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

by transgalactic
Tags: prove, rational, root, square
 P: 1,399 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?
Mentor
P: 19,715
 Quote by 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?
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.
 HW Helper Sci Advisor P: 1,276 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
P: 81

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

 Quote by Mark44 For your second question $$\sqrt{5} + (-\sqrt{5}) = 0$$ Those are both irrational numbers, but their sum is rational.
Somehow, that just seems like cheating, but it does fit the problem we were given.
HW Helper
P: 1,276
 Quote by Chaos2009 Somehow, that just seems like cheating, but it does fit the problem we were given.
Any mathematical statement can be disproved by a counterexample. The simpler the counterexample, the better.
 P: 1,399 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?
 HW Helper Sci Advisor P: 1,276 As I suggested earlier, make use of the rational roots theorem.
 P: 1,399 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?
 Mentor P: 19,715 What did you set out to do in your post #6?
 P: 96 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.
P: 1,399
 Quote by 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?
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.

??
 Mentor P: 19,715 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.
 P: 1,399 you meant "irrational" (in the end) right?
 Mentor P: 19,715 I edited my reply. The last part should have said "there is no solution that is rational."
P: 1
 Quote by 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
how did you arrive at a^4 - 10a^2 + 1 = 0?
need help
HW Helper
P: 3,323
 Quote by jay17 how did you arrive at a^4 - 10a^2 + 1 = 0? need help
$$a=\sqrt{2}+\sqrt{3}$$

$$a^2=(\sqrt{2}+\sqrt{3})^2=2+2\sqrt{2}\sqrt{3}+3=5+2\sqrt{6}$$

$$(a^2-5)^2=(2\sqrt{6})^2$$

$$a^4-10a^2+25=24$$

$$a^4-10a^2+1=0$$

 Related Discussions Calculus & Beyond Homework 4 General Math 14 Calculus & Beyond Homework 2 Introductory Physics Homework 4 Brain Teasers 8