EDIT : This about explicit notation and clarity, usage of $\sqrt{(-x)^2}= x$ is for demonstrative purposes not for the problem itself.  $\sqrt {(-2) ^2} = \sqrt {2^2}$ causing confusion $-2=2$ which is nonsense.  What seems to be needed is somehow distinguishing between $-2=2$ and explicitly stating $2,-2$ are in the set $S=$same equivalence class of numbers $k$  s.t. $k^2$ having the same value.   Another observation is the order of operation is important, $\sqrt {(-2)^2} = \sqrt {2^2}$ can be evaluated as either $\sqrt {4} = \sqrt {4} =2$ or $ {-2} =  {2}$ , how ever implicitly it known that the square function performs first and only after that the square root function can operate, how could this implicit knowledge be worked into explicit notation?  Are there any notations that can be used to make order of operation explicitly visible or instead of ending up with $2=-2$, somehow end up with $2 \equiv-2$ with respect to their squares having same values?
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?
101781
Just out of curiosity I was trying to solve the equation $x^2=2^x$, initially I thought there would be just the two solutions $x=2$ and $x=4$, but wolfram shows that the two equations intersect at not 2 but 3 locations, the third being a negative value of $x$. The third solution isn't obvious like the other two, so I just have a few questions about the negative solution. Is it rational? is it commonly represented with a greek letter? If it is irrational is there a way to approximate it?
101103
We are given with  $$S_n=\{(x,y)|x^2+y^2=n, \text{where }x,y,n \in \mathbb Z\text{ and }n\geq0  \}$$ Now observe that $-\sqrt n\leq x\leq\sqrt n $ and similarily, $-\sqrt n\leq y\leq\sqrt n $. This means that $-\left \lfloor{\sqrt n}\right \rfloor\leq x\leq \left \lfloor{\sqrt n}\right \rfloor$ and  $-\left \lfloor{\sqrt n}\right \rfloor\leq y\leq \left \lfloor{\sqrt n}\right \rfloor$. Through this result one can easily write code and figure out the cardinality of set $S_i$ but I was wondering whether an expression can be found that would allow one to directly compute $|S_i|$.
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.
109313
In the book calculus made easy, page 22 the case of the negative power for   $y=x^{-2}$    $$\begin{align}  y+dy &amp; =(x+dx)^{-2}\tag{1}\\ \\ &amp; = x^{-2}\left(1+\frac{dx}{x}\right)^{-2}\tag{2} \end{align}$$  How do they get from $(1)$ to $(2)$?  And why isn't the binomial function applied to the second line but applied on the third line (the line that comes after line $(2))$?
36779
I found the following proof arguing for the irrationality of $\sqrt{2}$.     Suppose for the sake of contradiction that $\sqrt{2}$ is rational, and choose the least integer $q &gt; 0$ such that $(\sqrt{2} - 1)q$ is a nonnegative integer. Let $q'::= (\sqrt{2} - 1)q$. Clearly $0 &lt; q' &lt; q$. But an easy computation shows that $(\sqrt{2} - 1)q'$ is a nonnegative integer, contradicting the minimality of $q$.   Clearly, this is a vague proof in that it expects the reader to work out the math involved on his own. I took that as an exercise and worked out the following:  Assuming $\sqrt{2}$ to be rational, let $$\sqrt{2} = \frac{p}{q}$$ where $p$ and $q$ are least possible integers.  Clearly then, $\sqrt{2} - 1$ would also be rational, $$\sqrt{2} - 1 = \frac{p-q}{q}$$ where both $(p-q)$ and $q$ are also integers.  Multiplying both sides by $q$, we get $$(\sqrt{2} - 1)q = (p-q) = q'$$ where $q'$ is an integer. Using, $2&gt;\sqrt{2} &gt; 1$ , we get $$1&gt;\sqrt{2}-1&gt;0$$ proving that $q'$ is a non-negative integer.  As stated in the proof, computing $(\sqrt{2}-1)q'$ we get, $$(\sqrt{2}-1)q' = \frac{(p-q)^2}{q}$$which is not necessarily an integer (in fact necessarily not an integer). However, the proof claims it to be.  Am I missing something or something is wrong with how I proceeded?  Here is a link to the textbook where I found this proof: , (p 32, problem 1.14).  P.S. This is my first post here and my first time with TeX. I'd also like to know if there is anything wrong with the way my question is presented.
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 $x = 4$.
7010
I have read a few proofs that $\sqrt{2}$ is irrational.  I have never, however, been able to really grasp what they were talking about.  Is there a simplified proof that $\sqrt{2}$ is irrational?
27406
What is the $\sqrt{i^4}$?   $i^4$ = $(i^2)^2$ So is $\sqrt{i^4}$ = $\sqrt{(i^2)^2}$ = $i^2$ = $-1$?  Or is $\sqrt{i^4}$ = $\sqrt{1}$ = $1$?  When I plug it into my TI-89 Titanium, I get $1$.  Edit: I now see why $i^4 \neq -1$. $$\sqrt{x^2} = \left|{x}\right| $$ So$$\sqrt{(i^2)^2} = \left|i^2\right| = \left|-1\right| = 1$$
122604
How can I prove      $$\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}...}}}}=2$$   I don't know which method can be used for this?
21892
Several people have told me that $\sin(x)^2 = \sin(x^2)$. However, on several computing platforms, such as the TI-84 and Wolfram|Alpha, $\sin(x)^2 = \sin^2(x)$. Can I safely conclude that the notation $\sin(x)^2$ is ambiguous and should always be avoided in favor of $\sin^2(x)$ or $\sin^2 x$? I am having trouble finding any reference through Google or in textbooks (which, I presume, avoid notation like this).
