this is something that came up when working with one of my students today and it has been bothering me since.  It is more of a maths question than a pedagogical question so i figured i would ask here instead of MESE.  Why is $\sqrt{-1} = i$ and not $\sqrt{-1}=\pm i$?  With positive numbers the square root function always returns both a positive and negative number, is it different for negative numbers?
40496
What's the thing with $\sqrt{-1} = i$? Do they really teach this in the US? It makes very little sense, because $-i$ is also a square root of $-1$, and the choice of which root to label as $i$ is arbitrary. So saying $\sqrt{-1} = i$ is plainly false!  So why do people say $\sqrt{-1} = i$? Is this how it's taught in the US?
62044
$-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2}=\sqrt{1}=1$  This proof bugs me for the following reasons:   Mathematicians have defined the symbol $\sqrt{}$ (Named the principal square root) to mean 'take only the positive square root of the number under the radical' This makes it so that the $y = \sqrt{x}$ is a function and therefore for every x-value (input) there is only 1 y-value (output). Therefore $\sqrt{x^2}=|x|$ due to the above definition. But then this means $\sqrt{(-1)^2} = |-1| = 1!!!!$   I really don't understand what's wrong with the above proof, the only way this could make sense is to define $\sqrt{a} = \pm k$ where the symbol know gives rise to two solutions to the equation, but then y = $\sqrt{x}$ is not a function and is simply a relation.   If we have $\sqrt{a}=\pm k$ then $\sqrt{(-1)^2}$ can be equal to -1 as well and then $-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2} = -1 = -1$ and everything works out fine.  Side Note: Why do we even want y = $\sqrt{x}$ to be a function anyway? What's the harm in defining $\sqrt{x} = \pm k$ where y = $\sqrt{x}$ is not a function?     Proof for $\sqrt{x^2}=|x|$  Let k = $\sqrt{x^2}$ where k is a constant.  $k^2=(\sqrt{x^2})^2=(((x^2))^\frac{1}{2})^2$  $k^2=x^2$  $k^2-x^2=0$  $(k+x)(k-x)=0$  $k = x$ or $k=-x$  But $\sqrt{a}&gt;0$ for $a&gt;0$ by definition of the principal square root.  $\therefore k \neq-x$  In order to make this identity work properly, an absolute value sign is required.  $\therefore \sqrt{x^2}=|x|$    Btw, my question is not the same as why $\sqrt{-1\times-1}\neq(\sqrt{-1})^2$ as I am not using the property $\sqrt{a}\sqrt{b}=\sqrt{ab}$ here and I am not bringing up the concept of imaginary numbers.
123137
Simplify: $x^2 &gt; 1$.  My solution: Taking square root on both sides:  $±x &gt; ±1$  So my results are:   $x &gt; 1$ $x &gt; -1$ $-x &gt; 1$ $\implies$ $(-1 &gt; x)$ $-x &gt; -1$ $\implies$ $(1 &gt; x) $   But I strongly feel this is wrong. What is wrong here?  A step-by-step explanation will help me.
93546
I've recently started to think about this, and I'm sure a couple of you out there have, too.  In Algebra, we learned that $|x|\geq0$, no matter what number you plug in for $x$. For example: $$|-5|=5\geq0$$  We also learned that $x^2\geq0$. For example: $$(-5)^2=25\geq0$$ The exception for the $x^2$ rule is imaginary numbers (which we learn later on in Algebra II). Imaginary numbers are unique, in that their square is a negative number. For example: $$4i^2=-4$$ These imaginary numbers can be used when finding the "missing" roots of a polynomial equation.  My question to you is this: Is there any number whose absolute value is negative, and how could it be used?
6221
It is perhaps a bit embarrassing that while doing higher-level math, I have forgot some more fundamental concepts. I would like to ask whether the square root of a number includes both the positive and the negative square roots.  I know that for an equation $x^2 = 9$, the solution is $x = \pm 3$. But if simply given $\sqrt{9}$, does one assume that to mean only the positive root? And when simply talking about the square root of a number in general, would one be referring to both roots or just the positive one, when neither is specified?
116211
Why $\sqrt {12} = 2 \sqrt 3$? It is obvious? If we considered the function $f(s) = s^2 $ it is injective on positive numbers so we obtain the conclusion. But in  the same time it is an equality between irrational numbers. Suppose that we know just to compute the square roots.
