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Can someone explain to me why subtraction and division is both commutative? The reason I believe that they are commutative are as follows: $$ 3-2 = 3+(-2) $$ $$ 3+(-2) = (-2)+(3) $$ I've read a website that says this doesn't prove the above because I'm still "commutating the addition". But why is this a problem? What exactly differentiates this from commutating with a negative sign? Can you both explain to me intuitively why it makes a difference, and prove why this is true for all numbers?

The same with division: $$ 3*(1/6)=3/6 $$ $$ 3*(1/6) = (1/6)*3 $$

Can you explain to me intuitively why it makes a difference, and prove why this is true for all numbers?

Oh, and one final thing. I'm still a beginner. Can you explain it to me as simply as possible, using as few advanced math terms as possible?

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  • $\begingroup$ What you have done is replace subtraction by the addition of a negative number. That's in fact how subtraction is best defined - after you learn about negative. That addition is commutative. $\endgroup$ – Ethan Bolker Jun 22 '18 at 9:56
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A binary operation (like addition or subtraction, etc.) on a set $S$ can be thought of as a function from $S \times S$ to $S$. For example, subtraction on $\mathbb{R}$ is a function which takes each pair of real numbers $(x,y)$ to the real number $x - y$.

A binary operation $f$ is commutative if $f(x,y) = f(y,x)$ for all $x$ and $y$ in the set. Subtraction is not commutative because, taking your example, $f(3,2) = 3-2 =1$ and $f(2,3) = 2-3 = -1$.

When you write $3-2 = -2 + 3$, you are actually verifying the commutativity of addition, because what you’re actually showing is that $3+(-2) = (-2) + 3$, or in other words, $g(3,-2) = g(-2,3)$ where $g$ is the function that represents the binary operation that is addition.

The same goes for multiplication. If $h$ is the function representing division, then $h(x,y) = x/y$. And, taking your example, $h(3,6) = 3/6$ whereas $h(6,3) = 6/3$, which are not equal, so division is not commutative. What you are verifying is that multiplication is commutative, just as you did for subtraction.

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  • $\begingroup$ My question is though, then how do you explain the fact that 3-2 = 3 + (-2)? What exactly differentiates 3-2 from 3+(-2), and why exactly would this difference cause 3+(-2) to be commutative, but 3-2 to be not commutative? $\endgroup$ – Ethan Chan Jun 23 '18 at 3:09
  • $\begingroup$ @EthanChan $3-2 = 3 + (-2)$ by the definition of subtraction, actually. Subtracting two from three means adding three and the “additive inverse” of $2$. (The additive inverse of a number is that number which you have to add to it to get zero.) $\endgroup$ – Brahadeesh Jun 23 '18 at 3:15
  • $\begingroup$ Does this also help you see why adding 3 to the additive inverse of 2 and adding 2 to the additive inverse of 3 are two different things? $\endgroup$ – Brahadeesh Jun 23 '18 at 3:18
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You are right when you write that $3-2=3+(-2)$. But this doesn't turn subtraction into a commutative operation, since $2-3=2+(-3)\neq3+(-2)=3-2$.

The same thing applies to division.

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  • $\begingroup$ Is it possible, though, to prove that, for any few numbers, that addition is commutative, but subtraction is not, without making the assumption that subtraction is commutative? Thanks! This will help my understanding. $\endgroup$ – Ethan Chan Jun 23 '18 at 3:12
  • $\begingroup$ I don't understand your question. Why would anyone assume that subtraction is commutative? In which way would that help to prove that addition is commutative? $\endgroup$ – José Carlos Santos Jun 23 '18 at 6:57

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