claim: if n is even then n^2+3 is odd
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
The given proof is not valid.
In the proof, first, you correctly assume that n is even, meaning n can be represented as n = 2k for some integer k. However, you made a mistake during the simplification step.
When you substitute n = 2k into n^2 + 3, you incorrectly simplify it as (2k)^2 + 3 = 4k^2 + 3. This is where the error lies.
(2k)^2 should be expanded as 4k^2, but it should also be noted that (2k)^2 is always divisible by 4 since it contains a factor of 4. Therefore, (2k)^2 is always even.
Let's apply this correction and see if the claim holds.
n^2 + 3 = (2k)^2 + 3 = 4k^2 + 3
Observing the expression 4k^2 + 3, we can see that it has the form of an odd number, i.e., an even number (4k^2) plus an odd number (3). Therefore, n^2 + 3 is always an odd number regardless of whether n is even or odd.
To summarize, the claim that "if n is even, then n^2 + 3 is odd" is indeed true, but the given proof contains an error in the simplification step. The corrected proof shows that n^2 + 3 is odd due to the addition of an even number (4k^2) and an odd number (3).
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.
The provided proof is not valid. Let's go step-by-step to understand why.
1. Claim: If n is even, then n^2 + 3 is odd.
2. Assumption: Suppose n is an even number.
3. By definition, n = 2k for some integer k.
4. Substituting n in n^2 + 3, we get (2k)^2 + 3.
5. Simplifying, (2k)^2 + 3 becomes 4k^2 + 3.
6. At this point, the claim is that 4k^2 + 3 is odd. However, this claim needs to be proven.
7. The statement "4k^2 + 3 = 2(2k^2 + 1) + 1" is in the right direction, but not enough to conclude it is odd.
8. It is incorrect to assume that 2k^2 + 1 is an integer, as it is a squared term.
9. The correct way to prove the claim is by considering the parity of each term:
- The term 4k^2 is always even since it is the product of an even number (4) and any integer (k^2).
- The term 3 is odd as it is a constant.
- The sum of an even number and an odd number is always odd.
10. Therefore, n^2 + 3 is indeed odd when n is even.
In conclusion, the proof would be valid if you properly justified that the sum of an even number (4k^2) and an odd number (3) is always odd.