Rudin数学分析原理第一章答案
The Real and Complex Number
Systems
Written by Men-Gen Tsai
email: b89902089@ntu.edu.tw
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6. Fix b > 1.
(a) If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q , prove
that
(bm )1/n = (bp )1/q .
Hence it makes sense to de?ne br = ...
The Real and Complex Number
Systems
Written by Men-Gen Tsai
email: b89902089@ntu.edu.tw
1.
2.
3.
4.
5.
6. Fix b > 1.
(a) If m, n, p, q are integers, n > 0, q > 0, and r = m/n = p/q , prove
that
(bm )1/n = (bp )1/q .
Hence it makes sense to de?ne br = (bm )1 /n .
(b) Prove that br + s = br bs if r and s are rational.
(c) If x is real, de?ne B (x ) to be the set of all numbers bt , where t is
rational and t ≤x . Prove that
br = sup B (r )
where r is rational. Hence it makes sense to de?ne
bx = sup B (x )
for every real x.
(d) Prove that bx+ y = bxby for all real x and y.
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Proof: For (a): mq = np since m/n = p/q . Thus bmq = bnp .
By Theorem 1.21 we know that (bmq )1/ (mn ) = (bnp ) 1/ (mn ) , that is,
(bm )1/n = (bp ) 1/q , that is, br is well-de?ned.
For (b): Let r = m/n and s = p/q where m, n, p, q are integers, and
n > 0, q > 0. Hence (br + s )nq = (bm/n + p/q )nq = (b(mq + np ) / (nq ) )nq =
bmq + np = bmq bnp = (bm/n )nq (bp/q )nq = (bm/n bp/q ) nq . By Theorem 1.21
we know that (( br + s)nq )1/ (nq ) = (( bm/n bp/q )nq ) 1/ (nq ), that is br + s =
bm/n bp/q = br bs.
For (c): Note that br ∈B ( r ). For all bt ∈B ( r ) where t is rational and
t ≤ r . Hence, br = bt br - t ≥ bt 1r - t since b > 1 and r - t ≥ 0. Hence br
is an upper bound of B (r ). Hence br = sup B ( r ).
For (d): bx by = sup B (x ) sup B (y) ≥ bt x bt y = btx + t y for all rational
tx ≤ x and t y ≤ y. Note that t x + ty ≤ x + y and tx + t y is rational.
Therefore, sup B (x) sup B (y) is a upper bound of B (x + y), that is,
bx by ≥ sup B (x + y) = b( x + y ).
Conversely, we claim that bx br = bx + r if x ∈ R 1 and r ∈ Q. The
following is my proof.
bx + r = sup B (x + r ) = sup{ bs : s ≤ x + r, s ∈Q}
= sup{ bs- r br : s - r ≤ x, s - r ∈Q}
= br sup{ bs- r : s - r ≤ x, s - r ∈Q}
= br sup B (x)
= br bx .
And we also claim that bx+ y ≥ bx if y ≥ 0. The following is my proof:
2
(r ∈Q )
B (x) = { br : r ≤ x} ? { br : r ≤ x + y} = B (x + y ),
Therefore, sup B (x + y) ≥ sup B (x ), that is, bx + y ≥ bx .
Hence,
bx+ y = sup B (x + y)
= sup{ br : r ≤ x + y, r ∈Q}
= sup{ bsbr - s : r ≤ x + y, s ≤ x, r ∈Q, s ∈Q}
≥ sup{ sup B (x )br - s : r ≤ x + y, s ≤ x, r ∈Q, s ∈Q}
= sup B (x ) sup{ br - s : r ≤ x + y, s ≤ x, r ∈Q, s ∈Q}
= sup B (x ) sup{ br - s : r - s ≤x + y - s, s ≤x, r - s ∈Q}
= sup B (x ) sup B (x + y - s)
≥ sup B (x ) sup B (y)
= bx by
Therefore, bx + y = bx by .
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