-2

What is the negation of this mathematic expression: $(x + y) > z$.

I want to apply negation to the whole statement i.e. $\neg \big((x + y) > z \big)$

What will be the answer?

Jaideep Khare
  • 19,293
Namra Saheba
  • 13
  • 1
    Depends. Are $x,y,z$ numbers? – Thomas Andrews Apr 12 '17 at 18:41
  • 1
    The negation of this expression is $(x+y) \leq z$. – Reb Apr 12 '17 at 18:41
  • "Negation" and "negative" aren't the same thing. You probably mean the first. Please edit the title of your question accordingly. There's now a correct answer, too. for you to accept. – Ethan Bolker Apr 12 '17 at 18:43
  • If it is not true (a "negation"; not a negative btw) that x + y > z then $x+y \le z$. That's all. – fleablood Apr 12 '17 at 18:44
  • But won't the negation apply to (x + y) ? And yes they are all POSITIVE numbers. – Namra Saheba Apr 12 '17 at 18:45
  • Why is everybody down-voting the question? At least give reasons when you down-vote. I know the question might be silly to some but I have every right to ask questions until I get their answers. – Namra Saheba Apr 12 '17 at 18:46
  • The negation applies just to the logic; it has nothing to do with arithmetic. The negation of "greater than" is "not greater than". For numbers, "not greater than" is exactly the same as "less than or equal to". The fact that there's arithmetic on the left hand side is irrelevant. This is just what @BenjaminLindqvist 's correct answer says - which I see you've now accepted. – Ethan Bolker Apr 12 '17 at 19:02
  • People are downvoting your question because you have put no effort in it.(I am not downvoter) You must add something from your side. If you want us to take time and solve the question for you, you must also do the same. – Jaideep Khare Apr 12 '17 at 19:03
  • @EthanBolker Thanks a lot for clarifying! This is what I was confused about. Wish I could accept this comment as precisely my answer! – Namra Saheba Apr 12 '17 at 19:07
  • @JaideepKhare I am not much used to how people ask questions on this platform and unaware of how they judge your questions. Will be careful next time. – Namra Saheba Apr 12 '17 at 19:09
  • You're welcome. The accepted answer is just fine.Maybe I'll edit it to include my comment, since it helped you. – Ethan Bolker Apr 12 '17 at 19:11

2 Answers2

1

If $x+y$ is not greater than $z$ it must be either less than $z$, or equal to $z$, so $x+y \leq z$

Edited (in response to OP's comment) to clarify for future readers:

The negation applies just to the logic; it has nothing to do with arithmetic. The negation of "greater than" is "not greater than". For numbers, "not greater than" is exactly the same as "less than or equal to". The fact that there's arithmetic on the left hand side is irrelevant.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
Benjamin Lindqvist
  • 3,112
0

In this situation, I sometimes start to see expressions like:

$\neg (x < y) = \neg x \ge \neg y$

... as if this is some kind of DeMorgan operation. But clearly that is a mistake: $\neg x$ has no meaning when $x$ is a number. The negation of $x < y$ is simply $x \ge y$

Bram28
  • 100,612
  • 6
  • 70
  • 118