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When we say $$ \sqrt{9}= x $$ then $\;x = 3,\;$ right?
So why when we square both sides it becomes different: $$ (\sqrt{9})^2 = x^2$$ $$9 = x^2$$ Here $\;x =\pm 3.$

So, does $\;x = \pm 3\;$ in $\sqrt{9} = x?$

J. W. Tanner
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Moe
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    Welcome to Mathematics Stack Exchange. Usually $\sqrt 9$ denotes the principal square root – J. W. Tanner Jan 22 '21 at 18:38
  • Square root is not the same as $1/2$ power. Square root inherently takes the positive root (for positive reals anyway), but $1/2$ power is multi-valued in general. – Cameron Williams Jan 22 '21 at 18:38
  • I think that's wrong, Cameron. In complex analysis,1/2 power means "principal square root" - that is, the square root with nonnegative real part. Can you give me a context where $x^ \frac12$ is different to $\sqrt{x}\ $ ? – Adam Rubinson Jan 22 '21 at 18:52
  • Just because $x = A$ is one thing that does somehting, does not mean that $x =A$ is the ONLY thing that does something. Yes it is true that $3^2 = 9$ but that doesn't mean that $3$ is the ONLY this when squared is equal to $9$. – fleablood Jan 22 '21 at 18:55
  • There are lots of similar questions already; see for example https://math.stackexchange.com/questions/26363/square-roots-positive-and-negative and https://math.stackexchange.com/questions/1448885/square-root-confusion, and the linked questions there. – Hans Lundmark Jan 22 '21 at 18:57
  • Consider this. If you take the number $1$ and square it, subtract $3$ times it, and add $7$ you get. $1^2 - 31 + 7 = 5$. And if you take the number $2$ and do the same thing you get $2^2 -32 + 7 = 5$. Does that mean $1 = 2$? – fleablood Jan 22 '21 at 19:14
  • @CameronWilliams In the real numbers, $a^b$ is typically defined as $\exp(b\log a)$, in which case $x^{1/2}$ is single-valued. – Joe Jan 22 '21 at 19:17
  • @Joe .... which also implies $a > 0$ and in which $(-3)^2$ would be undefined. After all $(-3)^2 = \exp(2\log (-3))$ is undefined. – fleablood Jan 22 '21 at 19:39
  • Also an anology. If $x = 1$ then $(x-1) = 0$. But if we multiply $(x-2)$ and add $5$) we get $(x-1)(x-2)+5 = 0\cdot (x-2) + 5$ so $x^2 -3x +7 = 5$. But if we solve that then $x = 1$ but also $x = 2$. so when we have $x =1$ does that also mean $x=1$ or $x =2$? – fleablood Jan 22 '21 at 19:44
  • @fleablood $1^2 = 11$ but $2^2 i= 22$ the point I'm trying to make is when you power a number you are multiplying it to whatever it's value is, so $x=3$ when powered is $x^2=9$ what really happened is that we multiplied X side by X and 3 side by 3, and if we consider X to be a variable it wouldn't be correct to do that and would derive a totally different equation. – Moe Jan 22 '21 at 19:46
  • @fleablood Yes, if the base is negative or zero, then we have to revert back to the 'school textbook' definition of exponents. My point was that if we want $a^b$ to be a nice, continuous, and single-valued function, then it should never be negative. – Joe Jan 22 '21 at 19:49
  • " it wouldn't be correct to do that and would derive a totally different equation." Sure it is!. If $x = 3$ then $x\times x = 3\times 3$ and $x^2 = 3^2$. Nothing wrong with that. Anything you do with $x$ you can do with $3$ as $x$ and $3$ are the same thing. GIVEN that we know $x$ and $3$ are the same thing. If we DON'T know that $x$ and $3$ are the same thing we can't conclude from $x^2 = 3^2$ that $x = 3$. $x = y$ will always mean $x^2 -3x + 5 = y^2 -3y + 5$. But if we don't know $x=y$ we can't go the other way. $x^2 -3x + 5=y^2-3y+5$ does not mean $x=y$. – fleablood Jan 22 '21 at 20:18
  • @fleablood Can accept such clear answer, Thank you! – Moe Jan 22 '21 at 20:46

4 Answers4

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Just because $A$ is something that does $P$, does not mean that $A$ is the ONLY thing that does $P$.

