# What is wrong with my reasoning here?

1. Sep 13, 2014

### mpatryluk

I've just noticed that i have a fundamental lack of understanding in a specific aspect of math, and I can't find where my reasoning is flawed.

$$x = -5$$
$$x^2 = (-5)^2$$
$$x^2 = 25$$
$$x = 5$$

So in squaring the negative I remove it, and then when I root the number back to original magnitude, the negative has disappeared. What am I missing?

2. Sep 13, 2014

### DivergentSpectrum

when you square root a number, it can either be negative or positive because a negative times a negative is a positive, as well as a positive times a positive
(-1)(-1)=1=(1)(1)

3. Sep 13, 2014

### dkotschessaa

If you are presented with $x^2 = 25$ and asked to take the square root, your answer would be $x = \pm 5$.

If you are in some situation where you know the answer has to be positive or negative you would have to offer some clarification in writing (or explanation if you are demonstrating to someone). Sometimes, for example, in physical problems, you know you have a positive magnitude, so the negative answer might not have any physical meaning.

-Dave K

4. Sep 14, 2014

### HallsofIvy

Staff Emeritus
Strictly speaking, no. If you were asked to solve the equation $x^2= 25$ then the answer would be $x= \pm 5$. But if you are asked to take the square root of 25, the answer is 5. The square root is a function and, by definition of "function", every function, f, has to have a unique value, f(x), for any given x.

The real difficulty is that the "squaring function", $f(x)= x^2$, is NOT "one-to-one" and so does not have a true "inverse". In particular, while it is always true that $(\sqrt{x})^2= x$, it is NOT always true that $\sqrt{x^2}= x$. What is true is that $\sqrt{x^2}= |x|$.

5. Sep 21, 2014

### Staff: Mentor

What HallsOfIvy said is absolutely correct. In symbols, the three equations are the following:

$$x = 5$$
$$\Rightarrow x^2 = 5^2 = 25$$
$$\Leftrightarrow x = \pm 5$$

The squaring operation in the second equation is not one-to-one, and so not reversible, which is the reason for the "implies" one-way arrow. Another way to think of this is that the solution set of the first equation {5} is different from the solution set of the second equation, which is {5, -5}. The solution sets of the second and third equations are exactly the same, which is indication by the double-ended arrow.

The operations that are one-to-one, and hence reversible, are the arithmetic operations (add, subtract, multiply, divide - restricted to nonzero divisors), exponentiating (taking e to some power), taking the log, and so on.