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

Question

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

susheelh · Answer

IMO the answer is E. Stem : a and b and integers and their product is 2. Possible values of a,b = (2,1) or (1,2) or (-2,-1) or (-1, -2) Needed: Is a > b? I found it easier to keep track of the possibilities by using a table and Working through S1, S2 and S1+S2. Table is attached. From the Table we can see S1 : a and b could be (2,1) or (-2,-1). So a > b or a < b. Insuff S2: a^2 > b^2 => a^2-b^2 > 0 . a and b could be (2,1) or (-2,-1). So a > b or a < b. Insuff S1+S2: a and b could still be (2,1) or (-2,-1). So a > b or a < b. Insuff Even with both statements its not suff. I hope I am correct

Bang2919 · Answer

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.

chetan2u · Answer

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.. a1... 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 Follow any of above..

stonecold · Answer

Great question.
Here is what I did on this one -->

Since a and b are integers => a*b=2 => The possible cases are ==>
1,2
2,1
-1,-2
-2,-1

Lets see the 2 statements =>
a/b>1
Hence the two possible pairs are => 
(2,1)=> YES a>b is true
(-2,-1)=> NO a>b is false.

Hence not sufficient.

Statement 2 => 
a^2>b^2

Again => the two possible pairs are => 
(2,1)=> YES a>b is true
(-2,-1)=> NO a>b is false.

Not sufficient.

Combining the two statements => Two possible pairs are => 
(2,1)=> YES a>b is true
(-2,-1)=> NO a>b is false.

Hence not sufficient.

Hence E.

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

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