number theory

[shahar]

How it is possible to prove that the formula:
n(n-3)/2 is always even?

 

[Harry_the_cat]

It's not!
For example if n=5, n(n-3)/2 =5( 5- 3)/2 = 5 x 2 /2 = 5 which is odd.

 

[shahar]

It's not!
For example if n=5, n(n-3)/2 =5( 5- 3)/2 = 5 x 2 /2 = 5 which is odd.

OK. I have a typo, the correct version is (5(n-3))/2.
This is the formula of number of the diagnols of any polygon.

 

[Harry_the_cat]

OK. I have a typo, the correct version is (5(n-3))/2.
This is the formula of number of the diagnols of any polygon.

That's not always even either. Eg n=5.
A pentagon has 5 diagonals. That's not even.
The formula for the number of diagonals in a polygon was correct in your first post, ie n(n-3)/2.

 

[shahar]

That's not always even either. Eg n=5.
A pentagon has 5 diagonals. That's not even.
The formula for the number of diagonals in a polygon was correct in your first post, ie n(n-3)/2.

O.K. Here what I want to ask - why the solution is always whole number?

 

[Harry_the_cat]

n(n-3)/2

If n is odd, then n-3 is even, so n (n-3) is even, and so divisible by 2. Therefore n(n-3)/2 is a whole number.

If n is even, .... you can finish off the proof from here.

 

[khansaheb]

n(n-3)/2

If n is odd, then n-3 is even, so n (n-3) is even, and so divisible by 2. Therefore n(n-3)/2 is a whole number.

If n is even, .... you can finish off the proof from here.

I would only add a

condition n>3​

 

[Dr.Peterson]

I would only add a condition n>3

Actually, the formula works for $n\ge3$, since it correctly says that a triangle has no diagonals!

And since we usually don't consider a monogon or digon to be polygons, though the words do exist, even this isn't a serious restriction.

How it is possible to prove that the formula:
n(n-3)/2 is always even?

I take it that you meant to say it "always divides evenly". I imagine English isn't the only language in which "even" can be confusing that way!

 

[lookagain]

I would only add a

condition n>3​

shahar mentioned the expression is to equal a whole number. Zero in all of
the sources I checked list it as a whole number, so it should be $\displaystyle \ n \ge 3.$

 

[Steven G]

OK. I have a typo, the correct version is (5(n-3))/2.
This is the formula of number of the diagnols of any polygon.

Since 2 and 5 have no common factors other than 1, having the 5 in 5(n-3)/2 will not change whether or not 5(n-3)/2 is a whole number (that is the same as saying that 5(n-3) is even). If 5(n-3) is in fact even, then it is because n-3 is even. n-3 being even depends on the parity of n meaning that it is not always even.