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New to math. I'm looking for an explanation (proof, rule, relationship or property) that explains that the absolute value of the difference between two integers $x$ and $y$ are equal regardless of:

  1. the sign of $x$ or $y$
  2. the order of subtraction. meaning regardless whether $x$ is the minuend or subtrahend

$$x - y = |d| = y - x$$

Behrouz Maleki
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LogBar
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  • Something is not right here: $x-y\neq y-x$, but $x-y=|d|$ and $y-x=|d|$ is a correct way of looking at it... but I'm assuming that $x \neq y \neq 0$. –  Jul 14 '16 at 19:25
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    @KingDuken something is still not right. 2-1=1, so |d| must be 1. But 1-2$\neq$1. What is correct is $|x-y|=d=|y-x|$. – Polygon Jul 14 '16 at 19:29
  • Oops, yes. You're correct @Polygon –  Jul 14 '16 at 19:30
  • Some old school texts may use $x\sim y$ to represent the difference. – Ng Chung Tak Jul 15 '16 at 10:11

4 Answers4

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What you wrote, $x-y=|d|=y-x$ is incorrect. Take x=1 an y=2 for example. $1-2\neq2-1$

What you probably meant to say is this. ($d$ is positive)

$$|x-y|=d=|y-x|$$

Here is my explanation:

$$x-y=-(-x)+(-y)$$ $$x-y=-((-x)+y)$$ $$x-y=-(y-x)$$

Now, in simple terms, the absolute value of a a number just makes it posotive. Because of that, for any variable $n, |-n|=|n|$. Therefore:

$$|-(y-x)|=|y-x|$$ so $$|x-y|=|y-x|$$

Polygon
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  • If n = -3 then |-n| = -n, so I have to balk at your comment that absolute value removes negative signs. However for any x, |x|=|-x| so we can say "negative signs in abs value can be ignored". Thus |x -y| = |-(x-y)|=|y-x| which is about as short a proof as you can get. – fleablood Jul 14 '16 at 19:51
  • @fleablood wow. I knew my description was bad, but I didn't realize it was that bad. I'll change it ASAP. – Polygon Jul 14 '16 at 19:55
  • I didn't mean to imply it was bad. And it is accurate, though informal, to say abs value strips parity and renders everything positive. It's just that rendering positive doesn't mean stripping negative signs. If n is neg it means adding a negative sign (as the neg of a neg is pos-- whereas a neg by itself is simply neg) – fleablood Jul 14 '16 at 20:00
  • @Polygon: really appreciate how you spend the time show how you arrived there. – LogBar Sep 18 '16 at 18:04
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If $A \ge 0$ then $|A| = A $.

If $A < 0$ then $A = -|A|$ and $|A| =-A $.

So if $y > x $ then $y - x > 0$ so $|y-x| = y -x $, while $x - y < 0$ so $|x-y| = -(x - y) = -x + y = y - x = |y-x|.$

If $y < x $ then $y - x < 0$ and $|y-x|=-(y-x)=x - y $ while $x - y > 0$ so $|x-y| = x-y = |y-x|$.

And if $y=x $ then $y-x=x-y=0$ and $|y-x|=|x-y|=0$.

fleablood
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More colloquially:

An absolute value renders everything positive (or zero). If the number is positive to begin with then the absolute value is the original number. If the number is negative, then the absolute value is "the same size but it's positive opposite".

If you subtract a smaller number from a bigger number you get a positive number. If you subtract a bigger number from a smaller number you get a negative number. X - Y and Y - X will both be "the same size" but one will be positive and the other negative (assume X and Y are different numbers). So their "absolute size regardless of sign" will both be the same.

fleablood
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As I understand, you are asking why the absolute difference of two numbers is not dependent on the order in which those numbers are subtracted. In other words, you are asking why the following equation is true. $$|x-y|=|y-x|$$ The absolute value of a number, $a$, ignores the sign of that number, thus $$|a|=|-a|$$ In the above equation, if $$a=x-y$$ then we see that $$|x-y|=|-(x-y)|$$ which simplifies to $$|x-y|=|y-x|$$

pseudoeuclidean
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