Prf. :
Let x and y be real numbers. We can consider the following cases:
x and y are of the same
If x and y are both positive, then the statement is automatically true, similar to the case when both x and y are negative
one of x and y is negative
WLOG we let y = -n:
x-n <= x+n. If x-n is negative then it will be less than x+n, showing it to be true. If it were positive, then we would have to be greater than x+n, which tells us that x is greater than 0. Thus completing the proof.
For all real numbers x and y, |x+y| is less than or equal to |x|+|y|.
Looks as though the right idea is here, just some clarifications needed:
ReplyDelete(1) 'Parity' should be 'sign' in the first case considered
(2) The phrase "similar to when replacing x to -x and y to -y" is confusing. I think it is meant to cover the case when x and y are both negative, but could use some clarification.
(3) The second case considers the possibility that \(|x-n|\) is negative, but this cannot happen. Should this be \( x-n \) is negative?
Thanks for the post!