If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself?

For example the square of $2$ is $2^2=2 \cdot 2=4 $ .

But square root of $2$ is not $\frac{2}{2}=1$ .

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    $\begingroup$ This is like saying the opposite of zero is aero because z is the last letter and a is the first letter. $\endgroup$ Commented Jan 9, 2016 at 6:27
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    $\begingroup$ I don't see why this question has been downvoted. MSE is perfectly fit for elementary questions, and questions pertaining to understanding the reasons behind mathematical nomenclature seem to fall within the scope of acceptable topics. $\endgroup$ Commented Jan 9, 2016 at 6:33
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    $\begingroup$ This question has more thought put into it than posts that are just math problems without context or anything. I'd rather have this post. $\endgroup$
    – Em.
    Commented Jan 9, 2016 at 6:38
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    $\begingroup$ Downvoting does not mean someone thinks the question is off-topic (that would be close-votes). It means they think it's a poor question, for whatever reason. $\endgroup$ Commented Jan 9, 2016 at 10:29
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    $\begingroup$ if square = multiply two copies of self, then taking square root = divide by itself half of the times. $\sqrt{x}$ does equal to $x/\sqrt{x}$ for positive $x$. $\endgroup$ Commented Jan 9, 2016 at 11:06

26 Answers 26


taking square root means reversing the effect of squaring. Dividing a number by itself does not do that (but rather always returns 1 as you noted).

Compare your question to: if doubling a number means adding it to itself, shouldn't halving a number mean subtracting it from itself? Answer: obviously not.

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    $\begingroup$ Yep. This is the best answer. Upshot is, you can't push language around and expect it to be mathematically correct. +1. Would be +100 if I could do that. $\endgroup$
    – MPW
    Commented Jan 10, 2016 at 5:58
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    $\begingroup$ @MPW Thanks, though I never expected it to receive that much attention. I simply wrote it down the way I would have liked to hear it answered at some stage long time ago. Not necessarily a full, detailed explanation, but rather a hint in the right direction, with just enough kick to help me figure it out in my own terms. $\endgroup$
    – dxiv
    Commented Jan 10, 2016 at 17:18
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    $\begingroup$ @MPW you could use a bounty for the +100? $\endgroup$ Commented Jan 11, 2016 at 11:29
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    $\begingroup$ Squaring is multiplying a number by itself. The square root is really asking "What number should/did I multiply by itself to get this other number?", at least that's how I've always thought of it $\endgroup$
    – Jon Story
    Commented Jan 11, 2016 at 14:28
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    $\begingroup$ almost always returns 1. $\endgroup$
    – ClickRick
    Commented Jan 12, 2016 at 10:44

Squaring when explained in simple English, uses the word "itself". Here is an attempt to define the reverse process, finding square root, using the word "itself":

The square root of a number $N$ is that number $x$ such that when $N$ is divided by $x$ it gives itself (my grammar is poor, subject and object of this sentence. But I hope you get the drift)

Edit: this idea translated to an equation would give the following: if $N = 9$ then $x = 3$ and $N/x = 9$?? I guess itself in this context refers to $x$ and not $N$

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    $\begingroup$ (+1) Though dxiv's excellent answer concisely addresses the main mathematical issue, I think yours pins down the psychological origins of the question, namely, the use of "itself", and the two distinct roles "itself" plays when describing squaring and square-rooting. (For the same reason, some people intuitively believe increasing a number by $r$ percent then decreasing by $r$ percent "should" return the original number, and are struck when the procedure fails. The problem is resolved, of course, by consideration of "$r$ percent of what?", just as in your answer.) $\endgroup$ Commented Jan 9, 2016 at 13:15
  • $\begingroup$ @AndrewD.Hwang: pointing out the analogue in the concept of percentage will help in clearing many doubts of similar nature. $\endgroup$ Commented Jan 9, 2016 at 13:28
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    $\begingroup$ Put another way, to square, you multiply by the input. To square root, you divide by the output. Of course, output isn't something you have access to in the real world, but we're not talking about algorithms here, just definitions! $\endgroup$
    – Owen
    Commented Jan 11, 2016 at 6:17
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    $\begingroup$ I think it's more accurate to say "If we treat a number N as a square, then the square root X is the number from which the square N was formed, according to N = X^2". In your sentence, "itself" is ambiguous, and/or seems to refer to N when it should refer to X. $\endgroup$ Commented Jan 12, 2016 at 2:47

Since this question hinges directly on some fundamental ideas of math, this answer attempts to explicate those ideas in a similarly fundamental way.

