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My reasoning would be that since $x$ is negative, $x$ can be rewritten as $-1 \cdot y$ where $y$ is equal to $|x|$. Therefore we can write $a - x$ as $a - (-1) \cdot y = a + y$, which is not equal to $a + x$.

I'm asking because it seemed very natural to me to just turn the sign around, because we do that in school where for example $2 - (-4)$ is equal to $2 + 4$, but $4$ isn't equal to $-4$.

And if what I said is true, does it mean that I have to take this exact step while writing proofs as well, or is there a more elegant way to do it?

user729424
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iamtrying
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  • In $a+x$, $x$ carries the sign aswell. So in $2+4$, $x=4$. In $a-x$, the sames goes on, so in $2-(-4)$, $x=-4$. – Anthony Apr 25 '20 at 18:55
  • Usually, there's no reason to replace $x$ by anything. It's some number; might be positive of negative. Please use MathJax to format math expressions. – saulspatz Apr 25 '20 at 18:57
  • @Anthony, as far as I understand your comment you are saying that it is in fact wrong to simplify a - x to a +x. But how can someone write it in a proof, where you'd need to write a - x with a "+" in it? – iamtrying Apr 25 '20 at 19:04
  • @saulspatz in the example I've given I defined it to be smaller than zero. How can it be positive? And forthermore, how would you rewrite a - x if you'd need a "+" in it? – iamtrying Apr 25 '20 at 19:06
  • I was talking about the general situation. – saulspatz Apr 25 '20 at 19:07
  • Oh okay. And what would you do if you need to write a - x where x is negative as a + something? – iamtrying Apr 25 '20 at 19:08
  • You could do what you suggested in your question. What you wrote is correct. You could also write $+(-x)$. I'd like to see a specific case where you gain something by doing either of these. – saulspatz Apr 25 '20 at 19:10
  • @saulspatz for example if you wanted to show that $a - x$ is bigger than $a$, where $a$ is positive and $x$ is negative, you'd need to write that $a - x$ is equal to adding a positive value to $a$, and I was wondering how to do that more efficiently than in the description of my post. – iamtrying Apr 25 '20 at 19:46
  • $a-x>a$ because $x<0$. No more need be said. – saulspatz Apr 25 '20 at 19:55
  • @saulspatz okay, I guess everything's clear then. Thank you so much for your patience to explain me such fundamentals! – iamtrying Apr 25 '20 at 20:03
  • My pleasure. It's great to be able to interact with some else, locked down by myself as I am. – saulspatz Apr 25 '20 at 20:05

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