Confirming a easy proof: the product of two consecutive numbers is always even.

Question

Can someone confirm if my prove is right?

Theorem. The product of two consecutive integers is always even.

Proof. Define a number $n$ such that $n:=2k$ where $k\in\mathbb{Z}$, this ensures that $n$ is an even number. Define a second number $p$ such that $p:=n+1=2k+1$ this ensures that $p$ is an odd number and the numbers $n$ and $p$ are consecutive because they differ by $1$. There product is given by:

$$np=2k(2k+1)=2(2k^2+k)=4k^2+2k\tag1$$

Which is clearly a multiple of $2$.

Define a number $h$ such that $h:=2m+1$ where $m\in\mathbb{Z}$, this ensures that $h$ is an odd number. Define a second number $z$ such that $z:=n-1=2m$ this ensures that $z$ is an even number and the numbers $h$ and $z$ are consecutive because they differ by $1$. There product is given by:

$$hz=2m(2m+1)=2(2m^2+m)=4m^2+2m\tag2$$

Which is clearly a multiple of $2$, this proves that $hz$ is an even number.

This proves this theorem.

If I am wrong how can I make my proof valid?

Answer 1

Style and sufficiency of proof generally depends on where you take the "floor of certainty" to be - what facts and statements can be taken as true and what needs to be demonstrated.

With a simple theorem like this you might say that we should set the floor a little lower, that is provide an exposition of even simple steps in the proof. For example, the fact that alternating integers are odd and even might need to be supported - or not.

Commenting on your proof, then, in the knowledge that this is a little subjective:

• Breaking into the cases of lower number even and lower number odd looks good, and you could do this more clearly.
• You don't need to define a separate number for $n{+}1$. It doesn't add clarity and could easily confuse the matter.
• You can continue to use $n$ and $n{+}1$ as the successive numbers in the two cases - again limit your new defined numbers, for example to where they will interact with each other; there's no such interaction between your cases.

So using your method I would write:

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Theorem. The product of two consecutive integers is always even.

Proof. Consider successive integers $n$ and $n{+}1$ with product $n(n{+}1)$ in two cases, (1) where $n$ is even and (2) where $n$ is odd.

Case 1, $n$ even:
Find $k\in\mathbb{Z}$ such that $n=2k$. Now the desired product is given by:

$$n(n+1)=2k(2k+1)=2(k(2k+1))\tag1$$

And since $k(2k+1)\in\mathbb{Z}$ this is thus a multiple of $2$ and even as required.

Case 2, $n$ odd:
Find $k\in\mathbb{Z}$ such that $n=2k+1$. Now the desired product is given by:

$$n(n+1)=(2k+1)(2k+2)=2(2k+1)(k+1)\tag 2$$

And since $(2k+1)(k+1)\in\mathbb{Z}$ the product is again a multiple of $2$ and even as required.

Since both cases demonstrate an even product, this proves this theorem.

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Answer 2

You proved that the product of an even number and the next number is even.

Since any number is even or odd, you could complete the proof of the theorem

by proving that the product of an odd number and the next number is also even.

Answer 3

Let $a$ and $b$ be two consecutive integers, with $a<b$. If $a$ is even, then $b$ is odd. Hence, $$ ab=(2n) \cdot (2n+1)=4n^2+2n=2(n^2+n) \, , $$ where $n \in \mathbb{Z}$ is the unique integer such that $2n=a$ and $2n+1=b$. If $a$ is odd, then $b$ is even. Can you finish the proof?

Answer 4

Consider all the products of two consecutive natural numbers, totally ordered by the minimum factor. The difference between two consecutive such products is even: $$ n\,(n+1)-(n-1)\,n=2n\,. $$ Therefore parity is preserved from one product to the next (if $a-2r=2s$ then $a=2(s+r)$), and the first product is $1\cdot 2=2$, so all the products are even.

As a bonus, we also get inductively that, since $1\cdot 2=2$, $$ n\,(n+1)=2\,(1+2+\cdots+n)\,, $$ and then $$ 1+2+\cdots + n=\frac{n\,(n+1)}2\,. $$