# How is showing x>5 proving the conclusion x≥4?

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.

• You are asked to show $x\ge4$. Since $x>5\Rightarrow x\ge4$, you are done. You don't need $x\ge4\Rightarrow x>5$. Commented Aug 13 at 3:29
• @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 Commented Aug 13 at 3:40
• @Srini As long as $x>5$, $x\ge4$ has to be true. It doesn't matter if $x=4$ cannot be true. Commented Aug 13 at 3:42
• The greater than relation is transitive. Commented Aug 13 at 19:44

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 $$x\gt 5 ~\implies ~ x\geq 4$$