Clarification on difference between the conditional and conjunction.

Question

My teacher told us that intuitively, beginners often find themselves thinking that → is the assertion that p and q are both true (that → is really just ∧ ). He showed us the following question to test whether we were actually understanding the conditional.

Test yourself: can you find a real number so that >1→<0 becomes a true statement?

He then told us that if we answered the question on the previous page negatively, we were probably looking for an that is simultaneously greater than 1 and less than 0, and that there is no such number. That we were effectively misunderstanding > 1 → < 0 as > 1 ∧ < 0.

I know that → mean if is true, then must be true, but I am still confused and I don't understand how that would apply in this situation. Can someone explain what the conditional means in this situation?

An answer I wrote a little earlier today, https://math.stackexchange.com/a/3806654/123905 , might be of use to you. — Eric Towers, Aug 28 '20 at 23:13
"can you find a real number so that >1→<0 becomes a true statement?" Let $x = -1$. The $x>1$ is false, but $x\lt 0$ is true. Therefore the implication is true, whereas $x>1 \land x\lt 0$ is false. Review the truth table for $p \rightarrow q$. It is true, unless $p$ is true, and $q$ is false. — amWhy, Aug 28 '20 at 23:20

score 1 · Accepted Answer · answered Aug 29 '20 at 00:43

The conjunction of two propositions is true only when both propositions are true. In other words, if both propositions are true, then the conjunction is true. In any other case, the conjunction is false.

The conditional of two propositions is true if the first proposition is true, then the second proposition is true also. It also holds true when the first proposition is false. Another way of understanding the conditional of two propositions 'p' and 'q' is to regard it as meaning the same as "not p or q". Thus, if 'p' is false, since "not p" is true, "not p or q" is true. And if 'p' is true, and 'q' is true also, then "not p or q" is true, since 'q' is true.

Thus, for understanding (( > 1) → ( < 0)), it will hold true whenever x <= 1, since then (x>1) is false, and an implication with a false first proposition or "antecedent" is true. The conjunction of the two propositions though is always false, since no real number is both greater than 1 and less than 0.

This really cleared things up for me, thank you! – Martin Aug 31 '20 at 08:16 — Martin, Aug 31 '20 at 08:16
@Martin You're welcome! – Doug Spoonwood Aug 31 '20 at 13:34 — Doug Spoonwood, Aug 31 '20 at 13:34

score 0 · Answer 2 · answered Aug 28 '20 at 23:25

Let $ p $ and $ q $ be two propositions such that

if $ p $ is true then $ q $ cannot be false, then in this case, we can write

$$p \implies q$$

we also say that the argument $$p$$ $$-----$$ $$q$$ is VALID.

but in the case where $ p $ is true and $ q $ not necessarily true, the implication is false and we write $$p \notimplies q$$

this is your case, with $$ p : x>1$$ and $$q : x<0$$

Clarification on difference between the conditional and conjunction.

2 Answers2