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Old April 25th, 2009, 01:57 PM
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Default [SOLVED] Proof - Can somebody please check my work

Hi,

I am to prove the following:

If a, b, c are real numbers s.t. a & c are NOT 0, then there is a unique number x s.t. x/a + b/c = 1.

Ok, this is what I have:

Let a, b, c be reals s.t. x/a + b/c = 1 and a & c are NOT 0.

(x * 1/a) + (b * 1/c) = 1

a[ (x*1/a) = 1 - (b * 1/c) ]

x = a [ 1 - (b *1/c)]

Since b*1/c is an element of R & 1-b/c is also and element of R,

then a(1-b/c) is also an element of R.

So x is an element of R.


This seems pretty straight forward, but I just want to make sure that I am not approaching this wrong or am missing something in my proof. Thanks.
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Old April 25th, 2009, 02:10 PM
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That works as far as it goes. You have shown existence.
Now show uniqueness: \frac{r}{a} + \frac{b}{c} = 1\,\& \,\frac{s}{a} + \frac{b}{c} = 1 \Rightarrow \quad r = s.
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Old April 25th, 2009, 02:29 PM
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will do! Thanks for help.
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