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.
 A: $$x=-1 \implies x^2 = 1$$
but
$$x=-1 \not\Longleftarrow x^2=1$$
because it could be that $x=1$.
A: 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 $1$. 
A: We have:
$$P \leftrightarrow Q \Rightarrow P \rightarrow Q$$
but not 
$$P \leftrightarrow Q \Leftrightarrow P \rightarrow Q$$
A: $x = 2 \implies x \ge 2$ is true.
$x \ge 2 \iff x = 2$ is false.
A: For example
it always true that
$$a\ge 2 \implies a^2\ge 4$$
but the following does not hold (eg $a=-3$)
$$a\ge 2 \iff a^2\ge 4$$
A: 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 $B$.
A: 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.
A: $$\mbox{[$(x_n)_n$ bounded sequence in $\Bbb R$] $\implies$ [$(x_n)_n$ has a convergent subsequence]}$$
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. 
A: 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.
A: $x=2$ implies that $x$ is even.
But $x$ being even does not imply $x=2$.
A: 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.
A: $x=0 \implies x\neq 1$, but
$x=0 \not \Longleftarrow x\neq 1$, because it is possible that $x=2\neq 0$.
