P: n is divided by 2. Q: n is an even number. Here n belongs to integer numbes only. How does it follow the truth table of 'if P then Q' ? In truth table, if truth value of P is false and Q is true then How the whole statement is true.
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Your question is extremely ill-phrased. Did you mean: $P(x)$ indicates whether or not $x$ is divisible by $2$, and $Q(x)$ indicates whether or not $x$ is an even number? – barak manos Jan 22 '17 at 17:21
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In any case, I guess that what you really want to ask is how can the expression "false $\implies$ true" can be true, right? – barak manos Jan 22 '17 at 17:23
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Yes, that's what I meant. – shashikant dwivedi Jan 22 '17 at 17:36
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In this case, $P $ and $Q$ are equivalent. We have both $P\to Q$, and $Q\to P$, so, when their truth values match, $(P\to Q)\land (Q\to P)$ is true only when each conjucnts are true, or both are false. – amWhy Jan 22 '17 at 17:49
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You're welcome, shashikant! – amWhy Jan 22 '17 at 17:55
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I'm assuming that the universe of discourse is all integers (or natural numbers?)
Note that any integer divisible by $2$ is even.
$$\forall n\in \mathbb Z(P(n) \rightarrow Q(n))$$ For all integers n, if $2\mid n$, then $n$ is an even number.
Also, we have, that every $$\forall n\in \mathbb Z(Q(n)\rightarrow P(n))$$
If you use the truth-table for $P\leftrightarrow Q$, you'll find that $P$ and $Q$ are either both true, or else both false.
In the end, we see that $\forall n \in \mathbb Z( P(n) \iff Q(n))$
To simplify matters, we will discuss $P$ to mean "$2\mid n$" and $Q$ to mean "n is even."
Because $P$ and $Q$ are equivalent, by definition, $P \leftrightarrow Q$.
If you use the truth-table for $P\leftrightarrow Q$, you'll find it is true only when $P$ and $Q$ are either both true, or else both false.
amWhy
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