Squaring a number can be thought of as a procedure. The particular procedure for squaring a number can use a template like the following:

$$ \Box \longrightarrow \Box\times\Box \longrightarrow \Box $$

We put the "input" value, for example, $2$, in the leftmost box, like this:

$$ 2 \longrightarrow \Box\times\Box \longrightarrow \Box $$

Next we make copies of the leftmost box and put them in the two boxes in the middle:

$$ 2 \longrightarrow 2 \times 2 \longrightarrow \Box $$

Notice that these two boxes must each contain the same number. Finally, we perform the indicated multiplication and write the result in the last box on the right:

$$ 2 \longrightarrow 2 \times 2 \longrightarrow 4 $$

To take a square root, we want to reverse the procedure, that is, work it backwards. So we take the "input" number, for example, $9$, and put it in the box on the right:

$$ \Box \longrightarrow \Box\times\Box \longrightarrow 9 $$

Now we have to decide what to put in the two boxes in the middle. We know we need the contents of the two boxes to be equal, and we know that when we do the multiplication the result has to be $9$. Suppose we guess the number in each box should be $3$. Then we have:

$$ \Box \longrightarrow 3\times3 \longrightarrow 9 $$

We can confirm that $3\times3$ does indeed give the result $9$, so all is good so far. Now we just need to deduce what number was in the leftmost box. We know the middle boxes were filled by copying that box, so it had to contain a $3$ as well. So we have

$$ 3 \longrightarrow 3\times3 \longrightarrow 9 $$

And that's why the square root of $9$ is $3$ rather than $9/9$. (Well, that and the fact that we refuse to put $-3$ in the two boxes in the middle, because life is better when we consistently follow a rule that says a "square root" must never be a negative number.)

We may later learn how to find square roots in a way that does not rely so much on making a lucky guess. But that's a matter of an algorithm for calculating a square root, not the definition of a square root.

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    $\begingroup$ I think leaving the arrows in the same direction when explaining square roots illustrates this better. In this case, I'm tempted to put $9 \times 9$ and obtain $81$. $\endgroup$
    – Seven
    Commented Jan 9, 2016 at 11:39
  • $\begingroup$ @Seven I wondered whether the arrows would make more sense all going in the same direction (so as to preserve the template). That makes at least three opinions in favor (yours, the upvote on your comment, and my second thoughts), so I'm making that change. $\endgroup$
    – David K
    Commented Jan 9, 2016 at 13:39
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    $\begingroup$ I suggest that it should be "$\square \to \square \times \square = \square$" otherwise it is slightly misleading. The procedure is only a single arrow, the rest are just simplification by equality. $\endgroup$
    – user21820
    Commented Jan 10, 2016 at 3:18
  • $\begingroup$ Thank you. Appreciate it. $\endgroup$ Commented Aug 24, 2022 at 12:40

An explanation with units, taking care of hidden dimensions

A square with sides of 1 meter has a 1 meter square area.

In short: taking the square of side length $l$ yields the area $l^2$. Taking the square root of something, interpreted as the area of a square $l^2$, yields the size of the side $l$.

Assume a single side of length $3$ meters (3 m). Squaring transforms the side into a geometrical square of area $3\times3$ square meters (m$^2$). The name "square" is very consistent. And units are squared too!

The reverse operation consists in, starting from a square of a certain area given in (m$^2$), finding the root, in other words the side (in m) that would have produced "this" square. If you divide the area by itself, you should divide units too, and you end up with $1$, a unit-less number, clearly not the answer (in meter).

