# Prove that the sum of the degrees in the interior angles of a polygon with $n$ sides is $180(n – 2)°$.

I would assume this question involves an inductive hypothesis.

1. Show $n=1$ is true.
2. Assume that if $n$ is replaced by $k$, the sum of the degrees in the interior angles of a polygon with $k$ sides is $180(n-2)$.
3. Assuming the assumption is true, I want to show that when $n=k+1$ that the sum of the degrees in the interior angles of a polygon with $k+1$ sides is $180((k+1)-2)$.

Problem I'm having is that what do i set $180((k+1)-2)$ to?

There is some confusion on your induction.

1. $n=1$ does not make sense, since polygons have at least 3 sides (a triangle).

2. This step, although not really wrong, is unnecessary. You are simply renaming $n$ as $k$.

3. This is ok, might you should be careful when renaming $n=k$ in step 2 (I'm being somewhat pedantic here).

Induction works as follows: take a property $P(n)$ about natural numbers (for example, "$n$ is even", or "The sum of the inner angles of any polygon with $n$ sides is $180^o(n-2)$"), which might be true or false. Fix a number $M$. Then we prove

1. $P(M)$ is true.
2. If $P(n)$ is true for some $n$, then $P(n+1)$ is true.

And the induction principle says that $P(n)$ is true for all $n\geq M$. In your case you need to show.

1. The sum of the angles of any 3-sided polygon (a triangle) is $180^o$.
2. Assume that the sum of the angles of any polygon of $n$ sides is $180^o(n-2)$. Then we need to show that the sum of the angles of any polygon of $n+1$ sides is $180^o(n-1)$.

To prove $2$, start with a polygon $P$ of $n+1$ sides. You need somehow to use the hypothesis, that is, make a polygon of $n$ sides appear, and calculate the angles somehow. You should do this by yourself.

• I've tried using the hypothesis to prove. What I did was using the hypothesis on $n+1$ by substituting $n$ with $180^o(n-2)$. That would leave me with $180^o(n-2)+1$=$180^o(n-1)$. Somehow, I still can't get them to equal. Am I doing something wrong? May 1, 2016 at 4:49
• @nyorkr23 Let's say you do know that the sum of the angles on a triangle is $180^o$. Only using this, try to prove that the sum of the angles of a square is $360^o$. Then try to do the same with a pentagon, hexagon, etc... You will probably come with a procedure which works on all polygons, where you "simplify" the problem of calculating the sum of angles to calculating the sum of angles on smaller polygons (which you then know how to do). May 1, 2016 at 5:21
• There are other more direct ways (instead of induction) to prove the claim.
– Mick
May 1, 2016 at 15:08

My understanding of math induction, is that you first show the initial case is true. Then you show that any generic place chosen after that, automatically leads to the next case in the sequence. It is sort of like dominoes, once the first one falls, all the others must then fall in sequence. So, here is what I am thinking. The initial case is a triangle, n = 3. We need to show that for the general case, the next figure in the sequence is 180 more than the previous, ie: D(n+1) = D(n)+180 = 180(n-2)+180

D(n) = 180(n-2)

1. D(3) = 180 (3-2) = 180 (1) = 180

2. D(n) = 180 (n-2)

D(n+1) = 180((n+1)-2) = 180(n+1-2) = 180(n-2+1) = 180(n-2)+180

I have to be honest here, math induction was never my strong suit. Perhaps someone who knows it better can give some input on my work.