Prove that the expression y(y + 1) always represents an even number, where y is any positive integer.

Thanks Heureka,

I'd just like to comment a little further

consider

an even number*an odd number.    

2 is a factor of the even number - by definition  (I think it is alright to say that.  )

so let the even number be 2n

Let the odd number be m

So an even* an odd = 2n*m AND since 2 is a factor  the resultant number HAS to be even!

Prove that the expression y(y + 1) always represents an even number, where y is any positive integer.

$$\small{\text{
y and y+1 are next numbers. So if y is even, y+1 must be odd and
}}
\\
\small{\text{
if y is odd then y+1 must be even.
\boxed{even \times odd = even} or \boxed{odd \times even = even}
}}$$

$$\small{\text{
$
\begin{array}{rc|rc}
y & even \quad & \quad y & odd \\
y+1 & odd \quad & \quad y+1 & even \\
\hline
y(y+1) & even \times odd \quad & \quad y(y+1) & odd \times even
\end{array}
$
}}$$

$$\boxed{
\begin{array}{c}
even \times odd = even \\
odd \times even = even
\end{array}
}$$

$$\small{\text{
Example:
$
\begin{array}{rc|rc}
4 & even \quad & \quad 7 & odd \\
5 & odd \quad & \quad 8 & even \\
\hline
20 & even \times odd \quad & \quad 56 & odd \times even
\end{array}
$
The products are always even \
}}$$