You can better interpret this as different dimensions, combined as powers to numbers and units. $3$ meters are $3^1$ m$^1$ in one dimension. It yields a square of area $3^2$ m$^2$ in two dimensions. So, you somehow multiply "powers" by $2$ when squaring. When taking the root, you divide the power by two. A 9 square meters square is for instance $9^1$ m$^2$, so if you divide powers by two, you get the correct answer: $9^{\frac{1}{2}}$ m$^{\frac{2}{2}}$.

Even if your problem seems unit-less, consider it has having $d$-unit" per dimension, and multiply and divide powers accordingly.

Another question is: can one imagine a square with negative sides? Because its area is the same as the one with positive sides. There are two square roots for $9$: $-3$ and $3$.

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    $\begingroup$ I like this approach, but the explanation seems overly complicated. More simply: Square of a number is the area of the square shape with each side the length equal to the magnitude of that number. Square root of a number is finding the length of a side of a square with an area equal to that number. $\endgroup$
    – jxh
    Commented Jan 9, 2016 at 13:17
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    $\begingroup$ Your version is very concise and sufficient. I do not know the math level of the OP, and I wrote it somehow as I would have told, with hints. Dimensional analysis seems important to check the consistancy of a reasoning. $\endgroup$ Commented Jan 9, 2016 at 13:31
  • $\begingroup$ @jxh and now, what about negative roots ;) $\endgroup$ Commented Jan 9, 2016 at 16:45
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    $\begingroup$ Ignored, since they are imaginary. $\endgroup$
    – jxh
    Commented Jan 9, 2016 at 18:32

The name square root comes from it being a root of this equation $x^2-A=0$ .

In this form it has nothing to do with division. In fact, we don't even need to know what division is, to formulate it.

Addition: So to explicitly answer the question - no, we should not mean to divide a number by itself when taking a square root, because it won't satisfy(solve) that equation. (except for A=1, to be entirely corect)

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    $\begingroup$ @user - What are you talking about?? It EXACTLY answers the question! In fact, it is IMO the best answer here. The problem the OP had was not understanding the meaning of the word "root". Diego's is the only answer here that clearly and concisely explains what the term means. After being disappointed in not having seen it in the other answers (Slipp's covers it, but not very clearly), I would have stated this myself had Diego not already supplied it. $\endgroup$ Commented Jan 10, 2016 at 22:56

That is not a useful function because it always equals 1 (except 0/0).
Also, the opposite of $2\to4$ is $4\to2$, where you divide by the number you started with, not by 4.

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    $\begingroup$ I don't know about "best" answer (as @KonstantinosGaitanas said), but definitely a worthy answer. Others address the nature of inverse operations, but none seem to have touched on a key feature of mathematics: the ability to answer "we don't do that" with "but what if we did?" What if we did define an operation on a number that divides it by itself? Well, we could, but it wouldn't be useful. $\endgroup$ Commented Jan 13, 2016 at 11:48

Consider the following question, which is just yours with simpler operations substituted. Then you will hopefully understand.

If doubling a number means adding that number to itself, then shouldn't halving a number mean to subtract a number from itself?

And if you've got this, here is some more to contemplate. Suppose your boss proposes to raise your salary by any percentage you like, with as only condition that at the end of the year it will be lowered again by the same percentage; which percentage would you choose? The highlighted question is relevant to the choice "100%".

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    $\begingroup$ its already included in dxiv's answer $\endgroup$
    – manshu
    Commented Jan 9, 2016 at 12:56
  • $\begingroup$ @manshu I admit I overlooked that (I probably wouldn't if I had my answers sorted by votes). But since I extended my answer, I think I'll leave it for now. $\endgroup$ Commented Jan 9, 2016 at 13:01
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    $\begingroup$ The answer to the last question is clearly “1,000,000,000%" — and then retire at the end of the year. $\endgroup$
    – mattdm
    Commented Jan 10, 2016 at 15:01

If squaring a number means multiplying that number with itself then shouldn't taking square root of a number mean to divide a number by itself ?

