0

Essentially, what I'm asking is what it means for two numbers to be equivalent.

Why I'm asking this:

  • If $a=b$ and $b=c$, then $a=c$. But if $a=5$, $b=\sqrt{25}$ and $c=-5$, obviously $a$ is inequal to $c$.

  • Also, is it correct to say that the square root of $25$ is equal to the fourth root of $625$?

Thanks!

Adi219
  • 109
  • 2
    You have to specify which square root, the positive or negative one. Otherwise $b$ is not well-defined. – Mathematician 42 Feb 18 '18 at 14:33
  • @Mathematician42 It's unspecified – Adi219 Feb 18 '18 at 14:34
  • 1
    If it is unspecified, then there's your problem: unspecified square roots are not well-defined, leading immediately to violations of the laws of equality. – Lee Mosher Feb 18 '18 at 14:35
  • That's the problem! (Don't get the downvote, this is a valid question) – Mathematician 42 Feb 18 '18 at 14:36
  • @Mathematician42 But how is equality defined? – Adi219 Feb 18 '18 at 14:36
  • That's a rather deep question. First of all, here we are talking about an equality of numbers. So the first question is how numbers are defined! A mathematically correct definition would take us off topic. – Mathematician 42 Feb 18 '18 at 14:39
  • @Mathematician42 Yeah, that's what I meant :) :) – Adi219 Feb 18 '18 at 14:41
  • The issue here is that you are actually using two different $b$'s since specifying $b$ as a square root of 5 is not uniquely defined. Both $b=5$ and $b'=-5$ are square roots of 25. You are not talking about only one number in that way. (Also, usually people mean positive square root when they say square root to avoid issues like this, indeed that's well-defined). – Mathematician 42 Feb 18 '18 at 14:41
  • Indeed, the definition of equality is a very deep question, and the answer is different depending on whom you talk to. In first-order logic, the notion of equality is purely syntactic: we say that two things $a$ and $b$ are equal if we can perform certain manipulations of symbols and validly obtain the string $a=b$. In constructive type theory, the equality of two objects is a type, and to prove that two things are equal is to exhibit a member of that type. But here, your problem is confusion over the definition of the function $x \mapsto \sqrt{x}$, as others have pointed out. – Patrick Stevens Feb 18 '18 at 14:44
  • Anyway, be careful with square roots, otherwise you get this:

    \begin{eqnarray} 1 &=& \sqrt{1}\ &=& \sqrt{(-1)^2}\ &=& \sqrt{-1}\sqrt{-1}\ &=& i\cdot i\ &=& i^2\ &=& -1 \end{eqnarray}

    Different issue though.

     – Mathematician 42 Feb 18 '18 at 14:46
  • 1
    @Mathematician42 I disagree strongly with "You have to specify which square root, the positive or negative one". By overwhelming convention, an unqualified $\sqrt{\cdot}$ indicates the positive one by default. – Patrick Stevens Feb 18 '18 at 14:51
  • @PatrickStevens: No, a square root of a number $a$ is a number $y$ such that $y^2=a$. The positive square root is called the principle square root of $a$. Wikipedia agrees with this as well. I agree that the symbol $\sqrt{a}$ indicates the principal square root of $a$ but that is not what a said. In fact, this is exactly what you did in your answer, you specified $\sqrt{\cdot}$ but not the definition of square root. – Mathematician 42 Feb 18 '18 at 14:55
  • @Mathematician42 I agree with everything you have said in that comment up until "that is not what a said", and indeed my answer below basically says the same thing. But $\sqrt{\cdot}$ never, ever indicates the negative square root, so it's simply wrong to assert that $\sqrt{25} = -5$. One may sometimes abuse notation to say that $\sqrt{25} = \pm 5$, but this is a bad abuse and it always requires clarification. – Patrick Stevens Feb 18 '18 at 14:57
  • @PatrickStevens: Aah, the question was edited, it said "square root of 25" first! Whew, got confused there :) – Mathematician 42 Feb 18 '18 at 14:58
  • @Mathematician42 It still does.... – Adi219 Feb 18 '18 at 14:58
  • Nono, now it says $\sqrt{25}=b$ as well, it didn't before (see the edit history). The edit of TPace changed the issue. – Mathematician 42 Feb 18 '18 at 15:00
  • @Mathematician42 My mistake, you're right :) – Adi219 Feb 18 '18 at 15:02
  • Similar confusion: https://math.stackexchange.com/q/2001412/9464 –  Feb 18 '18 at 15:09
  • @Jack Looks useful :) – Adi219 Feb 18 '18 at 15:12
  • 1
    To be short: given a positive real number $x$, "a square root of $x$", "the square root of $x$" and the notation "$\sqrt{x}$" could be three different things. –  Feb 18 '18 at 15:13
  • Let us write $\def\C{\operatorname{color}}\C(x)$ to mean the color of thing $x$. For example, $\C(\text{strawberry})$ is red. We have $\C(\text{strawberry}) = \C(\text{cherry})$ because they are both red. But that does not mean that strawberries are cherries. When you say $\sqrt{25} = \pm 5$, all it means is that $+5$ and $-5$ have the same square. That does not make them the same thing. And you could, I suppose, write $\sqrt[\text{color}]{\text{red}} = {\text{strawberry}, \text{cherry}}$, but that still doesn't mean that strawberries are cherries; it just means they are the same color. – MJD Feb 18 '18 at 16:06

