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Thank you for reading it. I know I made a lot of mistakes. This is my first ever proof that I have attempted. Another note is that I only have been studying proofs for about a week. Any advice will be helpful.

prove: $|x+y| ≤ |x| + |y|$
Case 1: ∀ values of x<0 and y<0, the function will decrease: $|x+y| \overset{x<0}= |y\pm x|$
$|x+y| \overset{y<0}= |-y+x)|$
$A=|-x+y|$ –-—-> $∂A/∂X=-1$
$B=|-y+x|$ $∂B/∂Y=-1$

Case 2: In the case of (x,y)>0, the two functions opposite of the inequalities are equal.
{|x+y|⇔ |x|+|y|: x>0 and y>0}
This is a normal property of the absolute value theorem.
Notation: {|x+y|∀ values of x and y = |x|+|y| ∀ for all values of x and y}

Case 3: Case 3 proves that the values of |x|+|y| are unaffected by values less than zero
$|x| = \begin{cases} x,&\text{if }x\ge 0\\ -x,&\text{if }x<0 \end{cases}$

$|y| = \begin{cases} y,&\text{if }y\ge 0\\ -y&\text{if }y<0 \end{cases}$

⇔$|X|+|y|>0$ when $(x,y)≠0$

Note: I don’t know if I properly stated the ∀correctly; however, I meant it as “for all“

Thank you for reading it. I know I made a lot of mistakes. This is my first ever proof that I have attempted. Another note is that I only have been studying proofs for about a week. Any advice will be helpful.

***Edited

EMathS
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2 Answers2

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I think that you are making it much too complicated.

There are only four cases: $x \ge 0$ or $ x < 0$ combined with $y \ge 0$ or $ y < 0$.

Prove the inequality for each of the four cases and you are done.

Concept such as "randomly chosen" values and partial derivatives of the variables are totally extraneous and just get in the way.

As you do more proofs, you will get a feel (from experience) which concepts are relevant and which are not.

marty cohen
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  • Indeed, it's even simpler. You really only need to address two cases: (1) $xy<0$, and (2) $xy\geq0$. – MPW Aug 20 '14 at 05:19
  • Thank you for the comments. Does my method still prove the conjecture if you will? – EMathS Aug 20 '14 at 18:11
  • I don't think that you should use partial derivatives at all. After all, they are a more advanced concept than absolute value. You can also use MPW's suggestion of only two cases: x and y have different signs, and x and y have the same sign. – marty cohen Aug 25 '14 at 04:44
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The first rule is to make sure the sentences are grammatically correct. Your first sentence starts "As values of x and y are randomly assigned negative values independent of the other variable..." implies that there is another variable beyond just x and y. You then used the implies sign, $\implies$, which doesn't fit in the parenthetical statement. The (x, y)<0 doesn't make mathematical sense. Only real numbers use the sign <, and (x, y) is not a real number. Maybe you meant x times y. Perhaps you could re-structure the first sentence and we could continue afterwards. For a first try you are doing fine.

Paul Sundheim
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  • How should I notate "For all values of X and Y less than zero"? – EMathS Aug 20 '14 at 18:12
  • @EMathS If you want to say that $x$ and $y$ are both positive, there are many ways. You could say $x,y>0$ (notice, no brackets), or $(x,y)\in\mathbb{R}_{++}^2$ (if they are positive reals; however, this notation is unnecessary as there is no need to ever group $x$ and $y$ as a pair), or simply $x>0$ and $y>0$ which is perhaps the most clear. Don't worry about being as notationally concise as possible when you are starting out -- put a higher priority on clarity. – Erik M Aug 20 '14 at 18:35
  • Instead of "the function will decrease", which you haven't specified, you could say "$|x+y|$ is decreasing". Then you avoid referring to something that hasn't been defined. Also I don't think, if x < 0, the statement $|x+y|=|x\pm y|$ is true. Try $x = -1$ and $y = -1$ then the LHS gives you 2 while the right hand side produces 0. I'm not sure what you mean there. Maybe we are steering in the wrong direction. If you square each side individually, you can then establish a relationship between the two sides and work backwards. Let me know if you would like to see more details about this. – Paul Sundheim Aug 20 '14 at 21:19
  • I should work on notating certain arguments. Paul Sundheim, how would you have gone about in trying this proof? It is interesting to see different methods. – EMathS Aug 23 '14 at 04:52