It is true that $3^2 = 9$ but $3$ is not the only thing that when squared will equal $9$.

so $3^2 =9$ and $w^2 = 9$ does not in any way mean that $w = 3$.

This is no more valid or reasonable than saying $(1)^2- 3(1) +7 = 5$ and $(2^2 - 3(2) + 7 = 5$ and therefor $1 = 2$.

Or saying if Fred is a member of the chess club, and Betty is a member of the chess club, then Fred and Betty are the same person.

....

Your confusion is based on what is the definition of $\sqrt{9}$.

The definition of $\sqrt{k}$ has TWO parts. If $\sqrt{k} = m$ then TWO things must be true: 1) $m^2 = k$.... It's a common mistake to think this is the only thing that must be true but we must also have 2) $m \ge 0$.

So although $3^2 = 9$ and $(-3)^2 = 9$ so both have 1) true. $3>0$ but $-3< 0$ so only $3$ has 2) true and $-3$ has 2) false.

So $\sqrt 9 = 3$ and $-3\ne \sqrt 9$.

Now just because $3 = \sqrt 9$. ANd $(\sqrt 9)^2 = 9$, it does not follow that $\sqrt 9$ is the ONLY thing that when squared equals $9$. $\sqrt 9$ is only one thing that when squared equals $9$.

Just like Fred isn't the only member of the chess club and Betty can also be a member; $\sqrt 9$ is not the only thing that when squared is equal to $9$. $-3$ when squared is also equal to $9$ when squared.

fleablood
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  • Answer is perfect so far But.. I don't understand why "m >= 0" is a rule? – Moe Jan 22 '21 at 19:36
  • @Mlotov: Because we want symbols to have precise meanings. The symbol $\sqrt9$ is supposed to denote a number. But if we only say that $\sqrt9$ is a number whose square is $9$, there are two numbers the symbol could denote. But we want to be able to point to one specific instance of such a number. And that's easier if the symbol $\sqrt9$ denotes only one of the two possibilities. We can then call the other possibility $-\sqrt9$ and all is well. – Vercassivelaunos Jan 22 '21 at 19:51
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    Because we SAY it is a rule. We know that $m^2 = k$ will usually have two values and only of them will be positive and one of them will be negative and we are interested in the positive one of them so we say "Hey, guys. We need some way to take the number $k$ and talk about the solutions to $x^2 = k$. since there are two such numbers its good enough to talk about just the positive one. Let's agree amongst ourselve to call that 'the square root of $k$' and lets write it with a weird crooked line, $\sqrt{}$ over then $k$. Does everyone like that idea? Okay, that's our definition now." – fleablood Jan 22 '21 at 19:53
  • @Mlotov Are you familiar with the concept of a function? If we want the 'square root' operation to be a function, then it must have a single-output. And it is more convenient to take the positive square root then the negative square root. – Joe Jan 22 '21 at 19:53
  • Clear enough answers, Thank you all! – Moe Jan 22 '21 at 20:54
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If $P$ implies $Q$, then this does not necessarily mean that $Q$ implies $P$. For example, all zebras are white-striped animals, but not all white-striped animals are zebras: $$ \require{cancel} \text{Zebra} \implies \text{Stripeyness} $$ but $$ \text{Stripeyness} \cancel{\implies} \text{Zebra} \, . $$ The same principle applies here. We have $$ x=3 \implies x^2=9 $$ but $$ x^2 = 9 \cancel{\implies} x=3 \, . \\ $$ This is because every positive number has two square roots, and so just from the fact that $x^2=9$, we can't deduce that $x=3$. However, we certainly can deduce from $x=3$ that $x^2=9$.

Joe
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Edited my answer.

As J. W. Tanner said in the comments, $\sqrt{9} = 3$. So $x = 3,\ $ not $-3$.

Therefore $x^2 = 9,\ $ because $3^2 = 9$.

