Proof by Induction (Teng Hui)
Inductive reasoning versus deductive reasoning
For inductive reasoning:
 Start with a specific conclusion (i.e. Shaq and Yao play basketball)
 Arrive at generalization (i.e. all tall players play basketball)
For deductive reasoning:
 Start with general facts (all birds have feathers)
 Specific fact (sparrow has feathers)
 Conclusion (sparrow is a bird)
False induction in math: it’s not valid to draw a conclusion by simply testing a few values
For example:
f(n) = n2 – 40n + 41 or an equation similar to it that seemingly yields prime numbers for
values of n, but doesn’t work for n values beyond a few terms
Inductive proof method, to prove a formula true for a particular variable for all nonnegative or positive integers:
1)
2)
3)
4)
Prove the equation works for n = 0 or n = 1 (base case)
Assume the equation works for an arbitrary integer k (inductive hypothesis)
Prove the statement is true for n = k + 1 (inductive step)
If k + 1 is valid, then the formula proves that the equation works for one value,
the value after, and so on.
Here's an example:
n
We'll prove that
i 
i 1
n(n  1)
for all positive integers n.
2
1
Base case. n = 1. LHS =
 i  1 , RHS =
i 1
1(1  1)
 1 , so the base case holds.
2
k
Inductive hypothesis: Assume for an arbitrary positive integer n=k that
i 
i 1
k 1
Inductive step: Prove for n=k+1 that
i 
i 1
k 1
k
i 1
i 1
 i  ( i)  (k  1)
(k  1)( k  2)
2
k (k  1)
2
k (k  1)
 k  1 , plugging into the inductive hypothesis
2
k
 (k  1)(  1)
2
k2
 (k  1)(
)
2
(k  1)( k  2)

, which proves the inductive step.
2

n
Thus, we can conclude that for all positive integers n,
i 
i 1
n(n  1)
.
2
In some sense this proof technique looks like a bit of magic. Let's investigate where it can
break down:
1) When the base case doesn't work.
2) When the inductive step can't be proved.
We will show the attempted proof of two false statements that illustrate both of these
problems.
n
1) Prove that
i 
i 1
n2 n
  1 , for all positive integers n.
2 2
We can see that plugging in n=1 to both the left and right hand sides does not yield a true
statement, so that is the problem in proving this statement. Interestingly enough however,
for this particular example, it turns out that you CAN actually prove the inductive step:
k
Inductive hypothesis: Assume for an arbitrary positive integer n=k that  i 
i 1
k 1
Inductive step: Prove for n=k+1 that
i 
i 1
k 1
k
i 1
i 1
(k  1)
(k  1)

1
2
2
2
 i  ( i)  (k  1)
k2 k
2k  2
 1
, plugging into the inductive hypothesis
2 2
2
k 2  2k  1 k  1


1
2
2
(k  1) 2 k  1


 1, proving the inductive step.
2
2

k2 k
 1
2 2
So for this problem, we see that if the formula were true for n=k, then it would also have
to be true for n=k+1. But, the problem is that there is no value of k for which the formula
is true, so proving the inductive step unfortunately proves nothing about the original
formula we had.
n
2) Prove that
 i  2n  1 , for all positive integers n.
i 1
Here the base case holds, since for n=1, both the left and right hand sides equal 1.
Now let's take a look at how trying to prove the inductive step breaks down:
k
Inductive hypothesis: Assume for an arbitrary positive integer n=k that
 i  2k  1
i 1
k 1
Inductive step: Prove for n=k+1 that
 i  2(k  1)  1  2k  1
i 1
k 1
k
 i  ( i)  (k  1)
i 1
i 1
 (2k  1)  (k  1)
 3k ,
At this point we are stuck, because it's impossible for us to get 3k to equal 2k+1. Most of
the time, this is false. Thus, it's impossible to show in the general case that if the formula
is true for k, that it is true for k+1. In this particular case, the formula is true for n=1 and
n=2, but then is false otherwise.
Hopefully these two examples solidify exactly how mathematical induction works and
why it properly can prove a general statement about all positive integers.