# What exactly do the signs <=> and => mean?

ehj
I don't seem to be able to find a clear answer to this question on the internet. My maths teacher today said that (simplified): b=0 => 0=0 where you multiplied with 0 on both sides. I thought that b=0 <=> 0=0 , but can't explain why, and don't know which is correct, because I don't know, exactly, what the signs mean.

Pere Callahan
A=>B means A implies B: if A is true then B is true;

A<=>B means A implies B and B implies A.

As for your examples, what is b..??

The statement b=0 => 0=0 doesn't make much sense to me; it's true if b actually is zero and false otherwise, but what are you trying to say with this formula?

The same goes for the statement b=0 <=> 0=0.

Staff Emeritus
Gold Member
The statement

P <=> Q

means the same thing as the pair of statements

P => Q
Q => P

The statement b=0 => 0=0 doesn't make much sense to me; it's true if b actually is zero and false otherwise, but what are you trying to say with this formula?
You are incorrect; that implication is, in fact, true. The truth of P=>Q does not require P to be true; it only requires that it is impossible for P to be true and Q to be false simultaneously. If you want to assert that both P and Q were true, then you would assert "P and Q".

Pere Callahan
You're right of course, thanks.
What could me make forget about "ex falso quodlibet".

approx
Basically, A=>B means if we know that A is true then B is true, also if we know that B is false we can show that A is false. If all we know is that A is false we aren't certain about B, and if all we know is that B is true we aren't certain about A.

A <=> B means they are either both true or both false.

ehj
Is it possible to show what you're saying with an example?
For instance (sin(x))^2 + (cos(x))^2 = 1 is true, so can I write

(sin(x))^2 + (cos(x))^2 = 1 <=> 0=0

?

kenewbie
The truth of P=>Q does not require P to be true; it only requires that it is impossible for P to be true and Q to be false simultaneously. If you want to assert that both P and Q were true, then you would assert "P and Q".

With Q being "0=0", wouldn't that mean that P can be anything you want, a part from a literal assignment of "0=1" or "0 != 0"?

7+7=14 => 0=0
7+7=15 => 0=0

Wouldn't a statement like that be totally meaningless?

k

Homework Helper
Is it possible to show what you're saying with an example?
For instance (sin(x))^2 + (cos(x))^2 = 1 is true, so can I write

(sin(x))^2 + (cos(x))^2 = 1 <=> 0=0

?
Since both sides are true and anything=> True, yes, that is a trivially true statement.

Staff Emeritus
Gold Member
With Q being "0=0", wouldn't that mean that P can be anything you want, a part from a literal assignment of "0=1" or "0 != 0"?

7+7=14 => 0=0
7+7=15 => 0=0

Wouldn't a statement like that be totally meaningless?

k
The statement
0=1 => 0=0​
is, in fact, a true statement about integers. You might not think it a useful statement, but that doesn't change the fact it's a true statement.

The utility becomes more obvious when you generalize: you know that
x = y => 2x = 2y​
is a true statement, right? It's an obvious (and easily proven) statement about integers! So, it must be true if it happens that x=0 and y=1.

That's why it's a "conditional" clause -- i.e. an "if-then" statement. Suppose that we know P=>Q is true. This tells us that, if P happens to be true, then Q must also be true. But if P doesn't happen to be true, it tells us nothing.

kenewbie
Many good points Hurkyl, thank you. I get why it is true.

But, assuming the OP didn't mangle his professors point in the process of posting this, why would he put a self evident truth or axiom (depending on how you look at it) as Q? (0 = 0, or in English; A thing is equal to itself). You cannot use math to prove 0 = 0. Since math requires that a = a in the first place, that would be a circular argument.

So, I don't see the point of putting 0 = 0 in Q.

k

mrandersdk
(sin(x))^2 + (cos(x))^2 = 1 <= 0=0

never seen such an elegant proof of (sin(x))^2 + (cos(x))^2 = 1. ;)