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A trivial answer is to set $X = \neg r$. In fact, the proposition $(r \implies X) \land (\neg r \implies \neg X)$ has the same truth value of $(r \iff X)$. So, the whole proposition becomes $$(p \iff \neg q) \land (r \iff X)$$ and, by setting $X=\neg r$, you obtain an easy contradiction $r \iff \neg r$ that makes the proposition a contradiction.

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Consider any point $x$ on real line. An open sphere centered at $x$ with radius $r$ is an open interval of radius $2r$. Let the open interval be $(a,b)$. Let the binary representations be $a=a_1,a_2,\cdots, a_n, \cdots$ $b=b_1,b_2,\cdots, b_n, \cdots$ For every such numbers we can find an intermediate binary representation having finite digits. This ...

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There might be simpler ways to prove this, but the only way I know how to do this is via the following theorem (which makes use of the pigeonhole principle): Theorem. Now show that given any positive integer $Q$ we may find positive integers $p, q$ with $1\leq q\leq Q$ and such that $|x-p/q|\leq 1/(qQ)\leq 1/q^2$. Proof. Let $Q$ be any positive integer. ...

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Try to show that for every real number $a$ and every positive integer $N$ there exist $p,q$ integers with $1 \le q \le N$ such that $|qa - p| \le 1/(N+1)$. This can be shown just using the pigeonhole principle in a clever way. The result is called Dirchlet's approximation theorem. In this way you could easily find more detailed information in case it is ...

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(i) A proof my contradiction works by supposing what is to be proven is false and reaching a contradiction by that assumption. (ii) $$\underbrace{(2m+1)}_{\text{odd}} - \underbrace{(2n)}_{\text{even}} = \underbrace{(2(m -n) + 1)}_{\text{odd}}$$

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The basic idea with a proof by contradiction is to start with something false: in this case assuming that $3n+2$ is even and $n$ is odd. Then we do only logically sound operations to what we start with. If you subtract $2$ from an even number, then the result is even, right? And if you subtract an odd number from an even number, you get an odd number. So we ...

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The result you want to prove is of the form $P\to Q$. The proof by contradiction consists in getting a contradiction from $P$ and $\sim Q$. The proof given is correct. When they say "If we add subtract an odd number from an even number, we get an odd number, so $3n-n=2n$", they mean that in general if you add or subtract an odd number from an even number, ...

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"State whether or not the equation is true for all values of the variable." Equation or inequality? If equation, the answer is usually no I think... e.g. $\frac{1}{x+5} = 2$ Domain: $\mathbb{R}$ \ $(-5)$ Solution Set: $-9/2$ SS $\neq$ Domain. Hence, no. Maybe you could justify the set inequality by stating some elements in one set that are not in the ...

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Yes, it works. The "proof" hinges on the premise that $\bigcup(F\cap G)$ is non-empty. Your added condition implies that $\bigcup(F\cap G)\neq\varnothing$, so the proof works out.

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