Trouble understanding only one way implication truth

Question

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.

Answer 1

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.

Answer 2

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 $1$.

Answer 3

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

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

Answer 4

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.

Answer 5

For example

it always true that

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

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

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

Answer 6

We have:

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

but not

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

Answer 7

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

Answer 8

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

Answer 9

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 $B$.

Answer 10

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

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.

Answer 11

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.}

Answer 12

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

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