Download HW 1 - Math TAMU

HOMEWORK 1 SOLUTION
From KW, Ch.1:
Problem 5.
(2 pts for each part)
(a) If c | a and c | b then c | ab.
Proof. Let h, k be integers such that a = ch, b = ck. Then ab = c2 hk = c(chk); since chk
is an integer, c|ab.
(b) If c | a and c | b then c2 | ab.
Proof. Let h, k be integers such that a = ch, b = ck. Then ab = c2 hk = c2 (hk); since hk
is an integer, c|ab.
(c) If c - a and c - b then c - (a + b).
This statement is false. Let c = 2, a = b = 1 to get a counterexample.
Problem 6.
(2 pts for each part)
(a) The sum of two even numbers is an even number.
Proof. Let a = 2h, b = 2k be two even numbers; then a + b = 2(h + k) that is even.
(b) The sum of two odd numbers is an even number.
Proof. Let a = 2h + 1, b = 2k + 1 be two odd numbers; then a + b = 2(h + k + 1) that is
even.
(c) The product of two even numbers is an even number and divisible by 4.
Proof. Let a = 2h, b = 2k be two even numbers; then ab = 2(2hk) = 4(hk) that is even
and divisible by 4.
(d) The product of two odd numbers is an odd number.
Proof. Let a = 2h + 1, b = 2k + 1 be two even numbers; then ab = 2(2hk + h + k) + 1 that
is odd.
(3 pts) Consider N = n2 − n = n(n − 1). If N is prime, then n = ±1 or
n − 1 = ±1. But n = 1 and n − 1 = −1 give N = 0 that is not prime. Therefore the only
solutions are n = −1 (giving N = 2) and n − 1 = 1, namely n = 2 (giving N = 2).
Problem 7.
(3 pts) Notice n2 + 1 = n2 − 1 + 2 = (n + 1)(n − 1) + 2. So (n + 1)|n2 + 1 if
and only if (n + 1)|2. The solutions are n + 1 = ±1, ±2 giving n = −3, −2, 0, 1.
Problem 19.
Problem 23.
(1 pt for each part)
(a) gcd(12, 36) = 12;
(b) gcd(3, 0) = 3;
(c) gcd(6, 7) = 1.
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Problem 31.
HOMEWORK 1 SOLUTION
(2 pts for each part)
(a) gcd(14, 100) = 2 and (−7) · 14 + (1) · 100 = 2;
(b) gcd(6, 84) = 6 and (1) · 6 + (0) · 84 = 6;
(c) gcd(182, 630) = 14 and (7) · 182 + (−2) · 630 = 14;
(d) gcd(1776, 1848) = 24 and (25) · 1848 + (−26) · 1776 = 24.
Problem 34.
(1 pt for (a) and 3 pts for (b))
(a) gcd(89, 55) = 1;
(b) By denition Fn+1 = Fn + Fn−1 ; in Euclidean Algorithm we obtain Fn = 1 · Fn + Fn−1 .
Therefore all quotients are 1 and the remainders coincide with the Fibonacci numbers
from Fn−1 to F0 = 1.
Problem 37.
(1 pt for each part)
(a) 12345 = 19410 ;
(b) 101012 = 2110 ;
(c) 11111 = 13310 .
Problem 38.
(1 pt for each part)
(a) 5432110 = 10552536 ;
(b) 100000010 = 111101000010010000002 ;
(c) 3141610 = 1604107 .
From HPS, Ch. 1
Problem 1.2.
(3 pts for each part)
(a) I think that I shall never see a billboard lovely as a tree.
(b) Love is not love which alters when it alteration nds.
(c) In baiting a mousetrap with cheese always leave room for the mouse.
Total is 50 points