Let $n$ be an integer. Prove that if $n^2+2n-1$ is even than $n$ is odd.
Proof: $$n^2+2n-1=2n$$ $$n^2-1=0$$ $$(n-1)(n+1)=0$$ $$n=-1,1$$
Which are odd.
Is this a complete proof? I feel like it only proves $n=-1,1$ not an odd number.
[Math] Odd and even number proofs
discrete mathematicsfoundationsproof-verificationproof-writing
Related Question
- Proof Verification – $3x + 1$ is Even if and Only if $5x-2$ is Odd
- [Math] Prove that graph with odd number of odd degree vertices does not exist
- [Math] Prove the proposition: $z$ is even if and only if $w$, $x$, and $y$ are even.
- [Math] mn is is even iff m is even or n is even
- [Math] Prove the product of an even integer and an odd integer is even
- [Math] Proving that an odd positive integer raised to a positive power results in an odd number.
Best Answer
You are making a mistake as follows: Suppose that $n^2+2n-1$ is an even number. This means it is two times some integer. But what you seem to be doing, is equating it to $2n$, which makes it a specific integer, since $n$ is fixed (once we calculate $n^2+2n-1$).
On the other hand, every even integer is given by $2k$, where $k$ is an integer.
To do this question, suppose that $n^2+2n-1 = 2k$, then transposing, $n^2-1 = 2(k-n)$. Hence, $(n-1)(n+1) = 2(k-n)$.
At this point, it is enough to see that if $n$ was even, both $n-1$ and $n+1$ are odd.
The right hand side is even (divisible by $2$), whereas the left hand side, being a product of odd numbers, is odd. This is a contradiction.
Hence, $n$ must be odd.