Page 22 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 1] NUMBERS 13

(b)Since no member of the set is greater than 1 and since there is at least one member of the set (namely 1)
which exceeds 1 for every positive number ,we see that 1 is the l.u.b. of the set.
Since no member of the set is less than 0 and since there is at least one member of the set which is
less than 0 þ for every positive (we can always choose for this purpose the number 1=n where n is a
positive integer greater than 1= ), we see that 0 is the g.l.b. of the set.
(c) Let x be any member of the set. Since we can always find a number x such that 0 < jxj < for any
positive number (e.g. we can always pick x to be the number 1=n where n is a positive integer greater
than 1= ), we see that 0 is a limit point of the set. To put this another way, we see that any deleted
neighborhood of 0 always includes members of the set, no matter how small we take > 0.
(d) The set is not a closed set since the limit point 0 does not belong to the given set.
(e) Since the set is bounded and infinite it must, by the Bolzano–Weierstrass theorem, have at least one
limit point. We have found this to be the case, so that the theorem is illustrated.

ALGEBRAIC AND TRANSCENDENTAL NUMBERS
p ffiffiffi p ffiffiffi
3
1.22. Prove that 2 þ 3 is an algebraic number.
3 3. 3 2.
p ffiffiffi p ffiffiffi p ffiffiffi p ffiffiffi 3
Let x ¼ 2 þ Then x 3 ¼ Cubing both sides and simplifying, we find x þ 9x 2 ¼
p ffiffiffi 2 6 4 3 2
3 3ðx þ 1Þ. Then squaring both sides and simplifying we find x 9x 4x þ 27x þ 36x 23 ¼ 0.
p ffiffiffi p ffiffiffi
Since this is a polynomial equation with integral coefficients it follows that 3 2 þ 3, which is a
solution, is an algebraic number.
1.23. Prove that the set of all algebraic numbers is a countable set.
n
Algebraic numbers are solutions to polynomial equations of the form a 0 x þ a 1 x n 1 þ þ a n ¼ 0
where a 0 ; a 1 ; ... ; a n are integers.
Let P ¼ja 0 jþja 1 jþ þja n jþ n. For any given value of P there are only a finite number of possible
polynomial equations and thus only a finite number of possible algebraic numbers.
Write all algebraic numbers corresponding to P ¼ 1; 2; 3; 4; .. . avoiding repetitions. Thus, all algebraic
numbers can be placed into 1-1 correspondence with the natural numbers and so are countable.

COMPLEX NUMBERS
1.24. Perform the indicated operations.
(a) ð4 2iÞþ ð 6 þ 5iÞ¼ 4 2i 6 þ 5i ¼ 4 6 þð 2 þ 5Þi ¼ 2 þ 3i
(b) ð 7 þ 3iÞ ð2 4iÞ¼ 7 þ 3i 2 þ 4i ¼ 9 þ 7i
2
(c) ð3 2iÞð1 þ 3iÞ¼ 3ð1 þ 3iÞ 2ið1 þ 3iÞ¼ 3 þ 9i 2i 6i ¼ 3 þ 9i 2i þ 6 ¼ 9 þ 7i
5 þ 5i 5 þ 5i 4 þ 3i ð 5 þ 5iÞð4 þ 3iÞ 20 15i þ 20i þ 15i 2
ðdÞ ¼ ¼ 2 ¼
4 3i 4 3i 4 þ 3i 16 9i 16 þ 9
35 þ 5i 5ð 7 þ iÞ 7 1
þ i
25 25 5 5
¼ ¼ ¼
2 2
2
2
3
4
2 2
i þ i þ i þ i þ i 5 i 1 þði ÞðiÞþði Þ þði Þ i i 1 i þ 1 þ i
1 þ i 1 þ i 1 þ i
ðeÞ ¼ ¼
i 1 i i i 2 i þ 1 1 1
þ i
1 þ i 1 i 1 i 2 2 2
¼ ¼ 2 ¼ ¼
q ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiq
2 2 2 2
ð f Þ j3 4ijj4 þ 3ij¼ ð3Þ þð 4Þ ð4Þ þð3Þ ¼ð5Þð5Þ¼ 25