To prove that if n is an integer and 2n+1 is odd, then n must be even, we will use proof by contradiction.
Assume that n is an odd integer. Then we can write n as 2k+1, where k is an integer. Substituting this value of n in the expression 2n+1, we get:
2n+1 = 2(2k+1)+1
= 4k+2+1
= 4k+3
Now, we know that an odd number can be written in the form 2m+1, where m is an integer. Therefore, we can write 4k+3 as 2(2k+1)+1, which means that 4k+3 is also an odd number.
But we have assumed that 2n+1 is odd. Therefore, we have:
2n+1 = 4k+3
Subtracting 1 from both sides, we get:
2n = 4k+2
Dividing both sides by 2, we get:
n = 2k+1
But this contradicts our assumption that n is odd. Therefore, our initial assumption that n is odd must be false.
Hence, if 2n+1 is odd, then n must be even.