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Remember that: $(a\implies b) \equiv (\neg a \lor b)$ Hence: $\therefore \neg (a\implies b) \equiv (a \land \neg b)$ The counter example of the implication is an event where $a$ happens and not $b$. ${\begin{array}{|c|c|c|c|}\hline a & b & a\to b & a \land \neg b \\ \hline\hline F & F & T & F \\\hline F & T & T & F \\ ...


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You can disprove the statement $a \implies \sim b$ using a counterexample if you can find one. However, this does not prove the statement $a \implies b$. This is because the negation of $a \implies \sim b$ isn't $a \implies b$. Recall that the negation of an implication is not an implication. $\sim (a \implies b) \equiv a \wedge \sim b$. In words, this ...


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Instead of having both $A$, $B$, $a$ and $b$, let's make some of them $p$ and $q$. Also, since you're talking about counterexamples, there must be a quantifier somewhere, so I guess you actually mean something like $$ A \equiv \forall x\,(p(x)\to q(x)) $$ $$ B \equiv \forall x\,(p(x)\to \neg q(x)) $$ Having a counterexample to $A$ means that we have a ...


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If $|a|>1$ then $$\quad0<\frac{1}{|a|}\text{ and } \frac{1}{|a|}<1.$$ So, $\frac{1}{|a|}$ is a number between $0$ and $1$. Since there is no integer between $0$ and $1$, we conclude that $\frac{1}{|a|}\notin\mathbb Z$. EDIT: Proof by contradiction, as you want: Suppose $\frac{1}{|a|}$ is an integer. Since $|a|>1$, we conclude that ...


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Suppose $\frac{1}{a} = b$ an integer. I multiply both sides of this equation by a, giving $ab = 1.$ This implies $a = b = \pm 1$ since the only integers who multiply out to $1$ are $\pm 1$ (since they are the units in the ring of integers) . But $|a|>1$. This is a contradiction.



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