3 Answers3

2

We say that $y$ is a square root of $x$ if $y^2 = x$.

We define a function $\sqrt{\cdot} : \mathbb{R}^+ \to \mathbb{R}$ ("the square root function") by $$\sqrt{x} := \text{the nonnegative number $y$ such that $y^2 = x$}$$

So you can see that $\sqrt{x}$ is a square root of $x$.

Not every square root of $4$ is equal to $\sqrt{4} = 2$. It turns out to be the case that $-\sqrt{4} = -2$ is also a square root of $4$.

When we refer to the square root of $x$, we mean $\sqrt{x}$; that is, the unique nonnegative number which squares to give $x$. When we refer to a square root of $x$, we mean any of the numbers which square to give $x$. It is a fact that there are usually two of these, and that one is the negative of the other; so in practice, we may refer to $\pm \sqrt{x}$ if we wish to identify all the square roots of a number. Only the positive one - that is, $\sqrt{x}$ - is the "principal" square root (or "the square root", or if it's really confusing from context, "the positive square root"); but both are square roots.

Patrick Stevens
  • 36,135
  • +1, but is equality defined then assuming that the roots are positive?? – Adi219 Feb 18 '18 at 14:52
  • @AdiC That's like saying "is the equality $5 \times \frac{1}{5} = 1$ defined assuming that $\frac{1}{5}$ is the multiplicative inverse of $5$?". The answer is "yes, but it's not really assuming anything; it's just the definition". – Patrick Stevens Feb 18 '18 at 14:54
1

Hint:

By definition, in the real numbers, $\sqrt{25}=+5$ (the positive number such that its square is $25$).

And the same for any root of even index.

Note that if we define:

$b=\{$a real number such that $b^2=25\}$ than $b$ can have the two different values $b=\pm5$ and we cannot write an identity as $a=b$ or $c=b$.

Emilio Novati
  • 62,675
0

If you have an operator that yields more than one solution, then obviously the solutions are going to be different (otherwise there would only be one). But this does not mean they are equal.

Your example involves the function $f(x)=\sqrt x$, with $f(25)=\pm5$. Although both $-5$ and $5$ satisfy $\sqrt{25}$, $-5$ is clearly not the same as $5$. In other words, saying that the negative square root is equal to that of the positive makes no sense.

As suggested in a comment, to avoid this confusion, we use the principal square root: $f(x)=|\sqrt x|$

This may be extended to equations outputting two or more solutions. An example of this is solving the cubic $$x^3-2x^2-x+2=(x+1)(x-1)(x-2)=0$$ Here, $-1$, $1$ and $2$ are solutions, but $-1\neq1\neq2$.

ə̷̶̸͇̘̜́̍͗̂̄︣͟
  • 27,005