4
$\begingroup$

Is $-1$ a perfect square?

We know that $i^2 = -1$. Does that mean $-1$ is a perfect square because $i$ is not an irrational or decimal number?

$\endgroup$
7
  • 7
    $\begingroup$ In the ring of Gaussian integers $\mathbf Z[i]$, yes. $\endgroup$
    – Bernard
    Jan 22 '17 at 21:52
  • 7
    $\begingroup$ What is your definition of perfect square? The usual definition says it's the square of an integer. $\endgroup$
    – Crostul
    Jan 22 '17 at 21:53
  • 6
    $\begingroup$ Normally, we call a number a perfect square if it is the square of an integer. And with integer here we mean rational integer, i.e., no complex numbers allowed $\endgroup$ Jan 22 '17 at 21:53
  • $\begingroup$ @Crostul, I'm more asking if it is a perfect square using general terms, so whatever the generally accepted definition of a perfect square is. $\endgroup$
    – Travis
    Jan 22 '17 at 21:54
  • 3
    $\begingroup$ Every complex number is the square of some complex number, so a small modification would lead to "all complex numbers are perfect squares". In this case, the name "perfect square" will not make sense anymore. But if we restrict to $\mathbb Z[i]$, a definition would make some sense because not every number will be a square of some other number. $\endgroup$
    – Peter
    Jan 22 '17 at 22:02
21
$\begingroup$

The term perfect square is typically reserved for squares of integers, unless further context is specified, so in the regular usage of the term, $-1$ is not a perfect square.

More generally, you might want to refer to perfect squares in a ring (especially the ring of integers of an algebraic number field)—in such a context, we might say that $-1$ is a perfect square, but with reference to the ring. For example, $-1$ is a perfect square in $\mathbb{Z}[i]$, but is not a perfect square in $\mathbb{Z}$.

This situation is similar to how $i$ is not an irrational number—the term irrational number refers to real numbers which are not rational, not just any old mathematical entity which is not a rational number.

$\endgroup$
3
  • $\begingroup$ @BillDubuque, your link is to an ebay listing! Also, en.wikipedia.org/wiki/Irrational_number seems to agree with Clive's convention here that irrational numbers are necessarily real, so that $i$ is, by that convention, not irrational. $\endgroup$ Jan 23 '17 at 18:00
  • $\begingroup$ Your last sentence is inaccurate - see Is $i$ irrational? @Barry see the linked thread (link now fixed). The point is that the definition / convention depends on the context (just as for squares). $\endgroup$ Jan 23 '17 at 18:03
  • $\begingroup$ @BillDubuque, thanks for fixing the link. It's a nice answer there. I'm a little surprised the wikipedia entry doesn't make any mention of the alternative convention. $\endgroup$ Jan 23 '17 at 18:12

Not the answer you're looking for? Browse other questions tagged or ask your own question.