# 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?

whats the formal equation proof?

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Mark44
Mentor
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.

nicksauce
Homework Helper
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

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.

nicksauce
Homework Helper
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.

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?

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

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?

Mark44
Mentor
What did you set out to do in your post #6?

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.

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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.

??

Mark44
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.

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

Mark44
Mentor
I edited my reply. The last part should have said "there is no solution that is rational."

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

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

$$a=\sqrt{2}+\sqrt{3}$$

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

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

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

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