Thank you.
I actually proved it, but wasn't sure if it was correct, but i did the 4th case which is x>0 y<0 , I wasn't supposed to do that? I mean technically it's the same case with the second condition. By the way why do they always write greater or equal to zero? They always write less, and greater or equal. And Why shouldn't I think of zero as a different case?
Yes I am aware of that, it's just that I have exams coming up and there are a lot of questions, some of them I proved but wasn't sure if the proof is correct so I ask for the proof anyway for verification, and the ones that I couldn't, well I have to spend more time on them but now I have exams, I don't have enough time to think for solution for every question that I have... I am making effort, but it's not enough and it will change, but till then I am asking for your patience. Thank you.
The definition of |x| is "if $\displaystyle x\ge 0$ then |x|= x, otherwise x< 0." For the first problem, sgn(x) is defined as "if x> 0 sgn(x)= 1, if x< 0, sgn(x)= -1, sgn(0)= 0" so, yes, for that problem, you should treat x= 0 or y= 0 as separate cases.
I'm glad to hear that you have been making an effort- when you post questions, show what efforts you have made so we will know what kinds of hints will help you do the problems yourself.Yes I am aware of that, it's just that I have exams coming up and there are a lot of questions, some of them I proved but wasn't sure if the proof is correct so I ask for the proof anyway for verification, and the ones that I couldn't, well I have to spend more time on them but now I have exams, I don't have enough time to think for solution for every question that I have... I am making effort, but it's not enough and it will change, but till then I am asking for your patience. Thank you.
So I will have more cases in sign than the absolute value proofs, is that correct?
That is why I wasn't sure of my proof with sign because you can get 14 different cases (when they are both positive,negative,positive negative,zero, zero positive,zero negative)
And was my 4th case redundant in absolute value proof?
I am trying to make the proofs as simple as possible, because for every redundant thing they will take off points I guess in the exam, although I am sure they are not going to ask this.
I don't think that either of us can answer those questions.
It is purely a matter of style and the grader's preference. By this point you should know the likes and dislikes of you grader.
In effect the cases are when both are non-negative, both are negative, and one negative and one is non-negative.
Now as I implied, there are slight differences in the two expressions, |x| & sgn(x), (in the second x=0 is handled differently).
But none-the-less the three basic cases are the same.
Assume it’s not true and x=1 and y=-1, Then:
sgn(xy)=-1
sgn(x)sgn(y)=-1≠-1
EDIT: I think davidciprut asks great (educational) questions. I personally don't care how hard he tries to answer them, as long as we learn something along the way.
Assume what was not true? That sgn(xy)= sgn(x)sgn(y) is not true? All you have shown is that sng(xy)= sgn(x)sng(y) in the special case that x= 1 and y= -1. Giving an [b]example[b] is not proving.
Davidciprut, I would start "if either x or y is 0 then both sides of sgn(xy)= sgn(x)sgn(y) are 0 so it is true." Then go ahead and do the other cases.