41023
Possible Duplicate:             Well, I remembered this after having Algebra II a year ago, is it possible that this is a valid proof that $1 = -1$?  $$ 1 = \sqrt{1} = \sqrt{-1\cdot-1} = \sqrt{-1} \cdot \sqrt{-1} = i \cdot i = i^2 = -1 $$  $$ \therefore 1 = -1 $$  So is this actually fully valid? Or can it be disproved?
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.
31200
Today I learned that two roots of an irreducible polynomial are "algebraically indistinguishable."   In $\mathbb{Q}(\sqrt{2})$, define the conjugate of $a+b\sqrt{2}$ as $\overline{a+b\sqrt{2}} = a - b\sqrt{2}$.   My understanding/intuition on "algebraically indistinguishable" is that if $P$ and $Q$ are algebraic expressions (meaning using the ring operations only) in $\mathbb{Q}(\sqrt{2})$ with $P$ = $Q$, then then we also have $\overline{P} = \overline{Q}$ where $\overline{P}$ and $\overline{Q}$ are the algebraic expressions $P$ and $Q$ except every member of $\mathbb{Q}(\sqrt{2})$ in the expression is replaced with its conjugate.  For example,  $2(3+\sqrt{2})(4+\sqrt{2}) - (6-\sqrt{2}) = 22 + 15\sqrt{2}$  and indeed,  $2(3-\sqrt{2})(4-\sqrt{2}) - (6+\sqrt{2}) = 22 - 15\sqrt{2}$.  And in $\mathbb{R}(i) = \mathbb{C}$, we're guaranteed the same phenomenon:  $2(3+i)(4+i) - (6-i) = 16 + 15i$  $2(3-i)(4-i) - (6+i) = 16 - 15i$  But what I find interesting is that once I involve familiar operations outside of multiplication and addition, this will no longer hold for $\mathbb{Q}(\sqrt{2})$, but it still holds for $\mathbb{R}(i) = \mathbb{C}$.  For example, $(-\sqrt{2})^{-\sqrt{2}} \neq \overline{\sqrt{2}^{\sqrt{2}}}$ while on the other hand,$(-i)^{-i} = \overline{i^{i}}$.  My initial answer to this is that exponentiation viewed as a binary operation is not closed in $\mathbb{Q}(\sqrt{2})$, so I must view this as an operation on $\mathbb{R}(i) = \mathbb{C}$, but this is a larger field than extending $\mathbb{Q}$ to include the roots of $x^{2}-2$, so in this field with exponentiation, I can now distinguish $\sqrt{2}$ and $-\sqrt{2}$. On the other hand, exponentiation on $\mathbb{R}(i) = \mathbb{C}$ is closed, so even exponentiation does not allow us to algebraically distinguish between $a+bi$ and $a-bi$.
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?
90126
There is something I do not understand in my lecture notes, mainly because I do not understand quotients very well. We consider a number field $K = \mathbb{Q}(\sqrt{d})$ where $d$ is squarefree.  If $\mathcal{O}_K$ denotes the ring of integers of $K$, it has a basis $\langle 1, \omega \rangle$ where $\omega$ is either $\sqrt d$ or $\frac{1+\sqrt{d}}{2}$ depending on $d \bmod 4$. We let $f(x) \in \mathbb{Z}[x]$ be the minimal polynomial of $\omega$ (so $f$ is monic of degree 2).  What I do not understand is first of all that we can write  $ \mathcal{O}_K = \mathbb{Z}[x] / f(x) $. Also, if $p \in \mathbb{Z}$ is prime, the notes then say $$ \frac{\mathcal{O}_K}{(p)} = \frac{\mathbb{Z}[x]}{(p,f(x))} = \frac{\mathbb{F}_p[x]}{f(x) \bmod p} $$  I do understand how to go from the first expression to the second(trivial when you know the above) but not how to go from the second to the third.  Also, the lecturer then gives the expression $\frac{\mathcal{O}_K}{(p)}$ when $p$ is ramified, split or inert. I do not really see why, in the sense how does it help to know that when $p$ is inert then  $$ \frac{\mathcal{O}_K}{(p)} = \frac{\mathbf{F}_p}{f(x)} = \mathbf{F}_{p^2} $$ for example ?  Thank you very much for your help !
92123
A friend is taking a college algebra class and they are teaching him that   $$-3^2 = -9$$  Their explanation is:  $$-3^2 = -(3^2) = -9.$$  It has been a long time for me but I thought that in the absence of any parenthesis that:  $$-3^2 = (-3) \times (-3) = 9.$$  They are even contradicting themselves because they teach the odd/even shortcut for exponents in another part of the book. i.e.:  if the exponent is even, the result it positive, and if  the exponent is odd the result is negative.    This is an actual picture of the book where they contradict themselves on the $-3^2 = -9$:    edit OK, since this has generated way more attention then I ever imagined I've updated here to respond to some of the comments.   1) I understand why the book is not contradicting itself in the picture specifically, or even in the "odd/even" exponent context, due to the fact that variable substitution always implies parens. The book and the teacher, from what my friend has said, do not do a good job of explaining that distinction though.   2) Yes, I understand why the answer is that $-3^2$ is NOT ambiguous is due to order of operations.   3) I made a comment about Khan Academy teaching it incorrectly, I realized I was wrong once I re-watched the video.