[CLOSED] Thanks  GoodDeeds and Henry [1] I understood the fundamental problem.  As √9 = ± 3    If , √A = √B Thus, ** ±a =  ±b** And so a = b; -a = b; and  a = -b;; Thus, √9 = +3 OR -3    Let, √A = ±a and, √B = ±b Is this reasoning correct?   If not how does this actually work?  [1] \bib{26369}{misc}{     title={Square roots -- positive and negative},     author={Henry ()},     note={URL:  (version: 2011-03-11)},     eprint={},     organization={Mathematics Stack Exchange} }
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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?
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Okay so I am in college and in the notes it shows this example:  // Example: Compute the square root of 3 modulo 143  3 modulo 143 = 3 mod (11*13)  Then he jumps to this:   √3 (mod 11) = ± 5  √3 (mod 13) = ± 4  Using the Chinese Remainder Theorem, we can calculate the four square roots as 82, 126, 17 and 61.  //  The lecturer never makes anything clear even though it is our first encounter with module arithmetic. I don't know were the 4 or the 5 come from or the √3 . Can anyone explain to be how he got  √3 (mod 11) = ± 5  √3 (mod 13) = ± 4   And also how does he use this information in conjunction with the Chinese Remainder Theorem arrive at the square roots. I would really appreciate if anyone could help me out here
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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.
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I am aware that 1/0 is undefined for two reasons:   If you would have to give an answer to this it is infinity which is not a number but a concept; The limit of 1/x for $x \to 0$ is either positive or negative infinity and since it has two limits we cannot state what its exact value is   How can we use this for square roots? It has multiple answers so why do we pick the positive one? if $x^2 = 16 \implies x = \sqrt{16} $ or $x = -\sqrt{16}$ for respectively the positive and negative solution. This implies that the square root function has a single answer and we must negate its answer to obtain the second solution. I understand this is extremely practical but it feels like 'cheating' since we know that there are 2 solutions, but in the function we simply ignore one of them. Is this correct?
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It is well known that $\sqrt{2}$ is irrational, and by modifying the proof (replacing 'even' with 'divisible by $3$'), one can prove that $\sqrt{3}$ is irrational, as well.  On the other hand, clearly $\sqrt{n^2} = n$ for any positive integer $n$.  It seems that any positive integer has a square root that is either an integer or irrational number.        How do we prove that if $a \in  \mathbb N$, then $\sqrt a$ is an integer or an irrational number?      I also notice that I can modify the proof that $\sqrt{2}$ is irrational to prove that $\sqrt[3]{2}, \sqrt[4]{2}, \cdots$ are all irrational.  This suggests we can extend the previous result to other radicals.        Can we extend 1? That is, can we show that for any $a, b \in \mathbb{N}$, $a^{1/b}$ is either an integer or irrational?
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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?
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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 $x^{\sqrt n}$?
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$$ a = b $$ $$ a^2 = ab $$ $$ a^2 - b^2 = ab - b^2 $$ $$ (a+b)(a - b) = b(a - b) $$ $$ a + b = b $$ $$ 2b = b $$ $$ 2 = 1 $$  Does the self-reference to the original formula make this path of argument invalid? I am confused as to what's going on here.  Would this be a better continuation from $a + b = b$:  $$ a + b = b \implies a = 0 \implies b = 0$$
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Q1 - Why is $\sqrt{-x}*\sqrt{-x}=-x?$  Q2 - I was thinking it would be:  $\sqrt{-x}*\sqrt{-x}=\sqrt{-x*-x}=\sqrt{x^2}$ but apparently not (why not?)  Q3 - What are the formal algebra rules to use? Can I calculate this without using i such as in: $\sqrt{-x}*\sqrt{-x}=i\sqrt{x}*i\sqrt{x}=-\sqrt{x^2}=-x$.
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I have the equation $x^2=x$.  If I divide $x$ from both sides I get $x=1$.  Yet clearly $x$ can also equal $0$.  What step in this process is wrong? It seems to me that there's only one step. And isn't dividing the same thing from both sides a valid step?  I hope this isn't a stupid question because I feel dumb asking about something so basic.  EDIT: To clarify, what I'm looking for is not only an explanation for why my methodology is wrong, but also a better methodology that will keep me from missing possible solutions in the future.
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I read this in a math textbook, stated as obvious fact. I cannot wrap my head around why this works. I understood the division lemma, and also got the algorithm to work: but the proof of why the algorithm works, just said "because of the equation" that I put in the question.  Could you please elaborate on why it works?
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$-1=(-1)^1=(-1)^\frac{2}{2}=((-1)^2)^\frac{1}{2}=\sqrt{(-1)^2}=\sqrt{1}=1$  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.
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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?
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Is there an error in this question?  Let u = $(1,0,2)$, v = $(2,1,0)$ and w = $(0,2,1)$. Compute  ||u + v - w.||v = v  This is exactly what is in the question (even the dot after w which I don't know what it means).  Thanks
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You learn in algebra that  $$\sqrt{ab}=\sqrt{a} \sqrt{b}$$ and that  $$\sqrt{\frac ab}=\frac {\sqrt a}{\sqrt b}$$ You also learn to never make the fatal mistake of thinking $$\sqrt{a+b}=\sqrt{a}+\sqrt{b}$$  However, I am wondering if there is a rule for $\sqrt{a+b}$. I would think it would be pretty complex, if it exists at all.  $$\sqrt{a+b}=\text{?}$$
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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: $$-1= (-1)^3 = (-1)^{6/2} = \sqrt{(-1)^6}= 1?$$
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Can you enter the rabbit hole recursively?  If the $ \sqrt{-1} = i $ then, what does $ \sqrt{-i} $ equal?