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#1 2007-10-15 01:49:11

xXxEmZyxXx
Member
Registered: 2007-10-15
Posts: 12

Proofs!!

Help me please dizzy

1) Prove that there do not exist natural numbers a, b with (a/b)^2=5

2) where does the proof go wrong if you try to use the same arguement to prove the (incorrect) statement that there do not exist natural numbers a, b with (a/b)^2=4?

Any help much appreciated!!!

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#2 2007-10-15 03:26:29

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: Proofs!!

Have you seen the proof for the square root of 2 is irrational?  It's pretty much just follow the same steps with a few twists.

However, an alternate method is to consider the prime factorization of each side.  One side will have an odd number of 5's, one side will have an even number of them.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2007-10-18 07:56:53

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: Proofs!!

When s is a positive integer and cannot be expressed as a square of an integer, e.g. 2, 3, 5, 6, 7, 8, 10...,

no integer a and b can satisfy:
(a/b)[sup]2[/sup]=s

Proof:
if so,
a[sup]2[/sup]=sb[sup]2[/sup]
So a[sup]2[/sup]|s

Because s cannot be expressed as a square of a integer, such as s=1=1², s=4=2², s=9=3²..., the only way to guarantee a²|s is a|s, then
a²|s²

a²=sb²

so b²|s
so b|s and b²|s²

To Ricky's arguement.


X'(y-Xβ)=0

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