120629
I am trying to help my daughter on her math homework and I am having some trouble on some equation solving steps. My current major concern relies on understanding why $\sqrt{8}/2$ equal to $\sqrt{2}$.  Thanks in advance
88828
Is my proof that the square root of a positive integer is either an integer or an irrational number correct?  The proof goes like this:  Suppose an arbitrary number n, where n is non-negative. If $\sqrt{n}$ is an integer, then $\sqrt{n}$ must be rational. Since $\sqrt{n}$ is an integer, we can conclude that n is a square number, that is  for some integer a.  Therefore, if n is a square number, then $\sqrt{n}$ is rational.  Suppose now that n is not a square number, we want to show that the square root of any non-square number is irrational.  We prove by contradiction. That is, we suppose that the square root of any non-square number is rational. So $\sqrt{n} = \frac{a}{b}$, where $a,b \in Z^+, b \neq 0$. We also suppose that $a \neq 0$, otherwise $\frac ab = 0$ , and n will be a square number, which is rational.   Hence $n = \frac {a^2}{b^2}$, so $nb^2 = a^2$.    Suppose $b=1$. Then $\sqrt n = a$ , which shows that n is a square number. So $b \neq 1$. Since $\sqrt n &gt; 1$, then $a&gt;b&gt;1$.  By the unique factorization of integers theorem, every positive integer greater than $1$ can be expressed as the product of its primes.  Therefore, we can write $a$ as a product of primes and for every prime number that exists in $a$, there will be an even number of primes in $a^2$. Similarly, we can express $b$ as a product of primes and for every prime number that exists in $b$, there will be an even number of primes in $b^2$.  However, we can also express $n$ as a product of primes. Since $n$ is not a square number, then there exist at least one prime number that has an odd number of primes.  Therefore, there exists at least one prime in the product of $nb^2$  that has an odd number of primes. Since $nb^2=a^2$ , then this contradicts the fact that there is an even number of primes in $a^2$ since a number can neither be even and odd.  Therefore, this contradicts the fact that $\sqrt n$ is rational. Therefore, $\sqrt n$ must be irrational.  Is this sufficient? Or is there any parts I did not explain well?
58567
i have a pretty nasty question. i was glancing through a few olympiad papers and stumbled upon this question:  prove that  $ i = \sqrt {-1}\ $.  i tried the conventional methods namely euler's formula  but could not figure what to do next. how do you actually proceed further? i have checked a few books and a proof is not available.are there any proof available for this
17288
Does there exist a square root of $-1$ in the ring $\mathbb{R}[x]/\langle(x^2+1)^2\rangle$?  Now, any element in the ring $\mathbb{R}[x]/\langle(x^2+1)^2\rangle$ is of the form $(a+ib)+e(c+id)$ where $i^2=-1$ and $e^2=0$. Now, setting $a=c=d=0$ and $b=1$, I think we would get a square of $-1$. Is this right? Any hints? Thanks beforehand.
56554
I apologize if this is too obvious, but I crave for a reasonable definition/explanation - why is  $(-1)(-1)=1$?  Something akin to the reason $a^{-1}=\frac{1}{a}$, I.e. because $a^{-1}a^1=a^{1-1=0}=1$, so we get $a^{-1}=\frac{1}{a^1}=\frac{1}{a}$ by dividing both sides by $a^1$.  That sort of stuff.
89441
Given the following matrix, I was asked to tell whether the matrix is the square of some matrix or not.  $$ \begin{pmatrix} 1 &amp; 0 &amp; 0\\ 0 &amp; -1 &amp; 0\\ 0 &amp; 0 &amp; -1 \end{pmatrix} $$  Since this matrix is diagonal, a square root is  $$ \begin{pmatrix} 1 &amp; 0 &amp; 0\\ 0 &amp; i &amp; 0\\ 0 &amp; 0 &amp; i \end{pmatrix} $$ Am I right?
5703
Can the expression $\sqrt{n} + \sqrt{m}$ be rational if neither $n,m \in \mathbb{N}$ are perfect squares? It doesn't seem likely, the only way that could happen is if for example $\sqrt{m} = a-\sqrt{n}, \ \ a \in \mathbb{Q}$, which I don't think is possible, but how to show it?
34782
There is a formula given in my module:  $$ \sqrt[n]{a^n} = \begin{cases}  \, a  &amp;\text{ if $n$ is odd } \\  |a| &amp;\text{ if $n$ is even } \end{cases} $$  I don't really understand the differences between them, kindly explain with an example.
73212
We know that $e^{\pi i} = -1$ because of de Moivre's formula. ($e^{\pi i} = \cos \pi + i\sin \pi = -1).$  Suppose we square both sides and get $e^{2\pi i} = 1$(which you also get from de Moivre's formula), then shouldn't $2\pi i=0$? What am I missing here?
91168
What would the graph of $(-1)^{x}$ look like?   I know that the value of the function alternates between $1$ and $-1$ when it is defined so I think it would just be points spread over the lines $x=1$ and $x=-1$.   Is this correct? Also, will there be any definite pattern of the points.
38052
I understand why $e^{i\pi} = -1$and as a result  $e^{i2\pi}=\left(e^{i\pi}\right)^2=1.$ These results can be confirmed using Euler's formula  But why does $e^{i\pi/3}\neq -1$ as we can write it $(e^{i\pi})^{1/3}$ and as such $(-1)^{1/3}$ the cubic root of $-1$ is $-1$ itself.  Yet using Euler's formula we get $e^{i\pi/3}=\frac{1}{2}+i\frac{\sqrt 3}{2}$.