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Jan 31st, 2007, 11:39 AM
#1
Thread Starter
Addicted Member
Proofs
Hi
Can anyone help me prove if a and b are rational numbers, then a(b^2) - 5 is a rational number.
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Jan 31st, 2007, 12:21 PM
#2
Re: Proofs
If a and b are rational, then they can be written as the quotient of 2 integer numbers, for instance:
a = p/q
b = v/w
where p,q,v,w belong to Z (the set of integer numbers)
So, then
ab2 - 5 = (p/q)(v/w)2 - 5 = (pv2 - 5qw2) / (qw2)
which in its turn is the quotient of integer numbers since only sums, substractions and products are involved in both the numerator and the denominator.
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
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Feb 3rd, 2007, 11:22 PM
#3
Hyperactive Member
Re: Proofs
Not to be picky would it still be rational if q=w=0? or q=0 or w=0?
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Feb 4th, 2007, 12:27 AM
#4
Re: Proofs
Neither q nor w can be zero, because a and b are given to be rational.
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Feb 4th, 2007, 03:56 PM
#5
Re: Proofs
If a=0 and b=0 then
ab2 - 5 = - 5
which is rational.
And 0 itself is rational. It can be expressed as a quotient:
0 = 0 / A
where A is any integer number different from 0.
Lottery is a tax on people who are bad at maths
If only mosquitoes sucked fat instead of blood...
To do is to be (Descartes). To be is to do (Sartre). To be do be do (Sinatra)
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