In a certain sense, repeated multiplication $(a^n)$ and repeated division $(a^{-n})$ are indeed opposite operations to one another, pretty much for the same reason that simple multiplication and simple division are also opposites. But, from a different perspective, repeated multiplication $(a^n)$ and root extraction $\big(a^{1/n}\big)$ are also opposites. How so ? Notice that, in the first case, by multiplying the two quantities, we get $1$ as a result, which is the neutral element for multiplication; i.e., $a^n\cdot a^{-n}=1.$ In the second case, however, by composing the two operations, we get the argument a itself as a result, i.e., $(a^n)^{1/n}=\big(a^{1/n}\big)^n=a.$


Your question has many analogies:

"If adding number to itself means multiplying it by 2, then why doesn't subtracting a number from itself mean dividing it by 2?" and so on

The problem you have noticed is not uncommon at all. It is the problem of specifying the logic that looks plausible for one particular case but that cannot be generalized. Try:

"If taking third power of a number means multiplying that number with itself two times then shouldn't taking a third root of a number mean (to do what) a number by itself?"

Although your question seems logical, you cannot extend it to any higher power.

You agree that by the squaring a square root, you need to come back to the original number. Obviously if we define it your way, you will not.

Multiplication and division are not the opposite operations. $a \cdot b = \frac{a}{b} \cdot b^2$ so $a \cdot a = \frac{a}{a} \cdot a^2$

There is indeed somewhere in the expression $ .. = \frac{a}{a}... $ but all other factors are missing.

When in doubt, try to generalize and you will notice why your logic is failing.


"If taking n-th power of a number means multiplying that number with itself (n-1) times then shouldn't taking an n-th root of a number mean dividing a number by itself (n-1)-times?"

$a/a/a/a/a/a/... (n-1)-times .../a=a^{2-n}$ and $a^{2-n}$ is not n-th root of a number.

Well $a^{\frac{1}{x}}=a^{2-x}$ has a solution $x=1$ which means that your logic is valid for $x=1$ turning your question into:

"If taking first power of a number means do not multiply that number with itself at all, then shouldn't taking a first root of a number mean do not divide a number by itself at all."

For $n=1$ and for $n=1$ only it does mean that. But $x^1$ or $\sqrt[1]{x}$ are no valid examples of the power function or root function, since they do not contain their characterizations in any particular sense.

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    $\begingroup$ Just saying: Taking the third power means multiplying a number with itself twice, not three times. $\endgroup$
    – gnasher729
    Commented Jan 9, 2016 at 23:23

Another explanation that I'm not seeing in the other answers:

The square root of a number $N$ ($\mathrm{4}$ in your example) is the number that if squared ($\mathrm{2}$) would come out to $N$ ($\mathrm{4}$).

Another way of thinking about an “inverse operation” (as others have noted) is a “what-if operation”. For the square root in your example, you're asking “What if I had a number that when squared is $\mathrm{4}$? What would my number be?”

Also, the name “square root” seems to be asking “what is the root of the square?” In the form $R^2 = N$, the $R$ part is called the root of the operation and the $^2$ is the square operation itself. Asking for the square root is asking “what root ($R$), if squared, would come out to our number ($N$)?”


Being more specific while translating from prosaic languages to the math language helps.

enter image description here

I really like this question because for me it summarizes general problems of math teaching.


You have to distinguish between ordinary language and technical language. For convenience, ordinary language is mixed with technical language, but you must guard against being misled by ordinary language. In particular, the ordinary-language term “opposite” is not well-defined. For example, is land travel the opposite of sea travel? or of air travel? So also in regard to your question: division is not the only “opposite” of multiplication. Consider the distributive property: a(b + c) = ab + ac. Applied from left-to-right, the distributive property is called multiplication, but applied from right-to-left, it is called factoring. Thus FACTORING is also an “opposite” of multiplication, and, it so happens, the one that applies in this case - that is, finding the square root of a number means finding “two equal factors” for the number.

The worst (or, best) example of a disconnect between ordinary language and mathematical language concerns divisors of 0. According to the definition of a divisor, 23 is a divisor of 0, as is 37. Therefore, based on ordinary language, the real number system has divisors of 0, but according to mathematical language the real number system does NOT have divisors of 0.

