Today I had an argument with my math teacher at school. We were answering some simple True/False questions and one of the questions was the following:

$$x^2\ne x\implies x\ne 1$$

I immediately answered true, but for some reason, everyone (including my classmates and math teacher) is disagreeing with me. According to them, when $x^2$ is not equal to $x$, $x$ also can't be $0$ and because $0$ isn't excluded as a possible value of $x$, the sentence is false. After hours, I am still unable to understand this ridiculously simple implication. I can't believe I'm stuck with something so simple.

Why I think the logical sentence above is true:
My understanding of the implication symbol $\implies$ is the following: If the left part is true, then the right part must be also true. If the left part is false, then nothing is said about the right part. In the right part of this specific implication nothing is said about whether $x$ can be $0$. Maybe $x$ can't be $-\pi i$ too, but as I see it, it doesn't really matter, as long as $x \ne 1$ holds. And it always holds when $x^2 \ne x$, therefore the sentence is true.


$x^2 \ne x \implies x \ne 1$: Is this sentence true or false, and why?

Sorry for bothering such an amazing community with such a simple question, but I had to ask someone.

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    $\begingroup$ This is true, as the contrapositive ($x = 1$ -> $x^2=x$) is obviously true. $\endgroup$ Commented Dec 5, 2011 at 14:20
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    $\begingroup$ They are wrong-the fact that $x\neq 0$ is also an implication doesn't mean anything. The statement: $x^2=x\implies x=1$ is false. $\endgroup$ Commented Dec 5, 2011 at 14:21
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    $\begingroup$ Also, their reasoning about "not excluding other values" is wrong. Today is Monday, which implies SO many things (!), but it is not false to just list one implication. Eg: If today is Monday, then tomorrow is Tuesday. Or, if today is Monday, then I have an appointment with the dentist. $\endgroup$ Commented Dec 5, 2011 at 14:23
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    $\begingroup$ @Chris: You could write your reasoning as answer. That way the site keeps working better (and you will get some upvotes). You could also refer your teacher to this site :-) $\endgroup$ Commented Dec 5, 2011 at 15:06
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    $\begingroup$ Your understanding of material implication ($\implies$) is exactly right, and as Jyrki said, your reasoning could stand as a perfectly good answer to the question. You could also point out that the teacher and class seem to be confusing $\implies$ and $\iff$: $x^2\ne x\iff x\ne 1$ is of course false precisely because $1$ isn't the only number that is its own square. $\endgroup$ Commented Dec 5, 2011 at 15:34

5 Answers 5


The short answer is: Yes, it is true, because the contrapositive just expresses the fact that $1^2=1$.

But in controversial discussions of these issues, it is often (but not always) a good idea to try out non-mathematical examples:

"If a nuclear bomb drops on the school building, you die."

"Hey, but you die, too."

"That doesn't help you much, though, so it is still true that you die."

"Oh no, if the supermarket is not open, I cannot buy chocolate chips cookies."

"You cannot buy milk and bread, either!"

"Yes, but I prefer to concentrate on the major consequences."

"If you sign this contract, you get a free pen."

"Hey, you didn't tell me that you get all my money."

"You didn't ask."

Non-mathematical examples also explain the psychology behind your teacher's and classmates' thinking. In real-life, the choice of consequences is usually a loaded message and can amount to a lie by omission. So, there is this lingering suspicion that the original statement suppresses information on 0 on purpose.

I suggest that you learn about some nonintuitive probability results and make bets with your teacher.

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    $\begingroup$ I like how the importance of cookies appears in surprising math/logic results! (+1). It might be time to show the teacher this question. $\endgroup$ Commented Dec 5, 2011 at 16:10
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    $\begingroup$ Excellent examples, thank you! And of course I already made bets :). $\endgroup$
    – Chris
    Commented Dec 5, 2011 at 16:29
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    $\begingroup$ Superb answer, which implies that I have up-voted it. $\endgroup$ Commented Jan 10, 2012 at 0:37

First, some general remarks about logical implications/conditional statements.

  1. As you know, $P \rightarrow Q$ is true when $P$ is false, or when $Q$ is true.

  2. As mentioned in the comments, the contrapositive of the implication $P \rightarrow Q$, written $\lnot Q \rightarrow \lnot P$, is logically equivalent to the implication.

  3. It is possible to write implications with merely the "or" operator. Namely, $P \rightarrow Q$ is equivalent to $\lnot P\text{ or }Q$, or in symbols, $\lnot P\lor Q$.

Now we can look at your specific case, using the above approaches.

