## rational numbers

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.
smara1403
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Joined: Sat Sep 21, 2013 5:07 pm
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### rational numbers

Hello and sorry if I'm not posting in the right section, but I have the following problem:
If a and b are rational numbers and sqrt from a+b does Not equal 0, how do I demonstrate that sqrt from a-b does not equal 0?

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
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### Re: rational numbers

smara1403 wrote:Hello and sorry if I'm not posting in the right section, but I have the following problem:
If a and b are rational numbers and sqrt from a+b does Not equal 0, how do I demonstrate that sqrt from a-b does not equal 0?

I'm not familiar with your terminology. Are you perhaps referring to square roots "of" a+b and a-b? If so, is the exercise along the lines of the following?

$\mbox{Let }\,a\, \mbox{ and }\, b\, \mbox{ be rational numbers such that}$

$\sqrt{a\, +\, b}\, \neq\, 0.\, \mbox{ Prove that }\, \sqrt{a\, -\, b}\, \neq\, 0.$

In particular, are you supposed to prove, or maybe DIS-prove with a counter-example (such as when b = a)?

smara1403
Posts: 2
Joined: Sat Sep 21, 2013 5:07 pm
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### Re: rational numbers

no, i'm sorry
i meant a+sqrt(b) !=0 and prove a-sqrt(b)!=0, where sqrt stands for square root and != for does not equal
i'm sorry for my terminology, but i'm not used to the english one

stapel_eliz
Posts: 1687
Joined: Mon Dec 08, 2008 4:22 pm
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smara1403 wrote:i meant a+sqrt(b) !=0 and prove a-sqrt(b)!=0, where sqrt stands for square root and != for does not equal

So is the exercise as follows?

$\mbox{Let }\, a\, \mbox{ and }\, b\, \mbox{ be rational numbers, and assume that }\, a\, +\, \sqrt{b}\, \neq\, 0.$

$\mbox{Prove that }\, a\, -\, \sqrt{b}\, \neq\, 0.$

If so, then what happens when a = 2 and b = 4?