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

[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?

 

[Mark44]

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]

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

 

[Chaos2009]

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]

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.

 

[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?

 

[nicksauce]

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

 

[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?

 

[Mark44]

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

 

[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.

 

[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 don't have a rational roots it doesn't mean that
"h" is always irrational,"h" could be a complex number.

??

 

[Mark44]

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.

 

[transgalactic]

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

 

[Mark44]

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

 

[jay17]

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]

Well since we've already summoned this thread from the grave...

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

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

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

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

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

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