claim: if n is even then n^2+3 is odd

Question

proof: suppose n is within natural numbers is even
then, by definition of even numbers, n=2k for some k included in integers
now by substitution and simplification
n^2+3=(2k)^2+3
n^2+3=4k^2+3
n^2+3=4k^2+2+1
n^2+3=2(2k^2+1)+1
which is odd because 2k^2+1 is included in integers, hence the claim

is this a valid proof? also can i just not write the last 2 lines of it since n^2+3=4k^2+3 is also always odd. also the final line says its odd because its included in integers but that alone doesnt mean its odd, its odd instead because your squaring an integer then multiplying it by 2 which will always give an even then adding a 3 which will make any even number odd

Answer 1

The given proof is not valid. Here are the issues with the proof:

1. The initial claim is not stated clearly. It should be "If n is an even integer, then n^2 + 3 is an odd integer."

2. The proof assumes that n is a natural number, but later uses k as an integer. It should be explicitly stated that n is an even integer.

3. The proof uses the wrong property of even numbers. The correct definition of an even number is n = 2k for some integer k, not natural number.

4. The proof correctly substitutes n = 2k, but it incorrectly simplifies the expression. (2k)^2 is not equal to 4k^2, it should be 4k^2.

5. The last two lines stating "n^2 + 3=2(2k^2+1)+1" and "which is odd because 2k^2+1 is included in integers" are unnecessary and incorrect. It is not a valid reason to claim the expression is odd just because 2k^2 + 1 belongs to the integers.

A correct proof could be as follows:

Claim: If n is an even integer, then n^2 + 3 is an odd integer.
Proof: Let n be an even integer. This means that there exists an integer k such that n = 2k.

We can substitute n = 2k into the expression n^2 + 3:
n^2 + 3 = (2k)^2 + 3
= 4k^2 + 3

Now, let's analyze the parity (evenness or oddness) of 4k^2 + 3. Every even number can be written as 2n for some integer n, and every odd number can be written as 2m + 1 for some integer m.

If 4k^2 + 3 is even, we can write it as 2n for some integer n. Then:
4k^2 + 3 = 2n
4k^2 = 2n - 3

The left side, 4k^2, is even since it is divisible by 2. We can write 4k^2 as 2m for some integer m. Substituting this into the equation, we get:
2m = 2n - 3
2m + 3 = 2n

The right side, 2n, is also even since it is divisible by 2. However, the left side, 2m + 3, is odd since it is not divisible by 2. This contradicts the assumption that 4k^2 + 3 is even.

Therefore, our assumption was incorrect, and 4k^2 + 3 cannot be even. This implies that 4k^2 + 3 must be odd. Hence, if n is an even integer, then n^2 + 3 is an odd integer.