0

I have been trying to learn binary operations and have not been able to understand how to prove that an operation is binary. For example:

Show that $*:\mathbb R×\mathbb R→\mathbb R$ given by $(a,b)\mapsto a+4b^2$ is a binary operation.

Now the answer in the book is:

Since $*$ carries each pair $(a, b)$ to a unique element $a + 4b^2$ in $\mathbb R$, $*$ is a binary operation on $\mathbb R$.

Firstly, I think that the reason they are giving is false as wouldn't $(a,-b)$ give the same answer under the same operation, thereby disproving the answer by the fact that $*$ carries each pair $(a, b)$ to a unique element?

Apologies if this is a newbie question, thanks.

Ѕᴀᴀᴅ
  • 34,263
JakePaul
  • 1
  • 3
    Carrying to a unique element means that $(a,b)$ isn't carried to two distinct elements; it doesn't mean nothing else maps to that element. – Malice Vidrine Mar 29 '19 at 05:05
  • A binary operation must be a function (not like a multifunction or relation) in the set theory sense. $\forall a \forall b \exists ! c: c = a \ast b$. – Henno Brandsma Mar 29 '19 at 05:21
  • Here "unique" means that the map is single-valued (vs. multi-valued). It is poor terminology as it invites confusion. – Bill Dubuque Mar 30 '19 at 01:54

1 Answers1

1

This is probably the result of a recently widespread misunderstanding (and not just in mathematics) of the word "unique". Let's be clear:

"unique" does not mean "different".

This confusion is very prevalent in web administrators claiming that their site has "123,456,789 unique visitors" when they really mean "123,456,789 different visitors".

Your operation $*$ carries each pair to a unique element, meaning that if $(a,b)$ is given, there is only one possible value of $a+4b^2$.

The operation does not carry each pair to a different element (or to put it more clearly, does not always carry different pairs to different elements), for exactly the reason you have stated. However this is not the same as the previous paragraph.

David
  • 82,662