  1. If $P$ is false, ie if $x^2 \neq x$ is false (so $x^2 = x$ ), then the statement is true, so we assume that $P$ is true. So, as a statement, $x^2 = x$ is false. Your teacher and classmates are rightly convinced that $x^2 = x$ is equivalent to ($x = 1$ or $x =0\;$), and we will use this here. If $P$ is true, then ($x=1\text{ or }x =0\;$) is false. In other words, ($x=1$) AND ($x=0\;$) are both false. I.e., ($x \neq 1$) and ($x \neq 0\;$) are true. I.e., if $P$, then $Q$.
  2. The contrapositive is $x = 1 \rightarrow x^2 = x$. True.
  3. We use the "sufficiency of or" to write our conditional as: $$\lnot(x^2 \neq x)\lor x \neq 1\;.$$ That is, $x^2 = x$ or $x \neq 1$, which is $$(x = 1\text{ or }x =0)\text{ or }x \neq 1,$$ which is $$(x = 1\text{ or }x \neq 1)\text{ or }x = 0\;,$$ which is $$(\text{TRUE})\text{ or }x = 0\;,$$ which is true.
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    $\begingroup$ Please pardon the (lack of) formatting- I typed this up on my phone! $\endgroup$ Commented Dec 5, 2011 at 15:36
  • $\begingroup$ Thanks, Brian! That looks great. $\endgroup$ Commented Dec 5, 2011 at 15:55
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    $\begingroup$ That's the proof I was looking for, thank you :) $\endgroup$
    – Chris
    Commented Dec 5, 2011 at 16:41

Thing to note. This is called logical implication.

$x^2≠x⟹x≠1$: Is this sentence true or false, and why?

We can always check that using an example. Let us look at this implication as $\rm P\implies Q$. Now we shall consider cases:

  • Case 1: If we consider $x = 0$, then $\rm P$ is false, and $\rm Q$ is true.
  • Case 2: If we consider $x = 1$, then $\rm P$ is false, and $\rm Q$ is false as well.
  • Case 3: If we consider each value except $x=0$ and $x = 1$, then both $\rm P$ and $\rm Q$ will be true since $x^2 = x \iff x^2 - x = 0 \iff x(x - 1) = 0$ which means that $x=0$ and $x = 1$ are the only possibilities.

Fortunately, our truth tables tell us that logical implication will hold true as far as we are not having $\rm P$ true and $\rm Q$ false. Look at the cases above; none of them has $\rm P$ true and $Q$ false. Thus Case 1, Case 2 and Case 3 are all true according to mathematical logic, so $\rm P\implies Q$ is true, or in other words: $x^2 \ne x \implies x \ne 1$ is true.

I apologize for being late, but I always have my two cents to offer... thank you.



You asked whether the following statement is true or not:

$x^2 \ne x \implies x \ne 1$

Another way to write the above is as:

if $(x^2 \ne x)$ then $(x \ne 1)$

The arrow symbol ($\implies$) is similar, but not quite the same, as the English phrase if....then....

The contra-positive of $(P \implies Q)$

Okay now: suppose that you have the statement of the form "If $P$ then $Q$"
$P$ and $Q$ can be any true or false statements of your choice.
For example, $P$ could be the statement "pillows are soft"
"If $P$ then $Q$" is also sometimes written as $(P \implies Q)$

There is a well known theorem in logic which states the following:

For any $P$ and $Q$ taken from the set $\{$ true, false $\}$, the following is true:
$(\text{not } P \implies \text{not } Q)$ if and only if $(Q \implies P)$

That means that the following two statements are equivalent:

if $(x^2 \ne x)$ then $(x \ne 1)$
if $(x = 1)$ then $(x^2 = x)$

In some sense, the statement says that $1^2 = 1$, which is true enough.


It considered to be very bad math to write things like the following:

Is it true or false that $(x^2 \ne x \implies x \ne 1)$?

Find $x$ such that $3 + (x-5)^{2} = 0$

This is because the above examples fail to explain what $x$ is.

  • Is $x$ a whole number, such as $1, 2, 3, \dots, 98, 99, 100$?
  • Is $x$ a decimal number, such as $\sqrt{2}$
  • Is $x$ a non-real complex number, such as $20 + 5*i$?

You are supposed to write things like this instead:

Is the following true or false?

  • For every real number $x, (x^2 \ne x \implies x \ne 1)$

Find every complex non-real number $x$ such that:

  • $3 + (x-5)^{2} = 0$

My last example uses the non-real number $i$
$i*i = -1$

I think if you think it will help if you start using "sets."

