# example of an induction proof

```Triangle Proof
by Kathy McDonald
section 3.1 #7
Prove: When dividing each side of
an equilateral triangle
into n segments
then connecting the division points
with all possible segments parallel
to the original sides, n&sup2; small
triangles are created.
Proof by induction:
Let S = {nN: f(n) = n&sup2;}
Show 1 S:
1
f(n) =n&sup2;
f(1) = 1 = 1&sup2;
Show 2 S:
when dividing each side into 2
segments
and connecting division points as
described,
4 small triangles are created.
f(n) =n&sup2;
f(2) = 4 = 2&sup2;
Show 3 S:
when dividing each side into 3
segments
and connecting division points as
described,
9 small triangles are created.
f(n) =n&sup2;
f(3) = 9 = 3&sup2;
Assume n S.
Assume when dividing each side
into n segments and connecting
division points as described, n&sup2;
small triangles are created.
Assume f(n) = n&sup2;.
Show n+1 S.
Show when dividing each side into
n+1 segments and connecting
division points as described, (n+1)&sup2;
small triangles are created.
Show f(n+1) = (n+1)&sup2;.
Consider a divided triangle
with n segments on each side.
When a segment equal in size to the
n segments is added to each side
and those endpoints are connected,
a space is created at the bottom of the
original triangle.
Also, a new, bigger equilateral
triangle has been created.
This new, bigger triangle has n+1
segments on each side.
n segments
+
1 segment
Now, the parallel dividing lines are
extended down
to the base of the new, bigger triangle.
More small triangles are created.
The n segments of the base of the
original triangle
correspond to n bases of the new,
small triangles created.
Also, the n+1 segments of the base of
the new, bigger triangle
correspond to n+1 bases of the new,
small triangles.
So, n+(n+1) bases
correspond to n+(n+1) new, small
triangles
By assumption, the original triangle
has n segments on each side
And n&sup2; small triangles inside.
By adding 1 segment to each side of
this triangle,
n + (n+1) small triangles are added.
The total small triangles of the new,
bigger triangle is:
n&sup2; + n +(n+1)
=n&sup2;+2n+1
=(n+1)(n+1)
= (n+1)&sup2;
This shows n+1 S.
By induction, S  N.
Dwight says, “that’s it.”
```