1. ## Integers (587) THANKS CAPTAINBLACK!

For all numbers x and y, let the operation @ be defined by x @ y=xy-y. If a and b are positive integers, which of the following can be equal to zero?

I. a @ b
II. (a+b)@b
III. a@(a+b)

a) I only
b) II only
c) III only
d) I and II
e) I and III

Can you please explain why? Thanks alot!!

2. Originally Posted by fabxx
For all numbers x and y, let the operation @ be defined by x @ y=xy-y. If a and b are positive integers, which of the following can be equal to zero?

I. a @ b
II. (a+b)@b
III. a@(a+b)

a) I only
b) II only
c) III only
d) I and II
e) I and III

Can you please explain why? Thanks alot!!
If a and b are any real numbers we have:

I. If a@b=0 then ab-b=0, which is the case when b=0 or a=1

II. If (a+b)@b=0 then (a+b)b-b=0, which is the case when b=0 or (a+b)=1

III. If a@(a+b)=0 then a(a+b)-(a+b)=0 which is the case when (a+b)=0 or a=1.

Now if instead a and b are positive integers I. reduces to a=1, II. has no possible solutions in positive integers (explain why), while III. again permits a=1.

RonL

3. Originally Posted by CaptainBlack
Now if instead a and b are positive integers I. reduces to a=1, II. has no possible solutions in positive integers (explain why), while III. again permits a=1.

RonL
Is it because for II. if (a+b)=1 then it either means that a=1-b or b=1-a. Then one of them (a or b) can't be positive because 1-a has to be negative.
Then you might say that let a=1. If a=1 then b has to equal 0 in order to satisfy (a+b)=1. But 0 is not a positive nor a negative. So (a+b)=1 has no possible solutions in positive integers. (: