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Addition and multiplication are commutative. Denoting $\circ$ as either such operation, we have $$x \circ y = z \Leftrightarrow y \circ x = z.$$ Subtraction and division have a similar property, where $$x \circ y = z \Leftrightarrow x \circ z = y.$$ Is there a name for this latter property?

I would like to teach my child, who has learned about commutativity and who has observed the second property casually. Giving the property a name would make it more explicit.

J W
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Reinstate Monica
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    When I was a child, one of the few things I disliked about math was that some school books would introduce useless, fancy terms like "subtrahend" and "diminuend." It might be nice to ask your kid to make up a name. –  Sep 15 '20 at 14:32
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    Kudos to your child for noticing this. I have not heard of a name for this property, but I would encourage your child to continue to notice properties and patterns in numbers. – Amy B Sep 15 '20 at 19:12
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    Is the use of $z$ important here? The fact that $x+y = y+x$ is called commutativity, and I think the fact that $x - y = -(y - x)$ is called anticommutativity. – Nick C Sep 15 '20 at 19:18
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    @NickC $z$ is important pedagogically because it creates similarity between the two equations. I could have stated the first as $x \circ y = y \circ x$ and the second as $x \circ (x \circ y) = y$ but then the similarity is not as apparent. – Reinstate Monica Sep 15 '20 at 19:33
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    If I were to make a name (as suggested here from individual to individual), I would've said $y$ and $z$ are complementary (parts) (w.r.t. $x$), so the sizes of what are traditionally called complementary angles would be "complementary with respect to 90 degrees". Another way to express this is that $z \mapsto x-z$ is an involution or of composition order 2 (of course, I didn't have this term when younger) – Vandermonde Sep 16 '20 at 02:37
  • Beware of the case when the operator is division and $x=z=0, y\neq0$. – Matt Sep 18 '20 at 15:23

4 Answers4

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This is "left involution". ("left" because it doesn't work when you try it on the right.) \begin{align*} x \circ y &= z & \\ x \circ (x\circ y) &= x \circ z & [\text{apply $x \circ -$}] \\ y &= x \circ z & [\text{simplify the involution}] \text{.} \end{align*}

I would be shocked to see anyone use that term outside of some very narrow niches. In ring theory, an involution is an antiisomorphism, which brings us around to the anticommutativity mentioned in comments and other answers.

Eric Towers
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I have never seen a name for this property specifically. When I was in grade school, I recall learning about Fact Families, which are generated by this property. The idea is that a fact family is all of the arithmetic equations generated by the same numbers.

This property in particular is really just a consequence that subtraction is the inverse of addition which is commutative. You can see this (using algebra, so probably not appropriate for a young child) by adding y to both sides. You then have your commutativity example above. This is not anti-commutativity as one comment said, although subtraction is anti-commutative.

As there is not a common name for this property (other than a reference to fact families), I agree with the suggestion @Ben Crowell to let your child give it a name.

Andrew Sansom
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    Why do you say it's not anti-commutativity? It seems to fit the wikipedia definition given. – Sue VanHattum Sep 15 '20 at 19:39
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    @SueVanHattum The question seems to be focused on the symmetry of the equations, not on the the fact that commuting the terms of subtraction gives the negative. This is clear since the x in the OP doesn't move. I included an edit to emphasize that subtraction is anti-commutative, even if that's not the original question. – Andrew Sansom Sep 15 '20 at 19:45
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    It does seem that the two concepts are the same though, which I didn't realize at the time – Reinstate Monica Sep 15 '20 at 19:46
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    @ReinstateMonica They both logically follow from the other, so they're equivalent in that sense, I suppose. But I'd hesitate to try and explain them as the same thing to a young child. They look very different, and without Algebra, that connection would be hard to make. – Andrew Sansom Sep 15 '20 at 19:50
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    If you're satisfied with things that are logically equivalent, then this is just logically equivalent to commutativity of addition, and you could stop there. –  Sep 16 '20 at 01:54
  • @SueVanHattum Because division is not anti-commutative. – Razvan Socol Sep 16 '20 at 03:24
  • @RazvanSocol As long as we're abstracting different operations with the same symbol in the question ($\circ$ for either addition or multiplication), then we could as easily do the same with subtract and divide, and allow the dash to alternately mean "negate" or "reciprocate" in $x\circ y = -(y \circ x)$. Then they both have the same appearance, which is what I think this problem is focusing on. – Nick C Sep 16 '20 at 03:47
  • @NickC: see https://en.wikipedia.org/wiki/Talk%3ADivision_(mathematics)#Division_is_anticommutative – Razvan Socol Sep 16 '20 at 08:18
  • @RazvanSocol Thank you for that link. – Nick C Sep 16 '20 at 12:24
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    Sounds like it's time for you and your son to write a research paper documenting this hole in the mathematical literature and give it a name! ;) – nick012000 Sep 16 '20 at 13:40
  • @RazvanSocol, you linked to wikipedia saying that division is anti-commutative, so I'm guessing that you either have a typo above, or changed your mind? – Sue VanHattum Sep 17 '20 at 00:26
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    @SueVanHattum I have pointed to a discussion page in which someone proposes changing the article to say that division is anti-commutative, but this is rejected by the other editors, with good arguments. – Razvan Socol Sep 25 '20 at 16:03
  • I see that people disagree. Thanks for the reference. – Sue VanHattum Sep 26 '20 at 21:19
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A helpful way to rewrite that statement would be (assuming subtraction for simplicity):

$x - y - z ⇔ x - z - y$

We are observing how swapping y and z does not change the value of the expression. While it may initially look like there is a useful property behind this, the example is showing an easy case of what you are allowed to swap. Here is a visual representation of what swaps are allowed.

Expression trees

These are expression trees. The filled in circles can be swapped with other circles that are filled in with the same color. The first tree shows the original case of $x - y - z$ and shows how $y$ and $z$ can be swapped. The next tree represents $(a - b) - (c - d)$. The third tree is another full binary tree, but with twice as many variables and represents $((a - b) - (c - d)) - ((e - f) - (g - h))$. You can normalize the first tree into the second by replacing $z$ in the original expression with $z - 0$.

If you just look at the first tree it may seem like you should be able to swap children on the right side, but once you start looking at the other trees, you should notice that the pattern is no longer quite as simple. One way to describe the pattern would be that a node's right child is able to swap with that node's other child's right child. This may be more restrictive than you initially expected when comparing this property to commutativity.

I do not know a name for this property, but it is not as useful as commutativity due to the number of restrictions on being able to swap nodes. If you run into this you may want to try and find a different approach that can use commutativity such as converting subtraction into addition of negative number, or by trying to make as many nodes of the tree swapable by for example not allowing parenthesis to be used.

Anonymous
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I don't know if this word is used specifically to describe this phenomenon, but the term "complement" is used in general to refer to two things that combine to make some third thing, so this applies here. When we subtract $b$ from $a$, we're basically asking what $b$'s (additive) complement with respect to $a$ is.

Another term that could be considered to apply is "conjugate".

Acccumulation
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