14 1. The Real Numbers
Induction step: If we assume the formula is true for a certain n, then multiplying
both sides of this formula by x + y yields
(x +
y)n+1
= x
n
k=0
n
k
xkyn−k
+ y
n
k=0
n
k
xkyn−k
=
n
k=0
n
k
xk+1yn−k
+
n
k=0
n
k
xkyn−k+1.
(1.2.7)
If we change variables in the first sum on the second line of (1.2.7) by replacing k
by k 1, then our expression for (x +
y)n+1
becomes
xn+1
+
n
k=1
n
k 1
xkyn−k+1
+
n
k=1
n
k
xkyn−k+1
+
yn+1
=
xn+1
+
n
k=1
n
k 1
+
n
k
xkyn+1−k
+
yn+1.
(1.2.8)
If we use the identity (to be proved in Exercise 1.2.17)
n
k 1
+
n
k
=
n + 1
k
,
then the right side of equation (1.2.8) becomes
xn+1
+
n
k=1
n + 1
k
xkyn+1−k
+
yn+1
=
n+1
k=0
n + 1
k
xkyn+1−k.
Thus, the binomial formula is true for n + 1 if it is true for n. This completes the
induction step and the proof of the theorem.
Exercise Set 1.2
In the first seven exercises use only Peano’s axioms and results that were proved in
Section 1.2 using only Peano’s axioms.
1. Prove that the commutative law for addition, n + m = m + n, holds in N. Use
induction and Examples 1.2.6 and 1.2.5.
2. Prove that if n, m N, then m + n = n. Hint: Use induction on n.
3. Use the preceding exercise to prove that if n, m N, n m, and m n, then
n = m. This is the reflexive property of an order relation.
4. Prove that the order relation on N has the transitive property: if k n and
n m, then k m.
5. Use the preceding exercise and Peano’s axioms to prove that if n N, then for
each element m N either m n or n m. Hint: Use induction on n.
6. Show how to define the product nm of two natural numbers. Hint: Use induc-
tion on m.
7. Use the definition of product that you gave in the preceding exercise to prove
that if n, m N, then n nm.
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