When we studied prime numbers we looked at the Euler phi-function ∅(n) = the number of integers between 1 and n with no factors in common with n = n – (the number of integers between 1 and n with a factor in common with n) Look at the special case where n = pq, with p and q being different primes. Let A = {the integers between 1 and pq which are divisible by p} and argue that |A| = q Let B = {the integers between 1 and pq which are divisible by q} and argue that |B| = p Why is |A ∩ B| = 1? Conclude that |A U B| =p + q -1 and that ∅(pq) = pq - (p + q -1) = (p-1)(q-1)
When we studied prime numbers we looked at the Euler phi-function ∅(n) = the number of integers between 1 and n with no factors in common with n = n – (the number of integers between 1 and n with a factor in common with n) Look at the special case where n = pq, with p and q being different primes. Let A = {the integers between 1 and pq which are divisible by p} and argue that |A| = q Let B = {the integers between 1 and pq which are divisible by q} and argue that |B| = p Why is |A ∩ B| = 1? Conclude that |A U B| =p + q -1 and that ∅(pq) = pq - (p + q -1) = (p-1)(q-1)
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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When we studied prime numbers we looked at the Euler phi-function
∅(n) = the number of integers between 1 and n with no factors in common with n
= n – (the number of integers between 1 and n with a factor in common with n)
Look at the special case where n = pq, with p and q being different primes.
Let A = {the integers between 1 and pq which are divisible by p}
and argue that |A| = q
Let B = {the integers between 1 and pq which are divisible by q}
and argue that |B| = p
Why is |A ∩ B| = 1?
Conclude that |A U B| =p + q -1
and that ∅(pq) = pq - (p + q -1) = (p-1)(q-1)
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