?

- May 16th 2010, 07:44 PMstatman101
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 - May 16th 2010, 07:49 PMroninpro
You have it right there. First, you are given that $\displaystyle \sqrt{41}$ is irrational. Then, you said that $\displaystyle \sqrt{41}=\frac{p}{q}-5$ is rational, since a rational subtract a rational is rational. Hence, $\displaystyle \sqrt{41}$ is rational and irrational.

Isn't this a problem? - May 17th 2010, 08:06 AMstatman101
ok partner, i can see it..thanks for sticking it out

- May 17th 2010, 08:24 AMundefined
If it's any use to you, it's also possible to write

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