I know this is a dumb question but I can't get the answer to another follow up question,

  1. What is the square root of 1?
  2. If the square root of 1 is itself then why does other square root of number not equal to themselves?
  3. Is there other square root of 1 aside from 1?

This question is somewhat a deconstruction of the fundamental logic of mathematical concepts but I can't find a good reason to answer this questions. Hoping for a good reason.

[Edit] if we're going to talk about principal square roots then therefore -1 would be included but i'm only referring to non negative numbers therefore 3.1. if there are other square root of 1 which could be between 0 and 1?


closed as too localized by Andrés E. Caicedo, Ross Millikan, Asaf Karagila, t.b., J. M. is a poor mathematician Apr 16 '11 at 18:15

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    $\begingroup$ no one has mentioned that $0 \cdot 0 = 0$. This can be attained from the equation $x^2=x \implies x(x-1)=0 \implies x=0 \, \mathrm{ or } \, 1$ $\endgroup$ – picakhu Apr 14 '11 at 16:44
  • $\begingroup$ The square of a positive number less than 1 is less than itself: $0\lt x\lt 1$ implies $0\lt x^2\lt x$ (multiplying through by $x$); so no number between $0$ and $1$ can be the square root of a larger number (in particular, of $1$). $\endgroup$ – Arturo Magidin Apr 14 '11 at 21:55
  • $\begingroup$ @sergiol No it doesn't: $-1^2=1\not=-1$. What you might be confused by: $-1$ does satisfy $x^2=\sqrt{x^2}$, but that's because square roots don't work the way they "should." $\endgroup$ – Noah Schweber May 21 '15 at 7:39

In a field such as $\:\mathbb Q\:,\ \mathbb R\:,\ \mathbb C\:,\:$ we have $\rm\ x^2 = 1\ \iff\ (x-1)\ (x+1) = 0\ \iff\ x = \pm 1\:.\: $ In rings that are not fields there can be more than two square-roots, e.g. modulo $15$ there are two additional roots $\rm\ (\pm\:4)^2\equiv 1\ (mod\ 15)\:.\:$ In some contexts authors define single-valued square-root functions that uniformly select one of the roots, e.g. the non-negative root, or principal branch, etc. In ring theory one can adjoin a generic square root of one $\rm\:s = \pm 1\:$ to a ring $\rm\:R\:$ where $\:2\:$ is invertible as follows $\rm\ R[s]\ \cong R[x]/(x^2-1)\ \cong R/(x-1) \times R/(x+1)\ \cong R^2\:,\ $ via $\rm\ f(x)\to (f(1),\ f(-1))\:;\:$ informally, do arithmetic in two parallel universes (rings), one with $\rm\ s = 1\ $ and one with $\rm\ s = -1\:.$

As to your second question concerning why there aren't other numbers that are their own square root, note that $\rm\ \sqrt{x} = x\ \Rightarrow x = x^2\ \Rightarrow\ x\ (x-1) = 0\ $ so $\rm\:x = 0\:$ or $\rm\:x = 1\:$ in a field (or domain). Elements satisfying $\rm\ x^2 = x\ $ are known as idempotents. Idempotents that are nontrivial ($\ne\: 0,\:1$) are intimately connected with nontrivial factorizations. In the above example $\:\mathbb Z/15\:,\:$ the integers modulo $\:15\:,\:$ we have the factorization $\ \mathbb Z/15\ \cong \mathbb Z/3 \times \mathbb Z/5\:,\:$ arising from the idempotent $\rm\:6\:,\:$ i.e. $ (0,1)\in \mathbb Z/3 \times \mathbb Z/5\:.\:$ Note that if $\rm\:e\:$ is idempotent then $\rm\:(1-2\:e)^2 =\: 4\:(e^2-e)+1 = 1\:,\: $ hence $\rm\:2\:e-1\:$ is a square-root of $1\:;\:$ it is a nontrivial square root $\rm\:2\:e-1\:\ne\: \pm 1\:$ if $2$ is cancellable. In our example $\rm\:e=6\:,\:$ so $\rm\:1-2\:e \equiv 4\:$ which does indeed square to $\rm\:1\ (mod\ 15)\:.\:$ We can quickly factor $15$ from this nontrivial square root $\rm\:s\:,\:$ namely $\rm\:gcd(15,s-1) = gcd(15,3) = 3\:;$ $\rm\ gcd(15,s+1) = gcd(15,5) = 5\:.\:$ In fact many integer factorization algorithms work by searching for nontrivial square-roots. This way of factoring by simple gcd computations works more generally. One can quickly factor $\rm\:n\:$ given any polynomial with more roots $\rm\:(mod\ n)\:$ than the degree of the polynomial - see my post here.

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    $\begingroup$ Dear Bill, sorry for going off-topic, but out of curiosity, why do you typeset all your variables in a roman (i.e. not italic) style? $\endgroup$ – Rahul Apr 14 '11 at 21:58
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    $\begingroup$ @Rahul: I find the default italic fonts far too ugly. Alas, they messed up the roman fonts too with the recent font changes. $\endgroup$ – Bill Dubuque Apr 14 '11 at 22:40
  • $\begingroup$ If on your system the default italic font here looks different from the typesetting that TeX produces, then that could be a website/browser incompatibility issue worth raising on Meta. (On the other hand, if it looks just like TeX output, and you just don't like the look of TeX, then never mind...) $\endgroup$ – Rahul Apr 15 '11 at 2:08
  1. The square root of $1$ is $1$.

  2. This is because the number $1$ has the special property $1\cdot 1=1$. This is not true for $2$, for instance: $2\cdot 2$ is not $2$.

  3. Yes, there is another number $x$ with $x^2=1$, namely $x=-1$. Usually, this solution is not called a square root of $1$, however, because you want "square root" to be a function, so that its value has to be unique.

Edit: To see that there are no further solution to $x^2=1$, notice that $x^2=1$ implies $(x+1)(x-1)=0$ and thus $x=\pm 1$.

  • $\begingroup$ aside from 1 or -1? $\endgroup$ – Richeve Bebedor Apr 14 '11 at 16:28
  • $\begingroup$ @rich I included an answer to this in my answer. $\endgroup$ – Rasmus Apr 14 '11 at 16:33
  • $\begingroup$ i have edited my description on the question to give you further details thanks for the reply $\endgroup$ – Richeve Bebedor Apr 14 '11 at 18:25
  1. The square root of $1$ is $\pm1$.

  2. $1$ is the only number which when multiplied by itself equals $1$.

  3. $-1$ multiplied by itself equals $1$.

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    $\begingroup$ "The" square root refers to the principal branch of the square root; to make square root into a function, we pick the nonnegative one, so "The" square root of $1$ is $1$, not $\pm 1$ (if it was really two things, it would be "the square root s" not "the square root"). You are confusing $\sqrt{1}$ with "solutions of $x^2=1$". The first clause of your second statement contradicts the second clause of that statement. $\endgroup$ – Arturo Magidin Apr 14 '11 at 17:23
  • $\begingroup$ Your point 2 is not right. (-1) multiplied by itself gives 1 as result $\endgroup$ – sergiol May 21 '15 at 7:20

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