Now the notation $t = \pm 3$ means that $\left(t=3\ \ or\ \ t = -3\right).$

So it is true that $x = \pm 3$ here because $\left(x=3\ \ or\ \ x = -3\right) $ is true.

Just because $x^2 = 9,\ $ doesn't mean that $x$ equals both $3$ and $-3$. It just means that $x = 3$ or $x = -3$ (or both). In this case, $x=3$ only, because $x$ was defined to be equal to $3$ only.

Adam Rubinson
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  • Is it even allowed to multiply both sides of an equation by an even number? $$ x = 3 $$ $$x^2 = 9$$ in the first equation x = 3 ONLY but in the second it's equal to + or - 3, So I'm thinking of this as multiplying a side by x and multiplying the other side by 3 $$ xx = 33$$ and this of course derives a whole new equation.

    and that's also why I said to raise the power of an even number because if it's odd then the equations will be the same: $x^3 = 81$ same as $x=3$.

     – Moe Jan 22 '21 at 19:28
  • Since $3\ne-3$, it can't possibly be the case that "$t=\pm3$ usually means that $t=3$ and $t=-3$". – Mark S. Jan 22 '21 at 19:29
  • @Mlotov If you prefer, you can multiply both sides by $3$ to get $3x=9$ and multiply both sides by $x$ to get $x^2=3x$, and then put them together to get $x^2=3x=9$, so $x^2=9$ is still a valid conclusion. – Mark S. Jan 22 '21 at 19:31
  • We can do whatever we want with $a = b$. If $a = b$ then $a^2 + 7a + e^a +\frac {\pi}a -\text{the third digit of } a= b^2 + 7b + e^b +\frac {\pi}b -\text{the third digit of } b$. But just because we can go one way doesn't mean we can go backwards. – fleablood Jan 22 '21 at 19:59
  • Image Fred Meyer loves penguins and Edna Vincay also loves penguins. Then Fred = Mr. Meyer so Fred's favorite animal=Mr. Meyer's favorite animal so penguins = penguins so Fred's favorite animal = Edna's favorite animals so.... Fred = Edna? Starting with Fred = Mr. Meyer we can do what every we want. If $3 = x$ then $3^2 = x^2$ and $3'\text{s favorite animal}=x'\text{s favorite animal}$. That's fine. But we can't go the other way. $3^2 = x^2 \not \implies 3=x$. ... Oh! But $\sqrt{3^2}=\sqrt{x^2}$. But $\sqrt{x^2} \ne x$. That just isn't true: $\sqrt{x^2}=|x|$; not $x$ itself. – fleablood Jan 22 '21 at 20:06
  • " Since $3\neq−3,$ it can't possibly be the case that "$t=\pm3$ usually means that $t=3$ and $t=−3$" ". I think you're right. So does $t=\pm3$ mean "$t = 3$ or $t = -3"?$ – Adam Rubinson Jan 22 '21 at 20:07
  • Usually the notation $t =\pm 3$ means "$t$ is a value that is either $3$ or $-3$ but we either don't know or don't wish to speculate which". The comment "$t=\pm 3$ usually means $t=3$ and $t=-3$" can't actually be true. $t =3$ AND $t=-3$ is impossible of course. – fleablood Jan 22 '21 at 20:10
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$a=b\implies a^2=b^2$ (merely by replacement of $b$ by $a$ in the right side of the implication, assuming the left side holds), but the converse is wrong in the sets $\mathbb R$, $\mathbb Z$, $\mathbb Q$, $\mathbb C$, and $\mathbb Z/p\mathbb Z$ for prime $p>2$. Proof by example: $a=1$, $b=-1$.

However, $a^2=b^\implies a=b$ in some other sets, including $\mathbb R_+$, $\mathbb N$, and the Booleans $\mathbb Z/2\mathbb Z=\{0,1\}$.

So, does $\;x = \pm 3\;$ in $\sqrt{9} = x$ ?

No in $\mathbb N$, $\mathbb Z$, $\mathbb Q$, $\mathbb R$, where a square root is non-negative by definition. Yes in sets where $-3=3$; or for some unusual definition of square root.

fgrieu
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