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### rudin数学分析原理答案解析 #### Problem 1.1: 实数与有理数的性质 **题目：** 若\(r \in \mathbb{Q} \backslash \{0\}\)（即\(r\)是有理数且不为0），\(x \in \mathbb{R} \backslash \mathbb{Q}\)（即\(x\)是无理数），证明\(r + x\)及\(rx\)均不属于\(\mathbb{Q}\)。 **解析：** 本题通过反证法进行证明。 1. **对于\(r + x\)：** - 假设\(r + x \in \mathbb{Q}\)，则根据\(\mathbb{R}\)和\(\mathbb{Q}\)的域性质，我们有\(x = (r + x) - r\)。 - 因为\(r \in \mathbb{Q}\)，所以\((r + x) - r \in \mathbb{Q}\)。 - 这意味着\(x \in \mathbb{Q}\)，与前提矛盾。因此\(r + x \not\in \mathbb{Q}\)。 2. **对于\(rx\)：** - 同样假设\(rx \in \mathbb{Q}\)。 - 根据题意\(r \neq 0\)，可以得到\(x = \frac{rx}{r}\)。 - 由于\(r, rx \in \mathbb{Q}\)，根据\(\mathbb{Q}\)的封闭性，\(\frac{rx}{r} \in \mathbb{Q}\)。 - 这与\(x\)是无理数相矛盾，因此\(rx \not\in \mathbb{Q}\)。 **结论：** 由以上证明知，如果\(r\)是有理数且\(x\)是无理数，则\(r + x\)及\(rx\)均为无理数。 --- #### Problem 1.2: 不存在满足\(x^2 = 12\)的有理数 **题目：** 证明不存在\(x \in \mathbb{Q}\)使得\(x^2 = 12\)。 **解析：** 本题同样采用反证法进行证明。 1. **假设存在这样的\(x\)：** - 设\(x = \frac{m}{n}\)，其中\(m, n \in \mathbb{Z}\)，并且\(\frac{m}{n}\)是最简形式。 - 由题意知\(x^2 = 12\)，即\(\left(\frac{m}{n}\right)^2 = 12\)，从而\(m^2 = 12n^2\)。 - 因为\(12n^2\)能被3整除，所以\(m^2\)也能被3整除。 - 由于3是质数，并考虑到\(\mathbb{Z}\)中的唯一分解定理，可以得出3一定能整除\(m\)。 - 由此可得\(m^2 = 9k^2 = 12n^2\)（其中\(k = \frac{m}{3}\)）。 - 进一步分析可知，\(12n^2\)能被9整除，这意味着\(n^2\)能被3整除，从而\(n\)也能被3整除。 - 这与最简分数\(\frac{m}{n}\)的定义相矛盾，因为\(m\)和\(n\)不能同时被同一个数整除。 **结论：** 不存在\(x \in \mathbb{Q}\)使得\(x^2 = 12\)。 --- #### Problem 1.5: 集合的下确界与上确界的性质 **题目：** 设\(A \subset \mathbb{R}\)非空且有下界，设\(-A = \{-a : a \in A\}\)。证明\(\inf(A) = -\sup(-A)\)。 **解析：** 首先证明\(-A\)非空且有上界。 1. **\(-A\)非空且有上界：** - \(A\)中至少有一个元素\(x\)，因此\(-x \in -A\)。 - \(A\)有下界\(m\)，那么\(-m\)就是\(-A\)的一个上界。接下来证明\(\inf(A) = -\sup(-A)\)。 1. **\(\inf(A) = -\sup(-A)\)的证明：** - 设\(\sup(-A) = b\)，需证明\(\inf(A) = -b\)。 - \(-b\)是\(A\)的一个下界，因为\(b\)是\(-A\)的上界。 - 要证明\(-b\)是最大的下界，假设\(c > -b\)，需要证明\(c\)不是\(A\)的下界。 - 由\(c > -b\)得\(-c < b\)，故\(-c\)不是\(-A\)的上界。 - 存在\(-a \in -A\)使得\(-a > -c\)，即\(a < c\)。 - 因此\(c\)不是\(A\)的下界。 **结论：** 综合以上分析，当\(A\)非空且有下界时，\(\inf(A) = -\sup(-A)\)。 --- ### 总结以上三个问题分别探讨了实数、有理数和集合的基本性质。通过严格的逻辑推导和反证法的应用，这些问题不仅加深了对这些基本概念的理解，还展示了数学证明的重要性和严谨性。这些问题的解答不仅对于理解数学分析的基本原理非常重要，而且也为进一步学习更高级的数学概念打下了坚实的基础。

