638 Appendix C 11 Answers & Hints for Selected Exercises
- k(3k 2 -1) + 3k + 1 -- 3k^2 +5k+2 2 - - (k+l)(3k+2) 2 - - (k+1)(3(k+l)-1) 2.
- Assume k^5 - k = 5m for some m E Z. Then (k + 1)^5 - (k + 1) =
k^5 + 5k^4 + l0k^3 + 10k^2 + 5k + 1 - k - 1 = ( k^5 - k) + ( 5k^4 + l0k^3 + 10k^2 + 5k)
= 5[m + k^4 + 2k^3 + 2k^2 + k]. - l+~+i+···+ 3k\1=l+~+i+···+ 3
1
k + 3k\l = [~ - ~ (f,o-)] + 3k\1
- 3 + - 2 2.3k+l 3 +2 -- 3 2 - 2 1 3k+ (^1) 1.
- 2k+^1 = 2k2:::; (k + 1)!2:::; (k + l)!(k + 2) = (k + 2)!.
- (1 + x)k+^1 = (1 + x)k(l + x) 2 [1 + kx + ~k (k - l)x^2 ](1 + x) since x 2 0.
=l+kx+~k(k -l)x^2 +x +kx^2 +~k(k-l)x^3
2 1 + (k + l)x + [~k (k - 1) + k ]x^2 = 1 + (k + l)x + [~k^2 + ~k ]x^2
= 1 + (k + l)x + ~(k + l)kx^2. - Assume 22 k-l + 1 = 3m for some m E N. Then 22 (k+l)-l + 1 = 22 k+l + 1
= 2222 k-l + 1 = 4(3m - 1) + 1 = 12m - 4 + 1 = 3(4m - 1).
19. xk+l -yk+l = xkx -yky = xkx -ykx +ykx -yky = x(xk -yk) +yk(x -y)
= x(x _ y)(xk-1 + xk-2y + xk-3y2 + ... + xyk- 2 + yk-1) + yk(x _ y)
= (x _ y)(xk + xk-ly + xk-2y2 + ... + xyk-1 + yk). ·EXERCISE SET 1.4- Let n E Z. n not divisible by 2 => 3k E Z 3 n = 2k+l => n^2 = 4k^2 +4k+l =>
n^2 not divisible by 2. - Suppose xis rational and y is irrational, and let z = x + y. Then y = z - x.
If z is rational than so is y, which would be a contradiction. - Let x be irrational. Since x+(-x) = 0, Exercise 5 says -x cannot be rational.
Since x(x-^1 ) = 1, Ex. 6 says x-^1 cannot be rational.
9. .J2. .J2 = 2.- 't/n E N, n+.J2 is irrational, by Ex. 6. Moreover, n+.J2 = m+.J2 => n = m,
so there are infinitely many such irrational numbers.
EXERCISE SET 1.5- x E Q => x = % for some a E Z , b E N. Then lal + 1 > l~I 2 % = x and
!al+ 1 EN. - Assume F has A.P., and a> 0. Then 't/x E F, ~ E F so by A.P. :Jn EN 3
n > ~· Since a> 0, this means na > x. - Suppose x E Archimedean F. First prove existence. If x is an integer, take
n = x + 1. If x is not an integer then x > 0 or x < 0. The first case is covered