Mathematical Induction Proof Technique

**Strings (S _{i}P_{j}A_{jk}) = S_{7}P_{2}A_{21} Base Sequence = 12735 String Sequence = 12735 - 2 - 21 **

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Mathematical Induction Proof Technique

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Mathematical induction proof technique is well suited for proof problems of the type:
*For a given population of integers, some event occurs*. An example of this type of proof problem is as follows:

For all integers n ≥ 1, ^{n}Σ_{k=1} = [n(n+1)]/2

(a) Prove, by induction, that, for every integer ≥ 5, 2^{n} > n^{2}.

(b) Prove, by induction, that any integer n ≥ 2 can be expressed as a finite product of primes.

**The strings**:
S_{7}P_{2}A_{21} (Identity - Physical Property).
**The math**:

Pj Problem of Interest is of type *identity* (physical property). Proofs establish truths. So they are *identity* problems.

Figure 121.2 illustrates the steps for proving by induction:

(1) Establish the truth of the statement for n = 1

(2) Assume the statement is true for n

(3) Establish the truth for n + 1.

(a) In this problem we have 5 as the lower bound instead of 1

So, for n=5, we have 2^{5} = 32 and 5^{2} = 25

So, 2^{5} > 5^{2}

Assume 2^{n} > n^{2}

We need to show that 2^{n+1} > (n+1)^{2}

2(2^{n}) > 2(n)^{2}

We need to show that 2(n)^{2} >(n+1)^{2} for n > 5.

2(n)^{2} - (n^{2} +2n -1) = (n-1)^{2}

(n+1)^{2} - (n^{2} +2n -1) = 2

So, for n > 5, 2(n)^{2} > (n+1)^{2}.

(b) Statement is true for n = 2.

Assume statement is true for all integers j, where 2 < j < n.

If n + 1 is prime, then statement is proved

If n + 1 is not prime, then it has a prime divisor, p.

So, there is an integer q, where 2 < q < n such that (n + 1) = pq.

But q can be expressed as a finite product of primes

So, n + 1 can be expressed as a finite product of primes.

The *point* **.** is a mathematical abstraction. It has negligible size and a great sense of position. Consequently, it is front and center in abstract existential reasoning.

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