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Prove that of sum square root of 2 and square root of 3 is not rational 
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#1
Nov2208, 12:27 AM

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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? 


#2
Nov2208, 12:38 AM

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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. 


#3
Nov2208, 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 


#4
Nov2208, 12:48 AM

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



#5
Nov2208, 12:50 AM

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#6
Nov2208, 01:31 AM

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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? 


#7
Nov2208, 01:55 AM

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



#8
Nov2208, 02:11 AM

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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
Nov2208, 05:19 PM

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



#10
Nov2208, 07:34 PM

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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. 


#11
Nov2308, 12:52 AM

P: 1,397

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
Nov2308, 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.



#13
Nov2308, 01:07 AM

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



#14
Nov2308, 01:16 AM

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



#15
Jun2209, 02:12 AM

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need help 


#16
Jun2209, 05:43 AM

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Well since we've already summoned this thread from the grave...
[tex]a^2=(\sqrt{2}+\sqrt{3})^2=2+2\sqrt{2}\sqrt{3}+3=5+2\sqrt{6}[/tex] [tex](a^25)^2=(2\sqrt{6})^2[/tex] [tex]a^410a^2+25=24[/tex] [tex]a^410a^2+1=0[/tex] 


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