# Trouble understanding only one way implication truth

I'm having a trouble understanding the concept. Can you give me a math example where $P\Rightarrow Q$ Is true but $P\Leftrightarrow Q$ is false? Thank you.

• $\,\big(1 = -1\big) \implies \big(2=2\big)\;$ is true, but the converse is false. – dxiv Feb 5 '18 at 20:57
• If the moon is made of cheese, then what @dxiv said is true. – Shufflepants Feb 6 '18 at 15:46
• @dxiv Yes, everything, be it true or false, follows from a falsehood (assuming the law of the excluded middle holds). – Dan Christensen Feb 6 '18 at 16:13
• Protip: The "official" name for this fallacy is "affirming the consequent." Google that and you'll get loads of examples, many of which have relatively little to do with math. – Kevin Feb 6 '18 at 22:26
• "If it is Tuesday then I will go to the gym." This is very different from "I only go the the gym on Tuesday". The important thing to remember in x => y is that it really only works when x is true. If x is not true then the statement doesn't apply so who knows whether y is true or not. – nurdyguy Feb 7 '18 at 15:26

## 12 Answers

Set theory example: let be $A\subset B$ with $A\ne B$. $$x\in A\implies x\in B$$ but $$x\in B\kern.6em\not\kern -.6em \implies x\in A.$$

Almost all the examples of the other answers are particular cases of this.

For real numbers:

$$x > 1$$

implies that $$x^2 > 1$$

But $x^2 > 1$ does not imply that $x > 1$. For instance, $(-2)^2 = 4 > 1$, but $-2$ is not greater than $1$.

$x = 2 \implies x \ge 2$ is true.

$x \ge 2 \iff x = 2$ is false.

An easy one: all squares are rectangles, but not every rectangle is a square.

The fallacy of the converse is something lots of students are guilty of. Remember, to say P implies Q means that if you have P, you have Q. The presence of Q doesn't mean P holds.

Another example: differentiability implies continuity. Every differentiable function is continuous on its domain's interior, but a continuous function need not have a derivative anywhere.

• Great answer. Just like Martín's set answer but with great examples – Skeleton Bow Feb 7 '18 at 14:13

$x=2$ implies that $x$ is even. But $x$ being even does not imply $x=2$.

$$x=-1 \implies x^2 = 1$$ but $$x=-1 \not\Longleftarrow x^2=1$$ because it could be that $x=1$.

We have:

$$P \leftrightarrow Q \Rightarrow P \rightarrow Q$$

but not

$$P \leftrightarrow Q \Leftrightarrow P \rightarrow Q$$

For example

it always true that

$$a\ge 2 \implies a^2\ge 4$$

but the following does not hold (eg $a=-3$)

$$a\ge 2 \iff a^2\ge 4$$

• No, this is not always true. $a = 1$, $b = -2$ means $a \ge b$ but $a^2 < b^2$. – hvd Feb 6 '18 at 6:36
• @hvd ops of course I had in mind b>0! – user Feb 6 '18 at 6:44
• why do I feel like this is copy of the very first answer? – Guy Fsone Feb 7 '18 at 8:30
• Maybe because you are wrong? We answered all togheter, please avoid useless considerations and think more to your own behavior here. – user Feb 7 '18 at 8:34

Imagine the following:

\begin{align} a&=1\\ \Leftrightarrow a^2&=a\\ \Leftrightarrow a^2-1&=a-1\\ \Leftrightarrow (a-1)(a+1)&=a-1\\ \Leftrightarrow a+1&=1\\ \Leftrightarrow a&=0 \end{align} Where's the mistake?

In general, $A\Rightarrow B$ means that, in order for $A$ to be true, it is neccessary that $B$ is true, but, nothing is neccessary in order for $A$ to be false.

On the other hand, $A\Leftrightarrow B$ means that, in order for $A$ to be true it is neccessary and sufficient the $B$ is true. So, when $A$ i true so does $B$ and when $A$ is false, so does $B$.

$$\mbox{[(x_n)_n bounded sequence in \Bbb R] \implies [(x_n)_n has a convergent subsequence]}$$

but the converse is failed just take $$u_{2n}= 2, ~~~and ~~~u_{2n+1} = n^2$$

the subsequence $(u_{2n})_n$ converges but $(u_{n})_n$ is unbounded.

In our beginner math classes we often used real world examples to grasp such concepts before starting with mathematical explanations. For this we used the following example:

$$snow \Rightarrow cold\\ cold \nLeftrightarrow snow$$

So when it's snowing outside it is definitely cold*. But if it's cold outside, you can not be sure that it is snowing.

* There are theoretically some exceptions to this statement, but for the sake of understanding the concept, they can be ignored.

$x=0 \implies x\neq 1$, but

$x=0 \not \Longleftarrow x\neq 1$, because it is possible that $x=2\neq 0$.