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MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1

Generally, a “solution” is something that would be acceptable if turned in in the

form presented here, although the solutions given are often close to minimal in this

respect. A “solution (sketch)” is too sketchy to be considered a complete solution

if turned in; varying amounts of detail would need to be ﬁlled in.

Problem 1.1: If r ∈ Q \{0} and x ∈ R \ Q, prove that r + x, rx 6∈ Q.

Solution: We prove this by contradiction. Let r ∈ Q\{0}, and suppose that r +x ∈

Q. Then, using the ﬁeld properties of both R and Q,wehavex =(r + x) −r ∈ Q.

Thus x 6∈ Q implies r + x 6∈ Q.

Similarly, if rx ∈ Q, then x =(rx)/r ∈ Q. (Here, in addition to the ﬁeld

properties of R and Q,weuser 6= 0.) Thus x 6∈ Q implies rx 6∈ Q.

Problem 1.2: Prove that there is no x ∈ Q such that x

= 12.

Solution: We prove this by contradiction. Suppose there is x ∈ Q such that

= 12. Write x =

in lowest terms. Then x

= 12 implies that m

=12n

Since 3 divides 12n

, it follows that 3 divides m

. Since 3 is prime (and by unique

factorization in Z), it follows that 3 divides m. Therefore 3

divides m

=12n

Since 3

does not divide 12, using again unique factorization in Z and the fact that

3 is prime, it follows that 3 divides n. We have proved that 3 divides both m and

n, contradicting the assumption that the fraction

is in lowest terms.

Alternate solution (Sketch): If x ∈ Q satisﬁes x

= 12, then

is in Q and satisﬁes

= 3. Now prove that there is no y ∈ Q such that y

= 3 by repeating the

proof that

√

2 6∈ Q.

Problem 1.5: Let A ⊂ R be nonempty and bounded below. Set −A = {−a: a ∈

A}. Prove that inf(A)=−sup(−A).

Solution: First note that −A is nonempty and bounded ab ove. Indeed, A contains

some element x, and then −x ∈ A; moreover, A has a lower bound m,and−m is

an upper bound for −A.

We now know that b = sup(−A) e xists. We show that −b = inf(A). That −b is

a lower bound for A is immediate from the fact that b is an upper bound for −A.

To show that −b is the greatest lower bound, we let c>−b and prove that c is not

a lower bound for A.Now−c<b,so−c is not an upper bound for −A. So there

exists x ∈−A such that x>−c. Then −x ∈ A and −x<c.Soc isnotalower

bound for A.

Problem 1.6: Let b ∈ R with b>1, ﬁxed throughout the problem.

Comment: We will assume known that the function n 7→ b

,fromZ to R,is

strictly increasing, that is, that for m, n ∈ Z,wehaveb

if and only if

m<n. Similarly, we take as known that x 7→ x

is strictly increasing when n is

Date: 1 October 2001.

MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1 3

also r =

and s =

, with nq > 0, so

=(b

)

1/(nq)

and b

=(b

)

1/(nq)

Now mq < np because r<s. Therefore, using the deﬁnition of c

1/(nq)

)

= b

=(b

)

Since x 7→ x

is strictly increasing, this implies that b

Now we can prove that if r ∈ Q then b

= sup(B(r)). By the above, if s ∈ Q

and s ≤ r, then b

≤ b

. This implies that b

is an upper bound for B(r). Since

∈ B(r), obviously no number smaller than b

can be an upp er bound for B(r).

