Confused about the difference between equivalence and implication

[torquerotates]

This is a really basic question that I don't know why I'm not getting. So from my understanding, a=b is defined as a=>b and b=>a.

So say a^2=c and a=c^(1/2)

so which implies which?

Say I start from a=c^(1/2)

I square both sides and I get a^2=c. So a=c^(1/2) => a^2=c.

But if I start from a^2=c, by taking the square root of both sides, I get a=c^(1/2) & -c^(1/2)

But since I got a=c^(1/2) among one of them, I have shown that a=c^(1/2) is still implied by a^2=c.

So a^2=c <=> a=c^(1/2)

Which is clearly not true. they are not the same b/c by taking the square root of both sides we get a=+ or -c^(1/2). Is there something wrong with my reasoning?

 

[CompuChip]

You did understand it correctly. Implication means, that something must logically follow from something else. For example: a > 1 implies that a > 0. If a > 1, you can reason logically and anyone (accepting your basic axioms and logical system, of course) could not deny the conclusion a > 0.
If two things imply each other, we call them equivalent. So if a => b and b => a we write a <=> b. I think a = b is a bad shorthand, because it makes you think of equality which it is not. The name "equivalence" is clear: the two things are either both true, or they are both false. If you know one of them, you know them both; therefore you can interchange them. For example, a > 0 is equivalent to (-a) < 0. I don't need to state them both, you can infer one from the other one.

Is there something wrong with my reasoning?

Yes, the error in your example is in this line:

But if I start from a^2=c, by taking the square root of both sides, I get a=c^(1/2) & -c^(1/2)

But since I got a=c^(1/2) among one of them, I have shown that a=c^(1/2) is still implied by a^2=c.

If you are trying to give a logically sound proof here, you should have written: a = c^(1/2) or a = - c^(1/2). It cannot be and, because they are different values and a cannot be equal to the two of them at the same time. And because it says "or" it doesn't follow that it is specifically one of the two (just that, if it isn't one, then it is the other).

 

[torquerotates]

Oh, ok. So because its a "or", it could be either so a^2=c does not => a=c^(1/2) ? And hence there is not equivalence?

 

[CompuChip]

Yep that's right.
The statement "a^2 = c" however is equivalent to the statement "a = \sqrt{c} \textbf{ or } a = -\sqrt{c}".

 

[evagelos]

This is a really basic question that I don't know why I'm not getting. So from my understanding, a=b is defined as a=>b and b=>a.

.......we can prove that...........

.....for all a,b [ a>=b & a<=b <=======> a=b]..........

so you do not have to define it

 

[CompuChip]

What do you mean by "a = b" ?
I assume it means "a if and only if b".

I don't like this notation, because it allows for statements like
1 = 2 = 3 = 4.

 

[HallsofIvy]

What do you mean by "a = b" ?
I assume it means "a if and only if b".

I don't like this notation, because it allows for statements like
1 = 2 = 3 = 4.

HOW does it allow for that? What does "1 if and only if 2" mean?

 

[CompuChip]

That's what I mean. The first and last "=" are ordinary equality symbols (if you insist on abstractness: it may be equality in first order logic, in which 1, 2, 3 and 4 are constants). The second "=" is then the equivalence symbol (meaning that the expression (1 = 2) takes the same truth value as (3 = 4)).

 

[snipez90]

If I am following the post correctly, a >= b b >= a <=> a = b is not correct since the second statement does not imply the first.

But the first statement does imply the second.

 

[CompuChip]

.......we can prove that...........

.....for all a,b [ a>=b & a<=b <=======> a=b]..........

so you do not have to define it

By the way, the notation ">=" instead of "=>" also suggests you are not talking about implication (\Rightarrow) but about inequality (\ge) here, in which case you are right.

This may get confusing.

 

[evagelos]

......thanks compuChip...........

 

[evagelos]

If I am following the post correctly, a >= b b >= a <=> a = b is not correct since the second statement does not imply the first.

But the first statement does imply the second.

Here you are:

a=b====> a>=b & a=b=====> b>=a hence a=b ====> a>=b & b>=a