# True or false? $x^2\ne x\implies x\ne 1$

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.

### TL;DR:

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

#### Solutions Collecting From Web of "True or false? $x^2\ne x\implies x\ne 1$"

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.

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.”

“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.”

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.

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.