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

by transgalactic
Tags: prove, rational, root, square
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transgalactic
#1
Nov22-08, 12:27 AM
P: 1,398
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
#2
Nov22-08, 12:38 AM
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P: 21,216
Quote Quote by transgalactic View Post
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 [tex]\sqrt{2} + \sqrt{3} = \frac{a}{b}[/tex]
where a and b are integers with no common factors.

For your second question [tex]\sqrt{5} + (-\sqrt{5}) = 0 [/tex]
Those are both irrational numbers, but their sum is rational.
nicksauce
#3
Nov22-08, 12:47 AM
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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
#4
Nov22-08, 12:48 AM
P: 81
Prove that of sum square root of 2 and square root of 3 is not rational

Quote Quote by Mark44 View Post
For your second question [tex]\sqrt{5} + (-\sqrt{5}) = 0 [/tex]
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
#5
Nov22-08, 12:50 AM
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Quote Quote by Chaos2009 View Post
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
#6
Nov22-08, 01:31 AM
P: 1,398
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
#7
Nov22-08, 01:55 AM
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As I suggested earlier, make use of the rational roots theorem.
transgalactic
#8
Nov22-08, 02:11 AM
P: 1,398
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
#9
Nov22-08, 05:19 PM
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What did you set out to do in your post #6?
Quantumpencil
#10
Nov22-08, 07:34 PM
P: 96
You don't need anything but parity to prove this.

Prove



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

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 [tex]\sqrt{6}[/tex] is irrational. So just show that if it equals p/q, p and q must both be even.
transgalactic
#11
Nov23-08, 12:52 AM
P: 1,398
Quote Quote by transgalactic View Post
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
#12
Nov23-08, 12:59 AM
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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
#13
Nov23-08, 01:07 AM
P: 1,398
you meant "irrational" (in the end) right?
Mark44
#14
Nov23-08, 01:16 AM
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I edited my reply. The last part should have said "there is no solution that is rational."
jay17
#15
Jun22-09, 02:12 AM
P: 1
Quote Quote by nicksauce View Post
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
#16
Jun22-09, 05:43 AM
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Well since we've already summoned this thread from the grave...
Quote Quote by jay17 View Post
how did you arrive at a^4 - 10a^2 + 1 = 0?
need help
[tex]a=\sqrt{2}+\sqrt{3}[/tex]

[tex]a^2=(\sqrt{2}+\sqrt{3})^2=2+2\sqrt{2}\sqrt{3}+3=5+2\sqrt{6}[/tex]

[tex](a^2-5)^2=(2\sqrt{6})^2[/tex]

[tex]a^4-10a^2+25=24[/tex]

[tex]a^4-10a^2+1=0[/tex]


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