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I am currently taking logic and proof techniques course.

I encountered this question, and I wonder if my approach is correct or not.

Result: Let $n ∈ \mathbb N$. Prove that if $n^3 − 5n − 10 > 0$, then $n ≥ 3$. (From Mathematical Proofs: A Transition to Advanced Mathematics Book.)

My proof:

Proof (By contrapositive): If $n<3$ then $n^3 − 5n − 10 <0$ where $n∈ \mathbb N$, the set of natural numbers that are less than $ 3$ is $\{1,2\}$

for 1: $$(1)^3 − 5(1) − 10=-1$$ for 2:$$(2)^3 − 5(2) − 10=-12$$ therefore the statement is true for all n where $n ∈ \mathbb N$ and $n<3$.

but thinking about it, if I approved that the contrapositive is true for $ n<3$, it still doesn't say anything about if its true or false for $n\leq 3$.

Parcly Taxel
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JackSymBol
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    You need not show $n\ge 3\implies n^3-5n-10>0$ , it is not an iff-statement , only an if-statement. – Peter Sep 23 '22 at 13:54
  • It does happen to be in this particular case that $n\geq 3$ does happen to imply that $n^3-5n-10>0$ but this is not what the problem was asking for. As for a hint on how to proceed there, show it is true for $n=3$ (here $3^3-5\cdot 3 - 10 = 27 -15-10=2$) and then proceed with induction. The amount you increase by each step greatly exceeds the amount you decrease by. To emphasize again though, this was not required to be proven based on the way the question was asked, and even if the particular function were such that there existed negative values for later $n$ it doesn't matter. – JMoravitz Sep 23 '22 at 14:09
  • For example, if the statement happened to be that $-0.0001\cdot n^4 + n^3-5n-10>0\implies n\geq 3$ that would still be true despite sufficiently large values of $n$ causing the expression in the inequality to be negative. Remember that there exist vacuous truths. – JMoravitz Sep 23 '22 at 14:11
  • @Peter but the originl statement is p(x)=>q(x), and ur statement is q(x) =>p(x) and they are not equivalent, and as far as I know if we wanted to start the solution from the right side we need to use the contrapostitve, to keep them equivalent. – JackSymBol Sep 23 '22 at 15:18
  • We start with the negation of the RIGHT side because $A\implies B$ can be proven by showing $not B\implies not A$. $A\implies B$ is only false , if $A$ is true and $B$ false. $not B \implies not A$ is only false , if not B is true (so $B$ false) and not A false (so $A$ true). Hence those implications are equivalent. This is why if-proofs work this way. – Peter Sep 23 '22 at 15:27

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