$-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2}=\sqrt{1}=1$  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.
43762
I know there is something wrong with this but I don't know where. It's some kind of a math fallacy and it is driving me crazy. Here it is: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
36423
Is $\frac{1}{\sqrt{2}}$ a fraction or not?   I just came across a Grade 8 student's class work where the teacher has given the above as an example to differentiate fractions from rational numbers. I think this is wrong fundamentally, as $\sqrt2$ is irrational while all fractions would be rational. Please advise. Thanks!
2500
How $\frac{1}{\sqrt{2}}$ can be equal to $\frac{\sqrt{2}}{2}$?  I got answer $\frac{1}{\sqrt{2}}$, but the real answer is $\frac{\sqrt{2}}{2}$. Anyway, calculator for both answers return same numbers.
51954
I'm currently learning about radicals and simplifying them, and I came across this problem on the internet and tried to solve it:  $$\sqrt{x+5} = x - 1$$  So I used this logic:  $$ \begin{align} \sqrt{x+5} &amp;= x - 1 \\ x + 5 &amp;= (x-1)^2 \\ x + 5 &amp;= (x-1)(x-1) \\ x + 5 &amp;= x^2 - 2x + 1 \\ 0 &amp;= x^2 - 3x - 4 \\ 0 &amp;= (x-4)(x+1) \\ \end{align} $$  Therefore, $x = -1$ and $x = 4$ satisfy the equation $0 = (x-4)(x+1)$.  But then I tried to plug them in the original problem $\sqrt{x+5}=x-1$:  $$ \begin{align} \sqrt{4 + 5} &amp;= 4 - 1 \\ \sqrt{9} &amp;= 3 \\ 3 &amp;= 3 \end{align} $$  So using 4 works as expected, but when using $-1$:  $$ \begin{align} \sqrt{-1 + 5} &amp;= -1 - 1 \\ \sqrt{4} &amp;= -2 \\ 2 &amp;\ne -2 \end{align} $$  At what stage am I going wrong?  And according the , the solution is $x = 4$.
65721
Why is ${x^{\frac{1}{2}}}$ the same as $\sqrt x $?   I'm currently studying indices/exponents, and this is a law that I was told to accept without much proof or explanation, could someone explain the reasoning behind this.  Thank you.
45673
What's wrong with this demonstration?:  $$A \iff 1 = 1^1$$ $$A \implies 1 = 1^\frac{2}{2}$$ $$A \implies 1 = (1^2)^\frac{1}{2}$$ $$A \implies 1 = ((-1)^2)^\frac{1}{2}$$ $$A \implies 1 = (-1)^\frac{2}{2}$$ $$A \implies 1 = (-1)^1 = -1$$  So finaly we get  $$1 = -1$$
3832
How $\sqrt{x}$ can be a function when $\sqrt{4}$ is equal to $-2$ and $2$?
88029
When $p$ is a prime, we know that Legendre symbol of $a$ is $$\left(\frac{a}{p}\right) = a^{({p-1})/{2}}$$  Suppose $a$ is a square, then $a = x^2$ for some $x$. Therefore, $a^{\frac{p-1}{2}} = x^{p-1} = 1$ by Fermat's little theorem. But if $a$ is not a square, then how to prove that $a^{\frac{p-1}{2}} = -1$?
106726
I noticed a large portion of general public (who knows what square root is) has a different concept regarding the surd of a positive number, $\sqrt\cdot$, or the principal square root function.  It seems to me a lot of people would say, for example, $\sqrt 4 = \pm 2$, instead of $\sqrt 4 = 2$. People even would correct a statement of the latter form to one with a $\pm$ sign. Some also claim that, since $2^2 = 4$ and $(-2)^2 = 4$, $\sqrt 4 = \pm 2$. Some people continue to quote other "evidences" like the $y=x^2$ graph. While most people understand there are two square roots for a positive number, some seem to have confused this with the surd notation.  From an educational viewpoint, what might be lacking when teaching students about surd forms? Is a lack of understanding to functions a reason for this misconception?    Now I have noticed another  that hinted that poster was confused. Following @AndréNicolas's comment below, might these confusion really come from two different communities using the same symbol?
