"$ab=0$ implies that $a=0$ or $b=0$", does this mean that we can have $a=b=0$? I dont understand if or means that if $a=0$ then $b \not= 0$?

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I think the "or" in this statement is inclusive; that is, it means (logically) that at least one of $a = 0$, $b = 0$ holds. –  Robert Lewis May 9 at 5:33
In mathematical speech, we say that $p \text{ or } q$ if either of $p$ and $q$ is true; that is, the only way that $p \text{ or } q$ can be false is if both $p$ and $q$ are false. –  T. Bongers May 9 at 5:33
@Mike, they are elements of a commutative ring, but it doesnt matter..im having troubles with the language...what does "or" mean? Can you see what i mean? –  Kolo May 9 at 5:36

In math, "or" always means inclusive or. In this case, "$a = b = 0$" is valid for "$a = 0$ or $b = 0$".

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$ab=0$ $\implies$ either $a=0,b=0 \text{ or } a=b=0$

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This is not true in all rings (= the context here). For example in $\Bbb{Z}_4$ we have the counterexample $a=b=\overline{2}$. The question is about the meaning of "or". –  Jyrki Lahtonen May 9 at 5:37
@JyrkiLahtonen true but the OP is not concerned in rings, just wanted to give a simple statement. –  Shobhit May 9 at 5:39
Read the OP's last comment under the main question. Commutative rings IS the context. –  Jyrki Lahtonen May 9 at 5:40
@JyrkiLahtonen Edit the answer to your liking, i would like to see the perfect answer. Thanks. –  Shobhit May 9 at 5:41
a OR b = x

a b | x
1 1 | 1
0 1 | 1
1 0 | 1
0 0 | 0


If a is $0$ or b is $0$ or they are both $0$, the statement is true.

$a=0$ then $b \not= 0$ This is and exclusive or then the table would look like this:

a XOR b = x

a b | x
1 1 | 0
0 1 | 1
1 0 | 1
0 0 | 0

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