Another good (bad) example of the disconnect between ordinary language and technical language is the difference in meaning between the formula for simple interest and the formula for compound interest: the formula for simple interest gives you exactly what it says, but the formula for compound interest gives you the total growth amount (so, to get the amount of compound interest, you have to subtract the principal from it). This example of the disconnect between ordinary language and technical language has the advantage of not requiring familiarity with ring-theoretic considerations.

Yet another good (bad) example of the disconnect between ordinary language and technical language is the fact that the naive reaction to hearing the phrase “the error involved in using the Trapezoidal Rule” is, “Well, if there is error involved in using the Trapezoidal Rule, then it’s pretty obvious that we shouldn’t be using it.”

Yet another good (bad) example of the disconnect between ordinary language and technical language is the defining of the notion of a “greatest” common divisor even though there is no (or, at least, not necessarily any) order relation defined (i.e., for an integral domain).

  • $\begingroup$ Interesting. According to wikipedia, there are two definitions of "divisor" (en.wikipedia.org/wiki/Divisor). Both specify that $m$ is a divisor of $n$ if there's an integer $k$ such that $n=km$, but one specifies that $m\neq 0$. Either way, (almost) every integer is a divisor of 0. On the other hand, in the ring theory sense, the reals have no "zero divisors". $\endgroup$ Commented Jan 15, 2016 at 0:14
  • $\begingroup$ comment I'm adding for tracking / lookup purposes: examples of a disconnect between ordinary language and technical language. $\endgroup$
    – Mike Jones
    Commented Mar 8, 2016 at 15:39

When you 'square' a number, you multiply a number by itself. But, when you derive the 'square root' of a number, you essentially find a number which, when 'squared', will give the number we're taking the 'square root' of.

I think geometrical analogy can help you conceptualise. Think of 'squaring' as finding the area of a square with a certain length of sides while 'taking square root' will refer to finding the length of the sides of a square with a certain area.


Lets see, we have:

  • square? ok then multiply
  • square-root? ok then divide

So far so good. Now it seems the multiplication is by the number itself. We have:

  • square? ok then use the number itself
  • square-root? ok then use ...?

It seems that in general, operations (like multiplication) and inverse operations (like division) are only valid inverses of each other when dealing with a given reoccurring value. The exact same value will have to be a given operand to both operations. Therefore, in this context where there is no such reoccurring operand the inverse relationship is no more valid.


I'll take a slightly different approach here and say simply because:

That is not how square roots were defined.

Mathematics is entirely built on definitions. That's not a square root because it by definition is not a square root.

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    $\begingroup$ This goes in the right direction, but the question that needs answering then is “Why was it not so defined?”. $\endgroup$
    – PJTraill
    Commented Jan 13, 2016 at 11:22

TL:DR? 'Root' has a specific meaning in mathematics. Taking the square root of a number means getting the 'root' of the equation that has the number on one side of the '=' sign, and a squaring operation on the other.

First, some terminology:

  • In its most simple terms, a mathematical expression is a statement that can represent a single number. eg. (5+6)x2, or (3+n)/(4+n) (where n represents some other number), etc.

  • In the latter example, the expression contains a variable, n. We say that this expression is a 'function of n', which we can write as f(n) (or g(n), or h(n), or myfunction(n), etc.). In this example we may write: f(n) := (3+n)/(4+n) (Note that := means 'is defined to be'.)

  • An equation is a statement of equality between two different mathematical expressions. eg. (5+6)x2 = 22, or (3+n)/(4+n) = 100, or 15-8=n.

If an equation contains one unknown, like the latter two examples above, then there should be one or more values that this unknown can take in order for the equation to be satisfied (i.e. for the equality to make sense). These are called the 'roots' of the eqaution.

By definition, the act of squaring a number, n, can be written as a function (let's call this 'square'). So, square(n) := n^2. If we know the answer to this, eg. 64, then we can write the equation:

 square(n) = 64,      or equivalently:      n^2 = 64.

The 'roots' of this equation are the values of n that fit. In this case, 8 and -8.

