Ok, so my college book is the worst book ever and I can only survive from this site and youtube. Could someone please explain the answer below? I really do not understand the answer and to me there is not sufficient information for me to get the conclusion.
Question:
Prove that if n is an integer and 3n+2 is even, then n is
even using
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.
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b) a proof by contradiction.Answer:
Suppose that $3n+2$ is even and that $n$ is odd. Since $3n+2$ is even, so is $3n$. If we add or subtract an odd number from an even number, we get an odd number, so $3n-n=2n$. But this is obviously not true.Therefore our supposition is wrong, and the proof by contradiction is complete.
Ok, so I understand I need to prove that the contradiction is wrong for the proof to be true. Hence, we try to prove that the results $3n+2$ is even and that $n$ is an odd number. But I do not understand what the process was to get there nor do I understand this line: "If we add subtract an odd number from an even number, we get an odd number, so $3n-n=2n$." Thanks in advance.
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$\begingroup$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 reach the conclusion that $2n$ is odd. But this is obviously false. $2$ times anything is even, so we have a contradiction. Hence what we started with has to be false, so $n$ is odd.
Does that make more sense? Let me know if you want me to clarify.
$\endgroup$ 3 $\begingroup$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, the result is odd (that's obvious), and since $3n-n=2n$ and $3n$ is even and $n$ is odd by hypothesis (that's $\sim Q$), $2n$ must be odd, which is a contradiction since any multiple of $2$ is even, not odd.
$\endgroup$ $\begingroup$(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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