proving irrationality

[rsnd]

how do you prove that sqrt(2) + sqrt(5) is irrational...

[matt grime]

by contradiction, suppose r=sqrt(2)+sqrt(5) and r is rational. play around with that and see what you can show this implies.

[rsnd]

by contradiction, suppose r=sqrt(2)+sqrt(5) and r is rational. play around with that and see what you can show this implies.
i did get it down to 7+2root10 = m^2/n^2 where i assumed root 2 + root 5 = m/n...and before that i proved root 10 to be irrational...so i am kinda stuck on the 7+2root10 = m^2/n^2 bit...how does that mean anything...

[matt grime]

why not do rearrange the equation before squaring it so that after squaring you only have, say sqrt(2) to deal with, which i presume you know is irrational.

i don't see why you're showing 7+ something is not of the from m^2/n^2, to be honest, either, and you certainly don't need to write r=m/n either.

[courtrigrad]

look at this (http://www.physicsforums.com/showthread.php?t=62220&page=1&highlight=courtrigrad%3A+irrational)
for √2 + √3, let it equal a...
a^2 = 5 + 2√6
a^2 - 5 = 2√6
a^4 - 10a^2 + 25 = 24
a^4 - 10a^2 + 1 = 0

The minimum polynomial of a is x^4 - 10x^2 + 1. (I haven't proven it actually is the minimum, but it will still suffice for this method of proof)

Now, the only possible rational roots of this are 1 and -1, and neither of these is √2 + √3, so it's irrational

[matt grime]

I think that's aiming way too high for this question, and besides, if you're going to go and use theorems like that then why not simply invoke the fact that sqrt(2)+sqrt(3) is an algebraic integer, and hence if it were rational it would be integral, which it clearly isn't.

r-sqrt(2) = sqrt(5)
square and rearrange to get sqrt(2) rational, contradiction.

[rsnd]

i had an argumetn like this


let root 2 + root 5 = m/n...i.e. assume they are rational...and we have already proved that root 10 is irrational...
squaring both sides:
2+5+2.root2.root5 = m2/n2
or 7+2.root10=m2/n2
this is where i got up to...
now i am thinking that if i put everything in terms of root 10 then:

root 10 = (m2/n2 - 7)/2

so we have irrational = "(m2/n2 - 7)/2"
so that implies that (m2/n2 - 7) is irrational and also m2/n2 is irrational and they cannot be irrationals and integers at the same time and therefore its a contradiction...makes sence??? may b not...im just insane!!!

[HallsofIvy]

Assuming you already know √(2) is irrational, matt grime's suggestion, "rearrange the equation before squaring it " gives this:

Suppose r= √(2)+ √(5) is rational. Then r- √(2)= √(5).
Squaring both sides, r2- 2√(2)r+ 2= 5 so 2√(2)= r2- 3. Then √(2)= (r2- 3)/2. Since the rational numbers are closed under multiplication, division, addition, and subtraction, the right hand side is rational, giving the contradiction that √(2) is rational.

[Curious3141]

Another way of doing it, which I think is slightly "nicer" than squaring is as follows :

Let \sqrt{5} + \sqrt{2} = x_1 and \sqrt{5} - \sqrt{2} = x_2

Assume x_1 is rational and can be expressed as \frac{p}{q}, where p and q are coprime integers.

\frac{1}{2}(x_1 - x_2) = \sqrt{2}, which is irrational, hence x_2 has to be irrational (since x_1 is rational by assumption).

But x_1x_2 = 3, giving x_2 = \frac{3}{x_1} = \frac{3q}{p}

This implies x_2 is rational, contradicting the above.

Hence x_1 is irrational (QED).

[HallsofIvy]

Yes, that is nice.

[Curious3141]

Yes, that is nice.

Thank you. :smile:

BTW, there is a small error in your explanation of the squaring proof. It should go like so :

r^2 - 2\sqrt{2}r + 2 = 5
2\sqrt{2}r = r^2 - 3
\sqrt{2} = \frac{1}{2}(r - \frac{3}{r})

And LHS is irrational, RHS is rational, contradiction. You forgot the 'r' when you typed it out. Just a minor point, doesn't change the substance of your proof at all. :biggrin:

[rsnd]

thanks to all! it kinda makes sence!