114713
If we define $f(x)$ as some arbitrary function,  then we can define $f'(x)$ or $f^{(1)}(x)$ as the first order and $f''(x)$ or $f^{(2)}(x)$ as the second order.  My question is:  Is there sure thing as a $f^{(-1)}(x)$ notation? Could it be an antiderivative or an integration? And no, I am not talking about $f^{-1}(x)$ that represents an inverse of $f(x)$.
54348
Is it wrong to say $$ \sqrt{x} \times \sqrt{x} =\sqrt{x^2}= \pm x$$ I am quite sure that  $\sqrt{(x)^2} = \pm(x)$  But, does $\sqrt{x } \times \sqrt{x} =- (x)$ doesn't holds in $\mathbb{R}$ but if we assume $\mathbb{C}$ it holds right?
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.
9551
Suppose I have a function defined as $$F(x)= \frac{x^3-4x^2+4x}{x^2-4}$$  Now I want to find the value of $F(2)$. I can do it in 2 ways:   Put $x=2$ and solve the function. It will give:  $$F(2)=\frac{0}{0}$$ which is not defined. Solve $F(x)$ first and then put $x=2$.   $$F(x)= \frac{x(x-2)^2}{(x-2)(x+2)}=\frac{x(x-2)}{x+2}$$  It will give $${F(2)=\frac{0}{4}}$$ which is zero.   How can zero equal not defined?
4300
I know there must be something unmathematical in the following but I don't know where it is: \begin{align} \sqrt{-1} &amp;= i \\\\\ \frac1{\sqrt{-1}} &amp;= \frac1i \\\\  \frac{\sqrt1}{\sqrt{-1}} &amp;= \frac1i \\\\  \sqrt{\frac1{-1}} &amp;= \frac1i \\\\  \sqrt{\frac{-1}1} &amp;= \frac1i \\\\  \sqrt{-1} &amp;= \frac1i \\\\  i &amp;= \frac1i \\\\  i^2 &amp;= 1 \\\\  -1 &amp;= 1 \quad !!? \end{align}
13294
Say you have a number $x^{\sqrt 2}$.  Is there any way to represent this number so that there's no root (or irrational) as the exponent (so that it's easier to understand for me)? I just can't wrap my head around this.  I was thinking something like $$x^{2^{1/2}}$$ and extending on that idea, but I don't know how.  Also, what about for $x^{\sqrt 3}$? Or $x^{\sqrt n}$?
123218
I am having a problem understanding this proof.   Suppose that $\sqrt{2}$ was rational.   Then there would exist positive integers $a,b$ such that $\sqrt{2}=\frac{a}{b}$. Consequently, the set $S=\{k\sqrt{2} : k \text{ and } k\sqrt{2}$ are positive integers $\}$ is nonempty set of positive integers. Therefore, by the well-ordering property, $S$ has a smallest element, say, $s = t \sqrt{2}$.   We have  $s \sqrt{2}-s = s\sqrt{2}-t\sqrt{2}=(s-t)\sqrt{2}$.  Because $s\sqrt{2} = 2t$ and $s$ are both integers,  $s\sqrt{2} - s = s\sqrt{2}-t\sqrt{2} = (s-t)\sqrt{2}$  must also be an integer.   Furthermore, it is positive, because $s\sqrt{2}-s = s(\sqrt{2}-1)$ and $\sqrt{2} &gt; 1$.   It is less than $s$, because $\sqrt{2} &lt; 2$ so that $\sqrt{2}-1 &lt; 1$. This contradicts the choice of $s$ as the smallest positive integer in $S$. It follows that $\sqrt{2}$ is irrational.  My main problem here is how they got the set $S$. My second confusion is the algebra part $s \sqrt{2}-s$. How did they come up with this?   If someone can explain this proof to me it would be great! thank you
112324
I recently saw someone say:     $\sqrt{x}$ is, by definition, positive   Which didn't sit easily with me as we always say $\pm$ for any square root.  Then it got me thinking that if, by definition, the symbol $\sqrt{x}$ meant the positive or the negative root of x, then the $\pm$ symbol would be redundant.  So, have mathematicians defined the root symbol only to include the positive root, thereby necessitating the $\pm$ symbol?  EDIT: This question is different from the one here:  because my question focuses principally on the meaning/convention of the square root symbol rather than on the topic of their being two solutions to $x^2=4$