Download Advanced Calculus I Fall 2004 Assignment #1 Due 9/3/04 Dr. D. P.

Assignment #1
Advanced Calculus I
Fall 2004
Due 9/3/04
Dr. D. P. Story
§1.1, pages 8–9.
Problem 1. Show that Q is closed under addition and multiplication.
Proof : Let x, y ∈ Q, then there exists integers p, q, r, s ∈ Z such that
p
r
x = , y = , q = 0, s = 0
q
s
Now, adding and multiplying these two rational together, we get,
x+y =
ps + rq
p r
+ =
,
q
s
qs
and
xy =
pr
p r
· = ,
qs
q s
ps + rq, qs ∈ Z, qs = 0
pr, qs ∈ Z, qs = 0
(1)
(2)
Lines (1) and (2) show that x + y ∈ Q and xy ∈ Q, respectively. Problem 2. Prove there is a rational number between any two distinct rational numbers.
Proof : Let x, y ∈ Q, x < y. Then there exists integers p, q, r, s ∈ Z such that
r
p
x = , y = , q = 0, s = 0
q
s
Then w = (x + y)/2 is a rational number between x and y. Indeed,
w=
x+y
p
r
ps + rq
=
+
=
,
2
2q 2s
2qs
ps + rq, 2qs ∈ Z, 2qs = 0
(1)
This shows that w ∈ Q.
To show that w is between x and y we see,
x=
x+x
x+y
x+y
y+y
<
=w=
<
=y
2
2
2
2
since x < y
and this completes the proof. Problem 6. Prove the reciprocal of an irrational number is an irrational number.
Proof : We shall prove the contrapositive. Let x ∈ R such that 1/x ∈ Q. Show x ∈ Q.
Indeed, since 1/x ∈ Q, ∃ p, q ∈ Z, q = 0 such that 1/x = p/q. Since 1/x ∈ Q, we see
that x = 0, hence p = 0. Finally
p
q
1
=
=⇒ x = , p = 0
x
q
p
Line (1) implies x ∈ Q, which proves the contrapositive. (1)
Page -2Problem 7. Prove that if x ∈ Q and y ∈
/ Q, then x + y ∈
/ Q.
Proof : Proof by contradiction. Assume x + y ∈ Q, then
y = (x + y) − x ∈ Q,
from Problem #1
Contradiction, since we are assuming y ∈
/ Q. Problem 9. If x ∈
/ Q, there exists a y ∈
/ Q such that xy ∈ Q.
Proof : Since x ∈
/ Q, we see that x = 0. Put y = 1/x, by Problem 6, y ∈
/ Q. Finally,
xy = x(1/x) = 1 ∈ Q
and this completes the proof. √
2−x∈
/ Q or 2 + x ∈
/ Q.
√
√
Proof : Proof by contradiction. Suppose x ∈ R such that 2 − x ∈ Q and 2 + x ∈ Q.
Since each of these two numbers is rational, their sum is rational (Problem 1):
√
√
( 2 − x) + ( 2 + x) ∈ Q
(1)
√
√
√
but ( 2 − x) + ( 2 + x) = 2 2 ∈
/ Q. This contradicts (1)! Problem 10. Show that for each x ∈ R, either
√
√
Problem 11. Suppose n ∈ N is not a perfect square, prove that n is irrational.
√
Proof : Proof
by
contradiction.
Assume
√n is rational. Then, there exists p, q ∈ N
√
such that n = p/q, or, in other words, q n is an integer. Let
√
S = {q ∈ N | q n ∈ N}
(1)
Now, S is a nonempty set of natural numbers, so by the Well Ordering Principle, S
has a least element, call this√q0 ∈ S.√Since q0 is the least element of S we see that
q0 ≥ 1. If q0 = 1 this means n = q0 n ∈ N, which it is not (since n is not a perfect
square). We deduce, therefore, q0 ≥ 2.
√
√
Since √n is not an integer, let
√ k ∈ N be the greatest integer less than n, and define
r = q0 ( n − k). Note that ( n − k) > 0 and so
√
√
r = q0 ( n − k) = q0 n − q0 k ∈ N
(r is an√integer since it is the difference of of
√ two integers, and positive since, as noted
above n − k > 0). Also note that since ( n − k) < 1, it follows that
√
r = q0 ( n − k) < q0
(2)
We have constructed a natural number r less than q0 , and now for the surprise: note
also that
√
√
√
√
r n = q0 ( n − k) n = q0 n − k(q0 n) ∈ Z+ = N
This shows r ∈ S. but from (2), r < q0 . Recall q0 is the least element of S, as defined
in (1)! Contradiction. Page -3Solution Notes√1 : Mr. King Yi was able to prove this using a slightly different argument. Write n = x/y, where x, y ∈ N, y = 0, where x/y is reduced to its lowest
terms. (In particular, y is as small as possible.) Use the division algorithm to write
x = qy + r, where q, r ∈ N and, noting that r = 0, we see that 0 < r < y. Finally, he
reasons,
ny 2 = x2 =⇒ ny 2 − qxy = x2 − qxy
=⇒ y(ny − qx) = x(x − qy)
x
ny − qx
=⇒
=
y
x − qy
ny − qx
x
=
since r = x − qy
=⇒
y
r
In other words, the fraction x/y has been reduced even further since r < y. Contradiction. Solution Notes 2 : It is possible to solve this problem using the Prime Factorization
Theorem, but this is a less desirable method of proof.