The following is an example of a set:

my_set $= \{1, 3, 6, 7, 22\}$

A set is like a suitcase full of clothing A set is also like a cookie jar, or cardboard box.
A set is a container.
A suitcase might contain a t-shirt. Well, my_set contains the numbers $1, 3, 6, 7,$ and $22$.


The number $3$ is like a t-shirt in the sense that the number $3$ is inside the suitcase.

I went to public school in the United States.
I did not see sets until I was in college.

Sets are basic, basic math. sets are more basic than knowing how to compute $4.5/0.3$

You are not allowed to write math like $x^2 \ne x \implies x \ne 1$ unless you first tell the reader what set $x$ comes from.

The following is a very ugly formula for a function named $WEIRD$:

$\text{WEIRD_FUNC}(X) = [1-w(x)]*\text{LEFT_PIECE}(X) + [w(x)]*\text{RIGHT_PIECE}(X)$

$\text{LEFT_PIECE}(X) = (x+10)*(x+5)$

$\text{RIGHT_PIECE}(X) = 3 + (x-7)^{2}$

$W(x) = \frac{tanh(10)}{2} + \frac{tanh(x)}{2}$

A plot of the weird function is shown below:


Suppose I asked you,

"Find all $x$ such that $WEIRD(x) = 0$"

The answers vary depending on which set $x$ is taken from.

The set of all integers $x$ such that $WEIRD(X) = 0$ is $\{-10\}$
The set of all real numbers $x$ such that $WEIRD(X) = 0$ is approximately $\{-10, -5.01\}$
The set of all complex numbers $x$ such that $WEIRD(X) = 0$ is approximately $\{-10, -5.01, 7+i*\sqrt{3}, 7-i*\sqrt{3}\}$

Analyzing your classmates argument regarding the number zero

Is the following statement true or not?

$x^2 \ne x \implies x \ne 1$

The following is your description of your teacher, and/or classmates, reasoning:

  1. When $x^2$ is not equal to $x$, $x$ also can't be $0$.
  2. $0$ is not excluded as a possible value of $x$
  3. some sentence, or another, is false.

That description is very difficult to understand.
I will say that the following three statements are equivalent to each-other.
Also, all statements are false:

$\forall x \in \mathbb{R}, x^2 \ne x$ for any decimal number $x$, $x^2 \ne x$ if $x$ is a real number, then $x^2 \ne x$

Here is a proof:

Line No. statement justification
0 $0$ exists axiom
1 $0$ is a decimal number. axiom
2 $0^{2} = 0$ axiom
3 not $(0^2 \neq 0)$ from line 2
4 there exists a decimal number $x$ such that not $(x^2 \neq x)$ from lines 0,1,3
5 not for every decimal number $x$, $(x^2 \neq x)$ from line 4

Notice that your teacher said, "possible value of $x$"

In mathematics, the symbol is sometimes used as short-hand notation for the word "possible"

All of the following are logically equivalent:

  • it is not possible for me to see a movie this weekend
  • $NOT$ for me to see a movie this weekend
  • It is necessary for me to NOT see a movie this weekend
  • for me to NOT see a movie this weekend

The statement "$0$ is not excluded as a possible value of $x$" can be written as:

  • not(not $x = 0$)
  • It is not the case that it is not possible that $x$ is zero.

The above is equivalent to "it is possible that $x = 0$"

I think that I can write a proof outline of what your professor and/or classmates were trying to say

Line No. statement justification
$0$ $x = 0$ axiom
$1$ $(x^2 \ne x) \implies$ not $x = 0$ axiom
$2$ $x^2 \ne x$ axiom
$3$ not $x = 0$ 1,2 modus ponens
4 $x = 0$ and not $x = 0$ lines 0, 3 conjunction
5 $NOT$ $(x^2 \ne x)$ from line 4, reject 2

I think that your teacher was saying that if $x = 0$ is possible, then it is not necessary that $x^2 \ne x$

$0 \in S \implies NOT(\forall x \in S, x^2 \ne x)$

One last Note

All of the following are equivalent:

  • (it is necessary that $x^2 \neq x$) implies that (it is necessary that $x \neq 1$)
  • (it is NOT necessary that $x \neq 1$) implies that (it is NOT necessary that $x^2 \neq x$)
  • (it is possible that $x = 1$) implies that (it is possible that $x^2 = x$)
  • For any set $S$, [(there exists $x$ in $S$ such that $x = 1$) implies that (there exists $x$ in $S$ such that $x^2 = x$)]

If you'd like a more informal understanding of why the statement given is indeed true, consider the argument of the proposition. It's purely stating that if $x^2 \neq x$, then $x \neq 1.$
This statement is true. It's not arguing that 1 is the only element in the domain of $x$ for which the statement is true.


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