So: the SQUARE ROOT of a number X is the one and only positive ROOT, n, of the equation where X is equal to the square function applied to n:

 square(n) = X.

(We take the positive root by convention.)


Consider these two functions:

  • $s$ takes items of food as input, and puts a slice of lemon on top of each.
  • $z$ takes drinks and adds 5% lemon juice to each.

Now suppose something has gone wrong and you need to reverse the procedure. Easy enough for $s$:

  • $s^{-1}$ takes items of food as input, and takes away what's sitting on top.

This is a lot like inverting multiplication by a constant factor: you have done some modification, and know that the result has some structure (like lemon sitting on top) so you can easily undo the multiplication, namely by division.

Not so in the second example. Here, the lemon juice has already mixed into the drink by the time you try to take it away. It's obviously no good to discard a spoonful from the surface of the drink. With the square root, you have an analogous problem: by self-multiplying a variable number, you forget the information of where exactly something was multiplied. You can't recognise a given shape of lemon slice that you could take off / divide away.

Arguably, it's in fact more like addition and substraction than like multiplication and division. But those two pairs behave the same, for all that matters for this question (mathematically speaking: they both form groups).

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    $\begingroup$ The OP does not know what a function is. $\endgroup$
    – Fattie
    Commented Jan 13, 2016 at 21:27
  • $\begingroup$ @JoeBlow: it shouldn't be necessary to know what exactly a function is, to understand where I'm going with this answer. (Though, ideally, I'd hope that the answer also serves as a slight introduction to the concept of functions and invertibility.) $\endgroup$ Commented Jan 13, 2016 at 21:32

When you square a number it flowers...

As you know when you square a number it gets really big, then bigger and bigger - you can think of it as branching out.

enter image description here

Conversely when you go "downwards" towards the root of a number, you are going down "inside" it. It gets dramatically smaller and smaller.

As you know multiplication and division are really just addition and subtraction. There is none of that branching power.

Setting aside mnemonics BTW the actual answer is simply:

In old fashioned language a 'root' means a 'solution'...

that's all there is to it.

In this case it's the "solution" to the "square".

So, x2 = 9, whats the "root" or "solution" to that equation.

(It's quite incredible only one answerer above pointed this out!)

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    $\begingroup$ The branching out illustrates exponential growth (multiplying by two each time), not squaring. $\endgroup$
    – user236182
    Commented Jan 16, 2016 at 17:06
  • $\begingroup$ Hi user - not really, trees are three dimensional, or you can think of four dimensional ones. Sure, the tree icon image I chose -- purely a suggestive mnemonic pedagogical device -- may be a poor image. (Thanks for that -- I changed the tree icon image.) The fact that this is the origin of the term root, is not a poor etymological image. $\endgroup$
    – Fattie
    Commented Jan 16, 2016 at 17:11

This is the same as saying, if multiplying a number by 2 means adding it to itself, then shouldn't dividing a number by 2 mean subtracting it from itself? Same thing here.


Another way to express $\sqrt{x}$ is $x^\frac{1}{2}$.

(The 2 in the exponent comes from the fact that it's a square root. If it were a cube root, like $\sqrt[3]{x}$, it would be $x^{\frac{1}{3}}$.)

When you square a number, it's $x^2$; when you take the square root of a number, it's $x^\frac{1}{2}$. So we can see that the operations actually ARE opposites, just not in the way you think.


Assuming $x > 0$


$$ \begin{matrix} x \cdot x = x^2 & \rightarrow & \sqrt{x \cdot x} = \sqrt{x^2} & \rightarrow & \sqrt{x} \cdot \sqrt{x} = x \\ \downarrow & & & & \downarrow \\ x^2 \div x = x & \rightarrow & \sqrt{x^2 \div x} = \sqrt{x} & \rightarrow & x \div \sqrt{x} = \sqrt{x} \end{matrix} $$

Visually (expanding on dkeck's answer):

enter image description here


Another way of wording the question is:

"If the curve for $y=x^2$ intersects with the line for $y = x \cdot z$ at $x=z$, then shouldn't the curve for $y=\sqrt{x}$ intersect with the line for $y=x \div z$ at $x=z$?"

