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If a and b are integers and a*b=2, is a>b? [#permalink]

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14 Aug 2017, 03:52

chetan2u wrote:

If a and b are integers and a*b=2, is a>b?

(1) \(\frac{a}{b}>1\) (2) \(a^2>b^2\)

self made

By using numbers we can arrive at E but I have a doubt in using algebra.

Since a*b=2, both a and b has to be same sign.

St.1: \(\frac{a}{b}>1\)

If a and b are positive, we can conclude a>b. If a and b are negative, \(\frac{-a}{-b}\), will those signs not cancel out and become just a/b ? Should we invert > sign even when we cancel sign in numerator and denominator?

By using numbers we can arrive at E but I have a doubt in using algebra.

Since a*b=2, both a and b has to be same sign.

St.1: \(\frac{a}{b}>1\)

If a and b are positive, we can conclude a>b. If a and b are negative, \(\frac{-a}{-b}\), will those signs not cancel out and become just a/b ? Should we invert > sign even when we cancel sign in numerator and denominator?

Please advise.

Hi.. When a and b are NEGATIVE, it does not become -a/-b, it remains a/b because NEGATIVE sign is within a and b..

So a =-x, b =-y... Now -x/-y>1....X>y.... Substitute in terms of a and B.. -a>-b.....Multiply by -1 and invert sign.. a<b..

Or a/b>1... \(1-\frac{a}{b}<0......\frac{b-a}{b}<0.....\) So b-a and b will be of different sign.. b is NEGATIVE, so b-a>0...b>a