Oh....now I get it!.......Correlation does not equal Causation

“If you cannot define what you are doing as a process, you do not understand what you are doing.”  -- W. Edwards Deming

Problem: Let p, q be distinct prime numbers.

 Prove that

pq−1 + qp−1 ≡ 1 (mod pq) .

Solution: Since p 6= q are prime numbers, we have gcd(p, q) = 1. By Fermat’s

Little Theorem, pq−1 ≡ 1 (mod q) . Clearly qp−1 ≡ 0 (mod q) . Thus

pq−1 + qp−1 ≡ 1 (mod q) .

Exchanging the roles of p and q in the above argument, we prove that

pq−1 + qp−1 ≡ 1 (mod p) .

In other words, pq−1 + qp−1 − 1 is divisible by both p and q. Since p and q are

relatively prime, we conclude that pq−1+qp−1−1 is divisible by pq, i.e. pq−1+qp−1 ≡

1 (mod pq) .

Problem 2. Let m, n be positive integers such that m|n. Prove that (m)| (n)

and that (mn) = m (n)

Solution: Since m|n, we can number the prime divisors of n such that

m = pa1

1 ...pas

s and n = pb1

1 ...pbs

s pbs+1

s+1 ...pbt

t ,

where t ≥ s, 0 < ai ≤ bi for i = 1, 2, ..., s and 0 < bi for i > s, and p1, ..., pt are

pairwise distinct prime numbers.

Now

(m) = (p1 − 1)pa1−1

1 ...(ps − 1)pas−1

s

and

(n) = (p1 − 1)pb1−1

1 ...(ps − 1)pbs−1

s (ps+1 − 1)pbs+1−1...(pt − 1)pbt−1

t .

It is clear now that (m)| (n). Moreover, mn = pa1+b1

1 ...pas+bs

s pbs+1

s+1 ...pbt

t and

(mn) = (p1 −1)pa1+b1−1

1 ...(ps −1)pas+bs−1

s (ps+1 −1)pbs+1−1...(pt −1)pbt−1

t = m (n).

counter-intuitive and unseen variables that are also correlated with other variables are virtually impossible to calculate, and causal relationships between seemingly antithetical components of the chemistry of Thinking serve to obligate us to confusion and the ultimate misdiagnosis of our efforts to understand all but the simplest process thinking.....
........ cum hoc ergo propter hoc ergo post hoc ergo propter hoc

                        THIS IS NOT AN ELEPHANT!


Beginning of a Journey by Yumi Kurosawa