Intuition behind Vacuous proofs
My book says we can quickly prove the conditional statement $P \implies Q $ when we know $P $ is false. This much I'm fine with as I can show it with a truth table. But then I'm asked to, using vacuous proofs, prove $P (0)$ is true, given if $n > 1$, then $n^{2}$ > $n $. How can $P (0)$ be true? 0 is not greater than 1. Or is there some other piece to vacuous proofs that I'm overlooking?
$\endgroup$ 32 Answers
$\begingroup$It looks like $P(n)$ is supposed to be the statement:
$n>1\implies n^2>n$.
When $n\le1,$ we have that $n>1$ is false. So, for example, $P(0)$ is true.
$\endgroup$ 2 $\begingroup$I'm not sure what your instructor wants.
"Whenever $0 > 1$ then $0^2 > 0$"
Well, this is vacuously true because $0$ is never greater than one.
Or maybe s/he wants this "Prove $P(n) = (n > 1 \implies n^2 > n)$". In which case, you'd say "If $n \le 1$, this is vacuously true and we don't need to show anything; if n > 1 then ...
Or, least likely, s/he wants.
If $0 > 1$ then $0^2 = 0*0 > 1*0 = 0$. Which would be true if 0 > 1. But I think s/he just wants the second interpretation.
$\endgroup$
