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ehj

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- Thread starter ehj
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- #1

ehj

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- #2

Pere Callahan

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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.

- #3

Hurkyl

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P <=> Q

means the same thing as the pair of statements

P => Q

Q => P

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".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?

- #4

Pere Callahan

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You're right of course, thanks.

What could me make forget about "ex falso quodlibet".

What could me make forget about "ex falso quodlibet".

- #5

approx

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

- #6

ehj

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For instance (sin(x))^2 + (cos(x))^2 = 1 is true, so can I write

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

?

- #7

kenewbie

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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

- #8

HallsofIvy

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Since both sides are true and

For instance (sin(x))^2 + (cos(x))^2 = 1 is true, so can I write

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

?

- #9

Hurkyl

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The statementWith 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

0=1 => 0=0

is, in fact, a true statement about integers. You might not think it a 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 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,

- #10

kenewbie

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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

- #11

mrandersdk

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

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