12

On a test I wrote an implication arrow "$\implies$" to show that I deduced one statement from the previous one, but I didn't get full score since it was more accurate to use an equivalence arrow "$\iff$". For example: $$ 2x = 4 \implies x = 2 $$ but it's also true the other way around: $$ 2x = 4 \impliedby x = 2$$ so it is more correct to write equivalence arrow: $$ 2x = 4 \iff x = 2$$ Given this i would assume that if $Q \implies P$ is true, then $Q \impliedby P$ is false.
Is this correct?


I don't want to check whether a statement only implies or is equivalent to another every time I do some operations to it.
So my second question is then: is there some other more loosely defined implication arrow that allows me to show that implication in one direction is true, without saying that implication the other direction is false? I also came across this picture, but i'm not entirely sure what the difference between those two definitions are.

Mathematics
  • 121
qerglkhqerlgkj
  • 159
  • 11
    Your teacher is just being a jerk. If you only need to use implication in a single direction to prove your claim, there is no reason to write the "equivalence arrow." The implication arrow certainly doesn't carry any information about the other direction in typical use. – David Mar 08 '17 at 22:41
  • 3
    One situation where people sometimes get the direction of the arrow wrong is when you work backward from the thing to be proved in order to find out how to prove it. That technique generally works only when the statements are all equivalent; and actually the only direction you need to actually show in the proof is the reverse direction, but people sometimes only write the forward arrow (the direction in which they found the statements). I don't see anything to indicate that you made that mistake, but it is something worth checking. – David K Mar 08 '17 at 23:42
  • 4
    The reason for the deduction might be this: If the problem asked you to solve an equation, then you need to show both directions for your claimed solutions. The direction $\Longrightarrow$ shows that you did not lose any solutions, the direction $\Longleftarrow$ shows that you did not find too many (i.e. that you did not miss a condition). So your usage of $\Longrightarrow$ is not wrong, but it might be that it is insufficient to solve the problem. – Eike Schulte Mar 11 '17 at 16:43

2 Answers2

4

As your own very example shows: just because the implication goes one way doesn't mean that it doesn't go the other way as well. In your case, it goes from left to right and from right to left, so we can write $P \Leftrightarrow Q$. But this does not mean that one of $P \Rightarrow Q$ or $Q \Rightarrow P$ is false. In fact, both would be true!

There is no commonly used symbol to say that you only have a one-way implication ... You'd have to explicitly say "$P \Rightarrow Q$ but not $Q \Rightarrow P$" ... Or $P \Rightarrow Q$ and $Q \not \Rightarrow P$

Bram28
  • 100,612
  • 6
  • 70
  • 118
1

"Given this i would assume that if Q⟹P is true, then Q⟸P is false. Is this correct?"

No that is not implied. However you may be thinking of Modus Tollens which goes:

P⇒Q, notQ hence notP

For example, if it rains (P) then it is wet outside (Q). Since it is not wet outside (notQ) it did not rain (not P).

Counterexample to the logic you questioned showing it's not correct:

Assume it is correct and apply to the rain to get:

If it rains then it is wet, hence it's not true that if were wet outside then it has rained.

That doesn't make any sense when you say it out loud, and that's not an accident. The logic is actually contradictory unless we add additional premises.

Our assumption that it was correct has been refuted.

The double arrow should be used whenever you have identified things either formally or by definition as one another.

For example:

You are a bachelor iff you are unmarried. (true by definition)

The linear transformation between two vector spaces is one to one iff that linear transformation only maps one element to 0. (requires formal proof in both directions)

Gordon Geringas
  • 11