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, 2018 at 20:57
• If the moon is made of cheese, then what @dxiv said is true. Feb 6, 2018 at 15:46
• @dxiv Yes, everything, be it true or false, follows from a falsehood (assuming the law of the excluded middle holds). Feb 6, 2018 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. Feb 6, 2018 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. Feb 7, 2018 at 15:26

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 Feb 7, 2018 at 14:13

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.

$x=2$ 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, 2018 at 6:36
• @hvd ops of course I had in mind b>0!
– user
Feb 6, 2018 at 6:44
• why do I feel like this is copy of the very first answer? Feb 7, 2018 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, 2018 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 $x=2\neq 0$.