"Numbers" in history have different meanings. In the past there'are only natrual numbers by counting, next there was negative numbers due to the concept of debt. Rational numbers seems to fill up the real continum but Pythagoreans has found irrational numbers. Finally the concept of real number continum is developed by Cantor and complex numbers appeared as powerful calculation tool.
Mathematical induction, processing natrual numbers, seems too "old fashioned" because there're far many problems concerning somewhere beyond natrual numbers, and we need different approaches. If MI can be extended over the reals, it's just fantastic. We will illustrate this by two examples: function equation and bionomial theorem over the reals. Before that, we will look at more different MI problems on A-level exam past papers.
Theorem. (Generalization of M.I.)
If (a) P(a) and P(b) is true implies that P(a+b), P(|a-b|) is true,
b) Exist least r that P(r) is true,
then P(k) is true is equivalent to k is a multiple of r.
6. N, Z, Q, R --- extented conclusion derived from MI
Example 16. f(x) is a everywhere continuous and differentiable function satisfying:
1)
2) Additive, i.e.,
3) f(1)=1.
Show that
Step 1: N
For the proprosition
Step 2: Z
Next
Now set the proposition "f(-n)=-n" and finish the induction.
Another approach is that
Step 3: Q
We use a double MI:
Theorem. The proposition P(a,b) is true for all a,b if:
1) P(1,1) is true.
2) If P(a,b) is true, then P(a,b+1) is true.
3) If P(a,b) is true, then P(a+1,b) is true.
This is quite powerful for rational numbers.
Here we use another approach of double MI that "For every b, P(a,b) implies P(a+1,b) is true".
For any given n,
Step 4: R
We know that irrational numbers can always be traced by a series of rational numbers.
Lemma. Every irrational numbers can be written in forms of sum of rational numbers.
Then if
P.S. Yet there's a simple method to do this question. Since the function is additive, we can show that the function has a constant slope, i.e., y' = m, so y = mx+c. Considering f(0)=0 and f(1)=1, we have f(x)=x.
Example 17. Extension of the binomial theorem.
Lemma. Vandermonde's Theorem:
The proof is trivial when you expand
Now consider the function
Now
For negative numbers, we have
For real numbers, we know that f is everywhere continous and differentiable in the interval
By comparing the coefficients we have the binomial coefficients over real numbers:
1) Show that
2) Show that
a) binomial substitution
b) the NZQR approach.
c) Compute
3) Show the following identities:
a)
b)
4) (2004 Pure I 5 modified) For real x, |a| < 1, show that: a)
Read the rest of this passage:
Part I
Part II
Part III