See for yourself, where $z=2$:


$y = \sqrt{x}$ intersects with $y = x \div z$ at $x = z^2$, not at $x = z$. $y = \sqrt{x}$ intersects with $y = x \div \sqrt{z}$ at $x = z$.

The question comes from the very subtle logical error of confusing the function $\lambda x \space \space x \cdot x$ with the function $\lambda a \space \space a \cdot x$. The above graph calls attention to the difference between these functions.

If squaring a number meant multiplying that number by $z$, and square root were defined as the inverse of square, then yes, taking the square root of a number would mean dividing it by $z$.


If squaring a number means multiplying it by itself, then taking the square root of a number means dividing the number by its square root.

Now as a definition that's a bit circular. Unsquaring this circle happens to not be expressible with fundamental arithmetic operations.

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    $\begingroup$ I don't see why people downvoted this. This is a perfectly accurate answer; it's essentially the idea of P Vanchinathan's answer, just stated a different way. $\endgroup$
    – Owen
    Commented Jan 11, 2016 at 6:20
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    $\begingroup$ Maybe because (in the first sentence) this answer again tries to push the language around and just this time pushing is done so that the result would match. In other words, it uses a defective logic to arrive at the right result. $\endgroup$
    – Cthulhu
    Commented Jan 11, 2016 at 15:09

What use is a term for dividing a number by itself? The result is always $1$.
We define the square root as the (positive) number, that when multiplied by itself, gives the desired number. This is a very useful expression.

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    $\begingroup$ I am not a down-voter of this answer. I think it's pretty good, except for the "feeding a troll" assumption, which I believe to be false. OP appears to be a "pretty passive" user here, so most chances are that he or she are not trying to "earn points" by posting senseless questions, but rather are genuinely puzzled by this issue. And even under the slightest chance that OP is indeed a "troll", I think that the question by itself has a point in the perspective of non-mathematician users, so it has a positive contribution. $\endgroup$ Commented Jan 9, 2016 at 10:45
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    $\begingroup$ I have removed your sentence that stated that you thought that this question was created by a troll. These things are rude and violating the Be Nice policy. $\endgroup$
    – wythagoras
    Commented Jan 9, 2016 at 15:03

As $\sqrt{n}\cdot\sqrt{n} = n$, then $\sqrt{n} = n\div\sqrt{n}$ and not $n\div n$


If you take a number, $N$ and multiply it times $x$ and then again times $x$, you would end up with $Nx^2$. To get $N$ back, you would divide by $x$ and then divide again by $x$.

The word root is generally used to represent one of the zeros of a function; that is a solution to $f(x) = 0$. In particular, a square root of N is a solution to the polynomial equation $x^2 - N = 0$. If $N$ is a non negative real number, then there is a positive square root and a negative square root of $N$. Usually we express the positive square root of $N$ as $\sqrt N$ and the negative square root of $N$ as $-\sqrt N$.


I suppose that , if you want dividing by $x$ in the definition of square root, then you can say that the square root of $N$ means finding the number, $x$, such that $\dfrac Nx = x$.

A comparison

\begin{array}{ccc} \text{action} & \text{equation} & \text{inverse} \\ \hline \text{doubling} & 2x = x + x & x = \dfrac 12(2x) \\ \text{squaring} & x^2 = x \cdot x & x = \sqrt{x^2} \\ \hline \end{array}

Notice that the inverse of doubling is not subtracting x and that the inverse of squaring is not dividing by x.


its a question of definition, in mathematics you can define whatever you want however you want, provided the definition is "well defined".

you CAN define squareroot to be dividing a number by itself, and for 0 define that to be 1 by continuity of the other cases.

but the definition which emerged from history is the inverse of squaring. With a pre-existing definition, you cannot just play around with patterns of words and arrive at truth!

even with perception, patterns are unreliable, eg optical illusions are based on our inbuilt pattern recognition being wrong!

