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Let $\star$ be a binary operation on the set $S=[0,1]$ defined to be $$\star : [0,1] \times [0,1] \to [0,1] $$

$$\text{where } a \star b = \text{min}\left(\frac12 a , \frac12 b\right) $$

From observation we can see that the set $S$ is closed under $\star$ and that each ordered pair $(a,b)$ is mapped to only one element in $S$.

For example, $1 \star 0.3 = 0.15$

But we also don't have every element in the codomain being hit. There doesn't exist any $(a,b) \in S^2$ such that $a \star b = 0.75$, for example.

Does this cause a problem at all? Is $\star$ still considered a binary operation on $S$? In class we were told all binary operations were surjective, but the textbook for the class states no such thing. And if it is not a problem, I am wondering if there are any more complicated or "elegant" examples. I am interested to see them if they are.

Thanks for any clarification on my confusion.

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    $\begingroup$ Binary operations do not need to be surjective. $\endgroup$ Commented Sep 5, 2018 at 16:02
  • $\begingroup$ Are you clear on the difference between codomain and range? $\endgroup$ Commented Sep 6, 2018 at 13:54
  • $\begingroup$ Yes, codomain is the set of elements that are being mapped to and the range or image is the subset of the codomain that actually has elements from the domain mapped to it $\endgroup$
    – WaveX
    Commented Sep 6, 2018 at 14:51

3 Answers 3

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Binary operations do not need to be surjective. Here is a natural example:

Let $\mathbb N = \{1,2,3,\dots \}$. Then $+: \mathbb N \times \mathbb N \to \mathbb N$ is not surjective because $1$ is not in the image.

Here is another natural, more interesting example:

Let $\mathbb N' = \{2,3,\dots \}$. Then $\times: \mathbb N' \times \mathbb N' \to \mathbb N'$ is not surjective because the prime numbers are not in the image.

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    $\begingroup$ Are these examples natural because they are not contrived, or because they involve $\mathbb N$? $\endgroup$ Commented Sep 5, 2018 at 22:47
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    $\begingroup$ @MishaLavrov, natural because the operations are basic ones, not invented. $\endgroup$
    – lhf
    Commented Sep 5, 2018 at 22:53
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In general binary operations are not surjective.

Note that, say, for some set $S$ and a fixed $s \in S$ the operation given by $a \times b = s$ for all $a,b \in S$ is a binary operation.

Just binary operation means really little. It's literally just a function from $S\times S$ to $S$.

However, if the binary relation has an identity element (or just a left-identity or a right-identity would also suffice), then it can be directly seen that it is surjective. This may or may not be the reason for the discrepancy you observed. It also explains why not few of the most common binary operations are in fact surjective (they have an identity), and further shows a way how to construct some somewhat natural ones that don't.

Note that in the two examples in lhf's answer they judiciously avoided to have the respective natural neutral element in the set.

Let me add some more examples:

  • The reals greater than $0.5$ with addition. This also works for the reals greater than $t$ for any fixed positive $t$, yet not for the positive reals.

  • The reals in the interval $[-0.5,0.5]$ with multiplication (works for any closed interval, even non-symmetric ones in $(-1,1)$ yet not for $(-1,1)$ itself).

  • The reals or also the complex numbers with absolute value greater than $2$ with multiplication (works for greater than $t$ for any fixed $t > 1$, yet not for greater than $1$).

  • The $n \times n$ (real) matrices with determinant greater than $2$ (works for any fixed $t>1$).

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  • $\begingroup$ However, if the binary relation has a neutral element - What do you mean by neutral element? $\endgroup$ Commented Sep 5, 2018 at 17:22
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    $\begingroup$ @taritgoswami Another term for "identity" element. $\endgroup$
    – Randall
    Commented Sep 5, 2018 at 18:09
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    $\begingroup$ @taritgoswami What Randall said. But maybe your point was that I glossed over the right left issue. I clarified the answer and added a link. $\endgroup$
    – quid
    Commented Sep 5, 2018 at 19:06
  • $\begingroup$ @quid Thank you :) $\endgroup$ Commented Sep 5, 2018 at 20:13
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    $\begingroup$ +1 for identifying the possible source of the confusion. $\endgroup$ Commented Sep 6, 2018 at 8:54
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I'll give a simple non-fancy example. Consider $f : \mathbb{R}^2 \to \mathbb{R}$ defined by $$f(x, y) = 0$$ think of this function as assigning the value $0$ to every point on the plane. Certainly $f$ is a binary operation, but it's as non-surjective as such a function can get.

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    $\begingroup$ Nor is it injective. Great example of a boring old function that just disproves a statement. $\endgroup$
    – user459879
    Commented Sep 5, 2018 at 19:40

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