I know how incredibly stupid this sounds, but bear with me.

Let's take any random $x$, say $3$, and any random $-x$, say $-3$. Let's plug it into $x^2$. They will both give the same result!

I know this conclusion can't be right, that because of the above, $3 = -3$. But how do we logically prove it wrong? I want to know what is logically flawed about the argument above?

I know from the instance above, we can draw 2 conclusions:

  1. $x$ is really equals to $-x$.
  2. Just because a function gives the same output for 2 separate numbers doesn't mean the 2 numbers are the same.

Can you please explain why conclusion 2 is the right one to come to?

As an extension, can you please also disprove conclusion 1 above? Why isn't the fact that both provide identical outputs when being plugged into the same function a legitimate reason to say that both inputs are the same? And when is this line of reasoning legitimate; when is it legitimate to say that because both inputs provide the same output, they are the same?

Can you explain all this as simply as possible? I'm still a beginner, and will struggle to understand any rigorous mathematical notation without explanation.

  • $\begingroup$ What exactly would be wrong? we have that $a^2=(-a)^2$ for any $a$. $\endgroup$ – user Jun 22 '18 at 9:26
  • $\begingroup$ Conclusion 1 applies if and only if your transformation ($x\mapsto x^2$ in your case) is an equivalence transformation (in other words: a bijective/one-to-one function). Squaring is not an equivalence transformation because it maps different numbers to the same value. Hence conclusion 2 must be considered. $\endgroup$ – M. Winter Jun 22 '18 at 9:38
  • $\begingroup$ To make things more explicit ask yourself the same question for the function $\mathbb R\to\mathbb R$ that is prescribed by $x\mapsto0$. Then $f(x)=0=f(3)$ for every $x\in\mathbb R$. Can we conclude that $x=3$ is true for every $x\in\mathbb R$??... $\endgroup$ – drhab Jun 22 '18 at 10:06
  • $\begingroup$ A function is characterized by the fact that no two different outputs can be "generated" by the same input, i.e. if $y_1 \ne y_2$, then it is not possible that $y_1=f(x)=y_2$. But this does not mean that two different inputs cannot generate the same output. Every human has a unique father, but this does not imply that a man can have more than one son. $\endgroup$ – Mauro ALLEGRANZA Jun 22 '18 at 14:11
  • 2
    $\begingroup$ How do you prove that cherries are not the same as fire engines? After all, the equation $$\text{color}(\text{cherry}) = \text{color}(\text{fire engine})$$ is true. $\endgroup$ – MJD Jun 22 '18 at 14:31

For any non injective function $f(x)$ we have that $a\neq b$ exist such that $f(a)=f(b)$, indeed $f(x)=x^2$ for $x\in\mathbb{R}$ is not injective and for any $a\in \mathbb{R}$ we have that $f(a)=f(-a)=a^2$.

  • $\begingroup$ I'm sorry, I'm still a beginner. You'll have to simplify this. What does a non injective function mean? And what does "∈" and "ℝ"mean? $\endgroup$ – Ethan Chan Jun 22 '18 at 9:26
  • $\begingroup$ @EthanChan A function is defined injective when $f(a)=f(b) \implies a=b$ en.wikipedia.org/wiki/Injective_function $\endgroup$ – user Jun 22 '18 at 9:27
  • $\begingroup$ @EthanChan in layman terms you can think of injectivity as one input having one and only one unique output. since for different $x$ we get the same $y$ ,$y=x^2$ is not injective $\endgroup$ – The Integrator Jun 22 '18 at 9:27
  • $\begingroup$ @EthanChan the symbol $\in$ indicate an element which belongs to a set, in this case the number a belongs to the real numbers. $\endgroup$ – user Jun 22 '18 at 9:29
  • 1
    $\begingroup$ @EthanChan It is good as an introduction and for applications, search here on MSE for some OP which suggest some good source also for a good framework on the theory. $\endgroup$ – user Jun 22 '18 at 9:42

Think of this as a case of mistaken identity. There are many functions that give the same output for two different inputs. For example, if two people have the same name, this does not mean that they are the same person.


The function that to each student assigns their student number is injective, as every student has a different number. Know the number, you can find the student.

The function that to each student assigns their cohort number is not injective, as there are more than one student in each cohort. Know the number, you only know a set of students the input lies in.

Even in the latter case we still call this a function, even though different inputs can give out the same output.


You start from $$x = 3,y=-x=-3$$ Then you try to prove that they are not the equal. This you do with a proof by contradiction. So you first assume they are equal: $$x=y$$ Then you square both. And here's the point. To square a number is not an "equivalent transformation". Meaning that some information just got lost (ie the sign).

Therefore you cannot argue that $$x^2=y^2 \Rightarrow x=y$$ Because $$\sqrt{x^2} = \pm x$$

Only bijective functions (ones that are both injective and surjective) are equivalent transformations. Just injective or just surjective is not enough.

  • $\begingroup$ just a detail. $\sqrt{x^{2}} \neq \pm x$, it is $\sqrt{x^{2}} = |x|$. you can say that the solutions to $x^{2}=a$ are $\sqrt{a}$ and $-\sqrt{a}$. $\endgroup$ – Grassy LittleRoot Jun 24 '18 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.