T(n) is an odd integer if and only if n is a perfect square. a(n)i on oda intese
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
7
![Proof. We have mentioned that the
12.
and
(b
u|p
u|p
13. P
the stated result follows immediately from Theorem 6.4.
14. I
PROBLEMS 6.1
Let m and n be positive integers and p1, P2,
least one of m or n. Then m and n may be written in the form
..
k1 k2
m = Pj P2
with k; > 0 for i = 1, 2, . , r
15.
%3D
..
n = p{' p.…· p
ji „j2
with j; 2 0 for i = 1, 2, . , r
16
...
Prove that
lcm(m, n) = p{ P2 P
d..dd = (u'u)pɔ
where u; = min {k¡, ji}, the smaller of k; and ji; and v; = max {k;, ji}, the larger oft.
and ji.
1.
1.
Use the result of Problem 1 to calculate gcd(12378, 3054) and lcm(12378, 3054).
Deduce from Problem 1 that gcd(m, n) lcm(m, n)
In the notation of Problem 1, show that gcd(m, n) = 1 if and only if kiji = 0 for
i = 1, 2, ... , r.
(a) Verify that t(n) = t(n + 1) = t(n +2) = t(n + 3) holds for n = 3655 and 4503.
(b) When n = 14, 206, and 957, show that o (n) = o (n + 1).
6. For any integer n > 1, establish the inequality t(n) < 2./n.
[Hint: If d | n, then one of d or n/d is less than or equal to An.1
7. Prove the following.
(a))t(n) is an odd integer if and only if n is a perfect square.
O g(n) is an odd integer if and only if n is a perfect square or twice a perfect square.
= mn for positive integers m and n.
%3D
%3D
%3D
%3D
%3D
%D
[Hint: If p is an odd prime, then 1+ p + p²+
8) Show that an 1/d = o (n)/n for every positive integer n.
If n is a square-free integer, prove that t (n) = 2, where r is the number of prime divisors
...+ )
+ p* is odd only when k is even.
of n.
10. Establish the assertions below:
(a) If n = p p.. p is the prime factorization of n > 1. then
(u))
(b) For any positive integer n,
P2
1-
>1+
23
(c) If n > 1 is a composite number, then o (n)
[Hint: See Problem 8.]
>n+n.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F66c7b497-dac7-4855-b923-2e60bbc73063%2Ff3dacc21-e136-4788-80dc-5745d1addc43%2Fd4cos5i.jpeg&w=3840&q=75)
Transcribed Image Text:Proof. We have mentioned that the
12.
and
(b
u|p
u|p
13. P
the stated result follows immediately from Theorem 6.4.
14. I
PROBLEMS 6.1
Let m and n be positive integers and p1, P2,
least one of m or n. Then m and n may be written in the form
..
k1 k2
m = Pj P2
with k; > 0 for i = 1, 2, . , r
15.
%3D
..
n = p{' p.…· p
ji „j2
with j; 2 0 for i = 1, 2, . , r
16
...
Prove that
lcm(m, n) = p{ P2 P
d..dd = (u'u)pɔ
where u; = min {k¡, ji}, the smaller of k; and ji; and v; = max {k;, ji}, the larger oft.
and ji.
1.
1.
Use the result of Problem 1 to calculate gcd(12378, 3054) and lcm(12378, 3054).
Deduce from Problem 1 that gcd(m, n) lcm(m, n)
In the notation of Problem 1, show that gcd(m, n) = 1 if and only if kiji = 0 for
i = 1, 2, ... , r.
(a) Verify that t(n) = t(n + 1) = t(n +2) = t(n + 3) holds for n = 3655 and 4503.
(b) When n = 14, 206, and 957, show that o (n) = o (n + 1).
6. For any integer n > 1, establish the inequality t(n) < 2./n.
[Hint: If d | n, then one of d or n/d is less than or equal to An.1
7. Prove the following.
(a))t(n) is an odd integer if and only if n is a perfect square.
O g(n) is an odd integer if and only if n is a perfect square or twice a perfect square.
= mn for positive integers m and n.
%3D
%3D
%3D
%3D
%3D
%D
[Hint: If p is an odd prime, then 1+ p + p²+
8) Show that an 1/d = o (n)/n for every positive integer n.
If n is a square-free integer, prove that t (n) = 2, where r is the number of prime divisors
...+ )
+ p* is odd only when k is even.
of n.
10. Establish the assertions below:
(a) If n = p p.. p is the prime factorization of n > 1. then
(u))
(b) For any positive integer n,
P2
1-
>1+
23
(c) If n > 1 is a composite number, then o (n)
[Hint: See Problem 8.]
>n+n.
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