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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?

Willie Wong
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Richeve Bebedor
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    no one has mentioned that $0 \cdot 0 = 0$. This can be attained from the equation $x^2=x \implies x(x-1)=0 \implies x=0 , \mathrm{ or } , 1$ – picakhu Apr 14 '11 at 16:44
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    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 $1$). – Arturo Magidin Apr 14 '11 at 21:55
  • @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."

     – Noah Schweber May 21 '15 at 07:39

3 Answers3

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In a field such as $\,\mathbb Q,\ \mathbb R,\ \mathbb C,\,$ we have $ \ 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 $ \ (\pm\,4)^2\equiv 1\pmod{\!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 we can adjoin a "sign" = generic square root of one $\,s = \pm1\,$ to a ring $ \,R\,$ as follows $ R[s]\cong R[x]/(x^2\!-\!1).\,$ If $\,2\,$ is invertible, CRT $\Rightarrow R[s] \cong R[x]/(x\!-\!1) \times R[x]/(x\!+\!1)\ \cong R^2,\, $ via $\, f(x)\to (f(1),f(-1));\,$ informally, this ring product implements arithmetic in two parallel universes (rings), where in the first world we have $\,s = 1\,$ , and $\, s = -1\,$ in the second.

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

Bill Dubuque
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    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? –  Apr 14 '11 at 21:58
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    @Rahul: I find the default italic fonts far too ugly. Alas, they messed up the roman fonts too with the recent font changes. – Bill Dubuque Apr 14 '11 at 22:40
  • 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...) –  Apr 15 '11 at 02:08
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  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 $x=\pm 1$.

Rasmus
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  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 $1$.

Community
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Slide Rule
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    "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. – Arturo Magidin Apr 14 '11 at 17:23
  • Your point 2 is not right. (-1) multiplied by itself gives 1 as result – sergiol May 21 '15 at 07:20