When proving implications, I always thought to prove it true I need to reach the same conclusion given in the implication. I encountered this problem (pg 56) that asked to prove
"for all $x\in\mathbb R$, if $x^2 - 7x + 10 \ge 0$ and $x > 3$, then $x \ge 4$." To which the proof ended when it was able to show $x > 5$. I understand if $x > 5$ then it must be greater or equal to $4$, but $x > 5$ doesn't cover the case when $x=4$, so they don't seem logically equivalent to me.
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7$\begingroup$ You are asked to show $x\ge4$. Since $x>5\Rightarrow x\ge4$, you are done. You don't need $x\ge4\Rightarrow x>5$. $\endgroup$– LucenapositionCommented Aug 13 at 3:29
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$\begingroup$ @Lucenaposition, I read your first implication statement in my mind in English as "x is greater than 5 implies that x is greater than OR equal to 4". Is that right? If so, I wonder why shouldn't it actually be "x is greater than 5 implies that x is greater than AND NOT equal to 4". Sorry if it's a dumb argument $\endgroup$– SriniCommented Aug 13 at 3:40
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$\begingroup$ @Srini As long as $x>5$, $x\ge4$ has to be true. It doesn't matter if $x=4$ cannot be true. $\endgroup$– LucenapositionCommented Aug 13 at 3:42
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$\begingroup$ The greater than relation is transitive. $\endgroup$– RyRy the Fly GuyCommented Aug 13 at 19:44
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How is showing $x\gt 5$ proving the conclusion $x\geq 4$?
Suppose $x\gt 5$
We have $x\gt 5$ and $5\gt 4$
By the transitivity of '$\gt$', we have $x\gt 4$.
Since $x\gt4$, we also have, as required, $x\geq 4$.
We conclude that $x\gt 5 ~\implies ~ x\geq 4$
answered Aug 13 at 16:14