For all integers x, the function f

Cydney Seigerman, Magoosh Tutor

Great question! :D
Yes, you're completely right that if a = b and that number were odd, then f(a) + f(b) would be even. However, since f(x) = x+1 for odd numbers, then for two odd numbers, f(a) + f(b) = a + 1 + b +1 = a + b + 2. In the case that a = b, we could replace b with a to come up with the final expression:
f(a) + f(b) = 2a + 2.
This does not equal a + b, which, if a = b, would be 2a.
And 2a is not equal to 2a + 2.
That shows us that both a and b cannot be odd.
I hope this helps! :)

Nov 3, 2015 • Reply

Francisco Olivas

Wonderful, now I get it! :-)

Oct 11, 2017 • Reply

Ramlah Merchant

Would this method be correct- I simply plugged in a+1+b-1 = a+ b to test out an even and an odd number. since the 1s would cancel out it became very clear that one number would have to be odd and the other even for the left hand side to emerge as a+b.

Nov 6, 2017 • Reply

Jonathan , Magoosh Tutor

Hi Ramiah,
The question asks with MUST be true, so if you plug in, you should be able to justify why the correct answer is correct for more than just the particular values you chose. If you thought "the only way to get a + b is if we have a + 1 + b -1 OR a - 1 + b + 1 so therefore a is odd and b is even or vice -versa" then you were correct :)

Nov 13, 2017 • Reply