So b

= sup(B(r)).

We now deﬁne b

= sup(B(x)) for every x ∈ R. We need to show that B(x)

is nonempty and bounded above. To show it is nonempty, choose (using the

Archimedean property) some k ∈ Z with k<x; then b

∈ B(x). To show it

is bounded above, similarly choose some k ∈ Z with k>x.Ifr ∈ Q ∩ (−∞,x],

then b

∈ B(k) so that b

≤ b

by Part (c). Thus b

is an upper bound for B(x).

This shows that the deﬁnition makes sense, and Part (c) shows it is consistent with

our earlier deﬁnition when r ∈ Q.

(d) Prove that b

x+y

= b

for all x, y ∈ R.

Solution:

In order to do this, we are going to need to replace the set B(x)abovebythe

set

(x)={b

: r ∈ Q ∩ (−∞,x)}

(that is, we require r<xrather than r ≤ x) in the deﬁnition of b

.(Ifyouare

skeptical, read the main part of the solution ﬁrst to see how this is used.)

We show that the replacement is possible via some lemmas.

Lemma 1. If x ∈ [0, ∞)andn ∈ Z satisﬁes n ≥ 0, then (1 + x)

≥ 1+nx.

Proof: The proof is by induction on n. The statement is obvious for n =0. So

assume it holds for some n. Then, since x ≥ 0,

(1 + x)

n+1

=(1+x)

(1 + x) ≥ (1 + nx)(1 + x)

=1+(n +1)x + nx

≥ 1+(n +1)x.

This proves the result for n +1.

Lemma 2. inf{b

1/n

: n ∈ N} = 1. (Recall that b>1andN = {1, 2, 3,...}.)

Proof: Clearly 1 is a lower bound. (Indeed, (b

1/n

)

= b>1=1

,sob

1/n

> 1.) We

show that 1 +x is not a lower bound when x>0. If 1 + x were a lower bound, then

1+x ≤ b

1/n

would imply (1 + x)

≤ (b

1/n

)

= b for all n ∈ N. By Lemma 1, we

would get 1 + nx ≤ b for all n ∈ N, which contradicts the Archimedean property

when x>0.

Lemma 3. sup{b

−1/n

: n ∈ N} =1.

Proof: Part (b) shows that b

−1/n

1/n

= b

= 1, whence b

−1/n

=(b

1/n

)

−1

. Since

all numbers b

−1/n

are strictly positive, it now follows from Lemma 2 that 1 is an

upper bound. Suppose x<1 is an upper bound. Then x

−1

is a lower bound for

4 MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1

1/n

: n ∈ N}. Since x

−1

> 1, this contradicts Lemma 2. Thus sup{b

−1/n

: n ∈

N} = 1, as claimed.

Lemma 4. b

= sup(B

(x)) for x ∈ R.

Proof: If x 6∈ Q, then B

(x)=B(x), so there is nothing to prove. If x ∈ Q,

then at least B

(x) ⊂ B(x), so b

≥ sup(B

(x)). Moreover, Part (b) shows that

x−1/n

= b

−1/n

for n ∈ N. The numbers b

x−1/n

are all in B

(x), and

sup{b

−1/n

: n ∈ N} = b

sup{b

−1/n

: n ∈ N}

because b

> 0, so using Lemma 3 in the last step gives

sup(B

(x)) ≥ sup{b

x−1/n

: n ∈ N} = b

sup{b

−1/n

: n ∈ N} = b

Now we can prove the formula b

x+y

= b

. We start by showing that b

x+y

≤

, which we do by showing that b

is an upper bound for B

(x + y). Thus let

r ∈ Q satisfy r<x+y. Then there are s

∈ R such that r = s

and s

<x,

<y. Choose s, t ∈ Q such that s

<s<xand t

<t<y. Then r<s+ t,so

s+t

= b

≤ b

. This shows that b

is an upper bound for B

(x + y).

