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

1. Nov 22, 2008

### transgalactic

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?

whats the formal equation proof?

Last edited: Nov 22, 2008
2. Nov 22, 2008

### Staff: Mentor

Try a proof by contradiction. Suppose that $$\sqrt{2} + \sqrt{3} = \frac{a}{b}$$
where a and b are integers with no common factors.

For your second question $$\sqrt{5} + (-\sqrt{5}) = 0$$
Those are both irrational numbers, but their sum is rational.

3. Nov 22, 2008

### nicksauce

Well clearly two irrational numbers can sum to give a rational number. For example (2+sqrt(2)) + (2 -sqrt(2)).

To show that the sqrt(2) + sqrt(3) is irrational, you can start with
a = sqrt(2) + sqrt(3)
Manipulate to get
a^4 - 10a^2 + 1 = 0

Then you can use the rational roots theorem, that shows that the only rational roots of this equation can be +- 1. http://en.wikipedia.org/wiki/Rational_root_theorem

4. Nov 22, 2008

### Chaos2009

Somehow, that just seems like cheating, but it does fit the problem we were given.

5. Nov 22, 2008

### nicksauce

Any mathematical statement can be disproved by a counterexample. The simpler the counterexample, the better.

6. Nov 22, 2008

### transgalactic

ok 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
what to do now?

7. Nov 22, 2008

### nicksauce

As I suggested earlier, make use of the rational roots theorem.

8. Nov 22, 2008

### transgalactic

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?

9. Nov 22, 2008

### Staff: Mentor

What did you set out to do in your post #6?

10. Nov 22, 2008

### Quantumpencil

You don't need anything but parity to prove this.

Prove

$$\sqrt{2}+\sqrt{3} = (p/q)$$
$$2 \sqrt{6}+5 = p^{2}/q^{2}$$

You know that the rationals form a field, which implies all the terms in the sum must be a rational.

Now all that remains is to show $$\sqrt{6}$$ is irrational. So just show that if it equals p/q, p and q must both be even.

Last edited by a moderator: Jun 22, 2009
11. Nov 23, 2008

### transgalactic

the root test showed me that there is no rational roots for this equation.
is it fair to consider "h" irrational because of it?

i was told that if we dont have a rational roots it doesnt mean that
"h" is always irrational,"h" could be a complex number.

??

12. Nov 23, 2008

### Staff: Mentor

If a number is complex, it certainly isn't a rational number. If h^2 - 10h + 1 = 0 doesn't have any rational roots, then there is no solution for h that is rational.

Last edited: Nov 23, 2008
13. Nov 23, 2008

### transgalactic

you meant "irrational" (in the end) right?

14. Nov 23, 2008

### Staff: Mentor

I edited my reply. The last part should have said "there is no solution that is rational."

15. Jun 22, 2009

### jay17

how did you arrive at a^4 - 10a^2 + 1 = 0?
need help

16. Jun 22, 2009

### Mentallic

$$a=\sqrt{2}+\sqrt{3}$$
$$a^2=(\sqrt{2}+\sqrt{3})^2=2+2\sqrt{2}\sqrt{3}+3=5+2\sqrt{6}$$
$$(a^2-5)^2=(2\sqrt{6})^2$$
$$a^4-10a^2+25=24$$
$$a^4-10a^2+1=0$$