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This is in the context of different ways of expressing P $\Rightarrow$ Q.

Some of you have suggested the similar question but it is not the same. He is asking for an explanation to why he casually dims down the statements, I'm asking for an explanation for the difference between the two aforementioned.

Thanks.

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    @MJD It's not exactly the same his question is slightly different. Thanks though – 2017-01-06
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    The usual math log "translation" is : $A \to B$ for "$A$ only if $B$" and $B \to A$ for "$A$ if $B$". – 2017-01-06
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    @MauroALLEGRANZA , isn't that the wrong way around? Shouldn't it be $A \rightarrow B$ for $A$ if $B$, and $B \rightarrow A$ for $A$ only if $B$ – 2017-01-06
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    Assuming the formalization $A \to B$ for "if $n^2$ is even, then $n$ is even", we may say that the presence of "only" in the natural language expression reverse the "point of view": it acts as a sort of contraposition. $A \to B$ can be read as : "if A, then B" and as "B, if A". An alternative is to use "when" : "When A, then B" and "B, when A". Consider now "B, when A": if A holds [$n^2$ is even], we are guaranteed that also B holds [$n$ is even]. With "B only when A", the "point of view" is reversed : B [$n$ is even] cannot hold without A [$n^2$ is even] holding also. – 2017-01-06

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"$n$ is even if $n^2$ is even" allows both even-odd and even-even.

$$\begin{matrix}&\text{even }n\\ \text{odd }n^2&\times\\ \text{even }n^2&\times \end{matrix}$$

"$n$ is even only if $n^2$ is even" only allows even-even.

$$\begin{matrix}&\text{even }n\\ \text{odd }n^2&\\ \text{even }n^2&\times \end{matrix}$$

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    That seems to be not correct. In the first case the combination (odd n/ odd n^2) is allowed, too, but you didn't make an x at the appropriate position. Similar problems with the second diagram. – 2017-01-06
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    @miracle173 well spotted, I would edit it but I can't. – 2017-01-06
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    @miracle173: I agree with you, fixing. – 2017-01-06
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    Anoher posibilithy to display this: "$n$ is even if $n^2$ is even": $$\begin{matrix}&\text{odd } n&\text{even }n\\ \text{odd }n^2&&\times\\ \text{even }n^2&\times&\times \end{matrix}$$ "$n$ is even only if $n^2$ is even": $$\begin{matrix}&\text{odd } n&\text{even }n\\ \text{odd }n^2&\times&\\ \text{even }n^2&\times&\times \end{matrix}$$ – 2017-01-06
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"n is even if $n^2$ is even" says that when $n^2$ is even then n is even.
"n is even only if $n^2$ is even" says that if $n^2$ is not even then n is not even.

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This shows the difference between $\Rightarrow$ and $\Leftrightarrow$.

In the first case, a $n$ could be even, if $n^2$ is odd. Only, if you have the "$n^2$ is even" information, you can determine the oddity of $n$. If $n^2$ is odd, there is nothing you can say about $n$.

The second case closes that gap, by stating, that $n$ will not be even, if $n^2$ is not.

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$n$ is even if $n^2$ is even:

$$n^2\equiv0\pmod2 \implies n\equiv0\pmod2$$

$n$ is even only if $n^2$ is even:

$$n\equiv0\pmod2 \implies n^2\equiv0\pmod2$$

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Let $p := \text{"$n$ is even"}$ and $q := \text{"$n^2$ is even"}$. Then "$p$, if $q$" means that $p$ is true if $q$ is, that is the implication $q \Rightarrow p$. On the other side "$p$, only if $p$" means $p \Rightarrow q$.