now with school maths they do play around with patterns, eg they will say $a^2=a * a$, $a^3=a * a * a$, and thus $a^2=a^3/a$, $a^3=a^4/a$, thus $a^0=a^1/a=1$. but this argument is dubious as it presumes $a^0$ to be pre-existing, in fact we need to define something first, then derive results. This argument is just a motivation to define $a^0$ to be 1. But if you want you could define $a^0$ to be 123.458.

we just prefer definitions which conform to more patterns and which lead to simpler formulae, with fewer exceptional cases, and eg I'd personally define $0^0$ to be 1.

similar to $a^0=1$ is that a union of zero sets is defined to be the empty set, and the intersection of zero sets is the set of all sets which isn't allowed!

probably most theoretical physicists dont know this particular problem!

with something like sines, 2 definitions have emerged from history, I think the original one must have been degrees, then people realised the maths was better with radians, and today it means the latter, and we put the degree sign if we mean degrees.

maths is based on definitions, and then logical steps. science is quite different and is based on observation. Popper said we cannot prove anything in science, only disprove by showing an example flouting an idea.

there is a classic example from history of the dangers of basing maths on observation, which is that $n^2 + n + 41$ is prime for 0, 1, 2, ..., 39, if one arrived at truth by patterns, one would conclude the function is always prime, but in fact for n = 40 it is composite.

This also is a danger with AI, that some forms of AI are pattern based truth, but in maths no amount of examples or patterns makes something true! only a logical argument, and even with that we can debate about say the Axiom of choice.

One other example is $\int x^n = 1/(n+1)x^(n+1)$, so the pattern then says $\int x^(-1) = 1/(-1+1)x^(-1+1) = -1/0 * 1$. In fact the integral is the natural logarithm. Patterns are at most suggestions and hints, they arent truth.

people not trained in maths tend to come to incorrect ideas via patterns, and paradoxes historically created a lot of confusion. gradually over the eras people arrived at MO which works.

a famous paradox is

0 = 0 + 0 + 0 + ....

= (1+(-1))+(1+(-1))+....

= 1 + ((-1)+1)+((-1)+1)+... (by rebracketing

= 1 + 0 + 0 + ....

= 1

thus 0 = 1. if you do pure maths at uni, you arent allowed to rebracket without appeal to a theorem or axiom. indiscriminate bracketing is not allowed otherwise you could rebracket (8/4)/2 to become 8/(4/2) and conclude that 1 = 4 !

at uni, you are taught what rebracketings are allowable, you MUST justify a rebracketing via some accepted principle, ultimately via a theorem.

even associativity of + is in fact a theorem, you cannot just assume this as a fact!

commutativity also is a theorem, ie $a+b=b+a$, which you have to prove in steps starting with the natural numbers via the Peano axioms.

Russell's paradox is a very famous example, where people were casually using sets, presuming these to pre-exist, and Russell's paradox caused major confusion and was major effort to get around. Russell's own attempt to resolve the paradox got outdone by simpler ideas of other people.

Note that square root has the problem that $(-2)^2=2^2=4$, so squaring doesnt have a well defined inverse in general, but we usually select out the non negative of the 2 inverses. with cube roots, a nonzero number has 3 inverses, but exactly 1 real inverse.

with modulo maths, a number could have many potential square roots, eg modulo 25, 0 has square roots 0, 5, 10, 15, 20

its ultimately about having a correct frame of reference, and is a problem of philosophy.

With general complex numbers, the squareroot isnt well defined, and you'd have to define which inverse you will use. logarithms also arent well defined for complex numbers, eg $e^(2*pi*i)=1$, thus 0 and $2 * pi * i$ and $4 * pi * i$ etc are different inverses of exponentiation. in fact exponentiation itself isnt well defined, because what about $e^i$ ?

$e^i = e^{i(1+2 * pi)} = e^{i(1 + 4 * pi)} =...$

thus $e^i$ could be $e^{-(1+2*pi)}$ or it could be $e^{-(1+4*pi)}$ etc!

these are matters of enormous confusion when you first learn maths, and even a top scientist will make plenty of blunders when it comes to maths! eg most physicists think maths is part of physics, in fact physics is based on observation, and maths is an external subject based on totally different principles used as a source of models for physics observations.


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