Prove every odd number is of the form 2k+5. Is every even number of the form 2k+6?

I know even numbers are of the form $n=2k, k \in \mathbb Z$, and that odd numbers are of the form $n=2k+1, k \in \mathbb Z$.

I'm being asked to prove that all odd numbers are of the form $n=2k+5, k \in \mathbb Z$ and asked if all even numbers are of the form $n=2k+6, k \in \mathbb Z$.

I was wondering if my proofs are right, and if not, if you could please guide me in the right direction.

Proof 1. If $n$ is an odd integer then $n=2k+1, k \in \mathbb Z$. Let $n=2l+5, l \in \mathbb Z$. Then, $n=2l+4+1 =2(l+2)+1$ Since $l+2$ is an integer, $n$ is odd.

Proof 2. If $n$ is an even integer, then $n=2k, k \in \mathbb Z$. Let $n=2l+6, k \in \mathbb Z$. Then, $n=2l+6=2(l+3)$. Since $(l+3)$ is an integer, then $n=2l+6$ is even.

Thank you all in advance.

Alex

4 Answers

You are trying to show that if $n$ is odd, then $\exists l \in \mathbb{Z}, n = 2l+5$ rather than the opposite.

If $n$ is an odd integer, then $ \exists k \in \mathbb{Z}, n = 2k+1,$.

$n = 2k+1 = 2(k-2)+4+1= 2(k-2) + 5$.

Since $k-2 \in \mathbb{Z}$, $n$ can be written in the form of $2l+5$ by letting $l=k-2$.

Also note that an even number can be written as an odd number $+1$.

You have the right idea but:

Proof 1. If n is an odd integer then $n=2k+1,k∈Z$.

Should I be fussy or let it go. flips a coin. I'll be fussy. You should specify that $n = 2k+1$ for some $k \in \mathbb Z$.

Let n=2l+5,l∈Z.

This is now a different number altogether because we dont know anything about the relationship of $l$ to $k$.

Then, n=2l+4+1=2(l+2)+1 Since l+2 is an integer, n is odd.

Okay, that proves $n = 2l+5$ is odd. But that wasn't what you were asked to prove. You were asked to prove if $n = 2k +1$ is an odd number, then $n$ can be written in the form $n = 2l + 5$.

Which is simply a matter of saying: $n = 2k + 1 = 2k + 1 -5 +5 = 2k -4 +5= 2(k-2)+5$. Let $l= k-2$ will put $n=2l + 5$ in the desired form.

Proof 1. If $n$ is an odd integer then $n=2k+1$, (for some) $k \in \mathbb Z$.(An important quantifier "for some" is missing.) Let $n=2l+5$, for some $l \in \mathbb Z$. (The existence of such $l$ is what you want to prove, not an assumption. Your proof breaks down here.) Then, $n=2l+4+1 =2(l+2)+1$ Since $l+2$ is an integer, $n$ is odd.

Proof 2. If $n$ is an even integer, then $n=2k$, ("for some") $k \in \mathbb Z$. Let $n=2l+6, k \in \mathbb Z$. (Typo here: $k$ should be $l$. Again, the quantifier is missing and more importantly, the existence of such $l$ is what you need to prove. The proof breaks down here.) Then, $n=2l+6=2(l+3)$. Since $(l+3)$ is an integer, then $n=2l+6$ is even.

To fix your Proof 1 (similarly for the second one), note that you correctly assumed that $n=2k+1$ for some integer $k$ but you had never used this assumption. Remember the goal is to write $n=2l+5$ for some $l$. Try to write $l$ in terms of $k$.

All odd numbers are of the form $2k + 5 : k\in \mathbb{Z}, 2k$ is even, $5$ is odd.

If you have an even number and add an even number, it will be even.

$$\text{i.e} \ \ \ \ \ \ \ \forall \{k, n\}\subset \mathbb{Z}, 2k + 2n = 2(k + n)$$ so this must be even.

This implies that $2k + 2n\pm 1$ is odd which implies that $2n\pm 1$ and $2k\pm 1$ is odd. Now we can make $n = 2$ since $n \in \mathbb{Z}$ and thus we have:

$$2k + 2\times2 + 1 = 2k + 4 + 1 = 2k + 5$$ which proves that $2k + 5$ is odd. And now if we want to make $2k + 2n\pm 1$ even, then we $\pm 1$ to bring $2k + 2n \pm 1 \pm 1$ which brings us $2k + 6$ which proves that $2k + 6$ is even.

But your proofs are correct. You don't actually need any guiding in the right direction because you are already there.