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It seems you have seen this question posed by many users. But still i want to clear my doubt Regarding Implication in Mathematical Logic which is $p \to q$
I have clearly understood these:
$1.$ If $p$ is True, $q$ is True, Then the Implication is True
$2.$ If $p$ is false, $q$ is False, Then the Implication is True
$3.$ If $p$ is True, $q$ is False, Then the Implication is False
But why If $p$ is False and $q$ is True, The Implication is treated as True.
One example I convinced myself with is:
If $x=2$, Then $x^2+1=5$
Where $p$ is the statement $x=2$ and $q$ is the statement $x^2+1=5$
Now if $p$ is false, Then still $q$ can be true, since $x=-2$ will help to make $q$ True.But there are infinite values of $x$ which makes $q$ false, hence majority of values of $x$ are making $q$ false. Hence the Implication is TRUE.
Now i started to think about a different example Viz:
If $x \gt 7$, Then $x \gt 5$
Now if $p$ is false There are infinite values of $x$ making $q$ True that is all numbers $ x \in (5 \:\: 7]$
Also $q$ is false for all $x \in (-\infty \:\: 5]$
Since density of numbers in second interval is more, The Implication Should be TRUE again.
Is this the Reason when $p$ is False and $q$ is True, Implication is considered TRUE?, since when $p$ is false the number of reasons for $q$ being true less than number of causes making $q$ false?