a) Generalize the result in part (a) of Exercise 196; that is, show that if p is a prime, the positive integers less than p , except 1 and p − 1 , can be split into ( p − 3 ) / 2 pairs of integers such that each pair consist of integers that are inverses of each other [Hint: Use the result of Exercise 17.] b) From part (a) conclude that ( p − 1 ) ! ≡ − 1 ( mod p ) whenever p is prime. This result is known as Wilson’s theorem. c) What can we conclude if n is a positive integer such that ( n − 1 ) ! ≠ − 1 ( mod n ) ?
a) Generalize the result in part (a) of Exercise 196; that is, show that if p is a prime, the positive integers less than p , except 1 and p − 1 , can be split into ( p − 3 ) / 2 pairs of integers such that each pair consist of integers that are inverses of each other [Hint: Use the result of Exercise 17.] b) From part (a) conclude that ( p − 1 ) ! ≡ − 1 ( mod p ) whenever p is prime. This result is known as Wilson’s theorem. c) What can we conclude if n is a positive integer such that ( n − 1 ) ! ≠ − 1 ( mod n ) ?
Solution Summary: The author explains that positive integers less than p can be split into (p - 3)/2 pairs such that each pair consists of inverses of each other.
a) Generalize the result in part (a) of Exercise 196; that is, show that if p is a prime, the positive integers less than p, except 1 and
p
−
1
, can be split into
(
p
−
3
)
/
2
pairs of integers such that each pair consist of integers that are inverses of each other [Hint: Use the result of Exercise 17.]
b) From part (a) conclude that
(
p
−
1
)
!
≡
−
1
(
mod
p
)
whenever p is prime. This result is known as Wilson’s theorem.
c) What can we conclude if n is a positive integer such that
(
n
−
1
)
!
≠
−
1
(
mod
n
)
?
A function is defined on the interval (-π/2,π/2) by this multipart rule:
if -π/2 < x < 0
f(x) =
a
if x=0
31-tan x
+31-cot x
if 0 < x < π/2
Here, a and b are constants. Find a and b so that the function f(x) is continuous at x=0.
a=
b= 3
Use the definition of continuity and the properties of limits to show that the function is continuous at the given number a.
f(x) = (x + 4x4) 5,
a = -1
lim f(x)
X--1
=
lim
x+4x
X--1
lim
X-1
4
x+4x
5
))"
5
))
by the power law
by the sum law
lim (x) + lim
X--1
4
4x
X-1
-(0,00+(
Find f(-1).
f(-1)=243
lim (x) +
-1 +4
35
4 ([
)
lim (x4)
5
x-1
Thus, by the definition of continuity, f is continuous at a = -1.
by the multiple constant law
by the direct substitution property
Chapter 4 Solutions
Discrete Mathematics and Its Applications ( 8th International Edition ) ISBN:9781260091991
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