Metatheory - University of Cambridge

1. Induction 5
Proof. Evidently, 0 = 2 × 0. So 0 has the induction property.
Now pick an arbitrary number, n, and suppose (for induction) that n has
the induction property. There are two cases to consider
• Suppose n is even, i.e. n = 2k, for some number k. Then n+1 = 2k +1,
i.e. n + 1 is odd.
• Suppose n is odd, i.e. n = 2k + 1, for some number k. Then n + 1 =
2k + 2 = 2(k + 1), i.e. n + 1 is even.
Either way, n+1 has the induction property. So if n has the induction property,
then so does n + 1.
This completes the proof of Proposition 1.1 by induction.
■
In addition to the use of induction, a couple of things should strike you about
this proof. In particular, you should note that it is not a formal proof in any
particular deductive system. It is a bit of informal, mathematical, reasoning,
written in mathematised English. That said, it is perfectly rigorous. (The
symbol ‘■’ at the end of the proof just indicates that the proof is complete;
other texts use other symbols, e.g. ‘QED’ or ‘□’.)
To cement ideas, let me offer two more examples of proofs by induction:
Proposition 1.2. 2 0 + 2 1 + 2 2 + . . . + 2 n = 2 n+1 − 1, for any number n.
Proof. Evidently 2 0 = 1 = 2 0+1 − 1. So 0 has the induction property.
Now, pick any arbitrary number, k, and suppose (for induction) that
Then we have:
2 0 + 2 1 + 2 2 + . . . + 2 k = 2 k+1 − 1
2 0 + 2 1 + 2 2 + . . . + 2 k + 2 k+1 = 2 k+1 − 1 + 2 k+1
= (2 k+1 × 2) − 1
= 2 k+2 − 1
So the induction property holds for k + 1 as well.
■
Proposition 1.3. 0 + 1 + 2 + . . . + n = n2 +n
2
, for any number n.
Proof. Evidently 0 = 02 +0
2
. So 0 has the induction property.
Now, pick any arbitrary number, k, and suppose (for induction) that
Then we have:
0 + 1 + 2 + . . . + k = k2 + k
2
0 + 1 + 2 + . . . + k + (k + 1) = k2 + k
2
+ (k + 1)
= k2 + k + 2(k + 1)
2
= (k2 + 2k + 1) + (k + 1)
2
= (k + 1)2 + (k + 1)
2