Addition of natural numbers/Proofs
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Mathematical proofs for addition of the natural numbers: additive identity, commutativity, and associativity. These proofs are used in the article Addition of natural numbers.
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[edit] Definitions
This article will use the definitions in addition of natural numbers, particularly [A1] and [A2]:
- a + 0 = a [A1]
- a + S(b) = S(a + b) [A2]
It is useful to define another natural number closely related to the successor function, namely "1". We define 1 to be the successor of 0, in other words,
.
Note that for all natural numbers a,
due to [A1] and [A2].
[edit] Proof of associativity
We prove associativity by first fixing natural numbers a and b and applying induction on the natural number c.
For the base case c = 0,
- (a + b) + 0 = a + b = a + (b + 0)
Each equation follows by definition [A1]; the first with a + b, the second with b.
Now, for the induction. We assume the induction hypothesis, namely we assume that for some natural number c,
- (a + b) + c = a + (b + c)
Then it follows,
- (a + b) + S(c)
- = S((a + b) + c) [by A2]
- = S(a + (b + c)) [by the induction hypothesis]
- = a + S(b + c) [by A2]
- = a + (b + S(c)) [by A2]
In other words, the induction hypothesis holds for S(c). Therefore, the induction on c is complete.
[edit] Proof of identity element
Definition [A1] states directly that 0 is a right identity. We prove that 0 is a left identity by induction on the natural number a.
For the base case a = 0, 0 + 0 = 0 by definition [A1]. Now we assume the induction hypothesis, that 0 + a = a. Then
- 0 + S(a)
- = S(0 + a) [by A2]
- = S(a) [by the induction hypothesis]
This completes the induction on a.
[edit] Proof of commutativity
We prove commutativity by applying induction on the natural number b. First we prove the base cases b = 0 and b = S(0) = 1 (i.e. we prove that 0 and 1 commute with everything).
The base case b = 0 follows immediately from the identity element property (0 is an additive identity), which has been proved above: a + 0 = a = 0 + a.
Next we will prove the base case b = 1, that 1 commutes with everything, i.e. for all natural numbers a, we have a + 1 = 1 + a. We will prove this by induction on a (an induction proof within an induction proof). Clearly, for a = 0, we have 0 + 1 = 0 + S(0) = S(0 + 0) = S(0) = 1 = 1 + 0. Now, suppose a + 1 = 1 + a. Then
- S(a) + 1
- = S(a) + S(0)
- = S(S(a) + 0) [by A2]
- = S((a + 1) + 0)
- = S(a + 1) [by A1]
- = S(1 + a) [by the induction hypothesis]
- = 1 + S(a) [by A2]
This completes the induction on a, and so we have proved the base case b = 1. Now, suppose that for all natural numbers a, we have a + b = b + a. We must show that for all natural numbers a, we have a + S(b) = S(b) + a. We have
- a + S(b)
- = a + (b + 1)
- = (a + b) + 1 [by associativity]
- = (b + a) + 1 [by the induction hypothesis]
- = b + (a + 1) [by associativity]
- = b + (1 + a) [by the base case b = 1]
- = (b + 1) + a [by associativity]
- = S(b) + a
This completes the induction on b.
[edit] See also
[edit] References
- Edmund Landau, Foundations of Analysis, Chelsea Pub Co. ISBN 0-8218-2693-X.
