# prove that of sum square root of 2 and square root of 3 is not rational

• Nov 22nd 2008, 12:30 AM
transgalactic
prove that of sum square root of 2 and square root of 3 is not rational
prove that the square root of 2 plus the square root of 3 is not rational?

does always the sum of two not rational numbers is a not rational number?

i know the proof 2 = a^2/b^2
i separately proved that square root of 2 and square root of 3 are irrational

how two prove that the sum of two such numbers is irrational too?

i will try to prove by contradiction:
suppose
(2)^0.5 + (3)^0.5 is a rational number
(if we multiply a rational number by a rational number we will get a rational number "h")
5+2*(2)^0.5 * (3)^0.5=h
24=h^2 -10*h +25

h^2 -10*h +1=0
by this rational roots theorem
the possible roots is +1 and -1

not one of the represent the actual roots of h^2 -10*h +1=0

what is the next step in the prove?
• Nov 22nd 2008, 12:42 AM
o_O
Quote:

does always the sum of two not rational numbers is a not rational number?
The sum of irrationals does not always equal a rational.

Consider: $a = 1 + \pi$ ... $b = 1 - \pi$

They're both irrational but $a + b = 2$ which is rational.

______________________________

As for your problem, you found out through the rational root theorem that the only possible rational roots are: $h = \pm 1$.

If you plugged them in, they are not roots to your quadratic and so there are no rational roots, i.e. $h$ is not rational. This leads to your contradiction.
• Nov 22nd 2008, 12:49 AM
transgalactic
so we should have roots +1 ,-1
but we get only irrational roots
then "h" is irrational

??
• Nov 22nd 2008, 01:18 AM
o_O
Careful how you word it. It is not correct to conclude that your roots to your quadratic are irrational simply because there are no rational roots. They may very well be complex. But in the context of our problem, we know that $h$ is real and not rational through our reasoning above. Thus, a contradiction to our assumption that $h$ is rational. So, $h$ must be irrational.
• Nov 22nd 2008, 03:49 AM
transgalactic
what words
after i find the rational roots :

what exactly do i need to write in the end of this proof ?
• Nov 22nd 2008, 09:09 AM
great_math
Quote:

Originally Posted by transgalactic
prove that the square root of 2 plus the square root of 3 is not rational?

Here is easiest way in my knowledge of doing it:

Let $\sqrt{2}+\sqrt{3}$ be rational
So $\sqrt{2}+\sqrt{3}=r$ where r is a rational number
Square it...
$(\sqrt{2}+\sqrt{3})=r^2$
$5+2\sqrt{6}=r^2=r_1$ as r is rational so $r^2$ is rational. let $r^2=r_1$
$\sqrt{6}=\frac{r_1-5}{2}=r_2$ where $\frac{r_1-5}{2}=r_2$ [same reason as above]

so $\sqrt{6}$ is rational which is a contradiction.

Hence $\sqrt{2}+\sqrt{3}$ is irrational

Try to prove that $\sqrt{6}$ is irrational or else if you can't i will give that..

OK?
• Nov 22nd 2008, 11:09 AM
transgalactic
Quote:

Originally Posted by o_O
Careful how you word it. It is not correct to conclude that your roots to your quadratic are irrational simply because there are no rational roots. They may very well be complex. But in the context of our problem, we know that $h$ is real and not rational through our reasoning above. Thus, a contradiction to our assumption that $h$ is rational. So, $h$ must be irrational.

you say that "h" is irrational
but why??

you said yourself that it doesnt have to be irrational because of the roots

then why else its irrational?
• Nov 22nd 2008, 12:47 PM
o_O
You let $h = \left(\sqrt{2} + \sqrt{3}\right)^2$ which is real, either rational or irrational.

When you established your equation $h^2-10h + 1 = 0$, we found out that through the rational roots test, $h$ is not rational (this is all the test gives us, that it is simply not rational). A contradiction to our assumption.

Since $h$ was either rational or irrational and assuming it was rational lead to a contradiction, then it must have been irrational.
• Nov 22nd 2008, 01:04 PM
transgalactic
how do i finish the proof

if i will say "the root test told us that we dont a rational root thats why "h" is
irrational"

but you told that these words are not correct
because if there is no rational roots then it doesnt say that "h" irrational.
it can be complex number(which is rational)

how would you finish the proof
?
• Nov 22nd 2008, 01:13 PM
o_O

I am saying that the rational root test simply tells us that $h$ is not rational.

But in the context of our problem, $h$ is either rational or irrational.

Since we found a contradiction when we assumed it was rational, it must be the other.
• Nov 22nd 2008, 01:26 PM
transgalactic
a number is always rational or irrational
there is no third way

even complex numbers are considered rational

??
• Nov 22nd 2008, 01:30 PM
o_O
Quote:

Originally Posted by transgalactic

even complex numbers are considered rational

Since when? Can you represent $3 + 5i$ as a fraction of two integers?
• Nov 22nd 2008, 01:41 PM
transgalactic
a complex number is rational ?