(Note that this does not work using B(x + y). If x + y ∈ Q but x, y 6∈ Q, then

x+y

∈ B(x + y), but it is not possible to ﬁnd s and t with b

∈ B(x), b

∈ B(y),

and b

= b

x+y

We now prove the reverse inequality. Suppose it fails, that is, b

x+y

. Then

x+y

The left hand side is thus not an upper bound for B

(x), so there exists s ∈ Q with

s<xand

x+y

It follows that

x+y

Repeating the argument, there is t ∈ Q with t<ysuch that

x+y

Therefore

x+y

= b

s+t

(using Part (b)). But b

s+t

∈ B

(x + y) because s + t ∈ Q and s + t<x+ y, so this

is a contradiction. Therefore b

x+y

≤ b

Problem 1.9: Deﬁne a relation on C by w<zif and only if either Re(w) < Re(z)

or both Re(w)=Re(z) and Im(w ) < Im(z). (For z ∈ C, the expressions Re(z)

and Im(z) denote the real and imaginary parts of z.) Prove that this makes C an

ordered set. Does this order have the least upper bound property?

Solution: We verify the two conditions in the deﬁnition of an order. For the ﬁrst,

let w, z ∈ C. There are three cases.

Case 1: Re(w) < Re(z). Then w<z, but w = z and w>zare both false.

Case 2: Re(w) > Re(z). Then w>z, but w = z and w<zare both false.

MATH 413 [513] (PHILLIPS) SOLUTIONS TO HOMEWORK 1 5

Case 3: Re(w)=Re(z). This case has three subcases.

Case 3.1: Im(w) < Im(z). Then w<z, but w = z and w>zare both false.

Case 3.2: Im(w) > Im(z). Then w>z, but w = z and w<zare both false.

Case 3.3: Im(w)=Im(z). Then w = z, but w>zand w<zare both false.

These cases exhaust all possibilities, and in each of them exactly one of w<z,

w = z,andw>zis true, as desired.

Now we prove transitivity. Let s<wand w<z. If either Re(s) < Re(w)

or Re(w) < Re(z), then clearly Re(s) < Re(z), so s<z.IfRe(s)=Re(w)

and Re(w)=Re(z), then the deﬁnition of the order requires Im(s) < Im(w)and

Im(w) < Im(z). We thus have Re(s)=Re(z) and Im(s) < Im(z), so s<zby

deﬁnition.

It remains to answer the last question. We show that this order does not have

the least upper bound property. Let S = {z ∈ C :Re(z) < 0}. Then S 6= ∅ because

−1 ∈ S,andS is bounded above because 1 is an upper bound for S.

We show that S does not have a least upper bound by showing that if w is an

upper bound for S, then there is a smaller upper bound. First, by the deﬁnition of

the order it is clear that Re(w) is an upper bound for

{Re(z): z ∈ S} =(−∞, 0).

Therefore Re(w) ≥ 0. Moreover, every u ∈ C with Re(u) ≥ 0 is in fact an upper

bound for S. In particular, if w is an upper bound for S, then w − i<wand has

the same real part, so is a smaller upper bound.

Note: A related argument shows that the set T = {z ∈ C:Re(z) ≤ 0} also has

no least upper bound. One shows that w is an upper bound for T if and only if

Re(w) > 0.

Problem 1.13: Prove that if x, y ∈ C, then ||x|−|y|| ≤ |x −y|.

Solution: The desired inequality is equivalent to

|x|−|y|≤|x −y| and |y|−|x|≤|x − y |.

We prove the ﬁrst; the second follows by exchanging x and y.

Set z = x − y. Then x = y + z. The triangle inequality gives |x|≤|y| + |z|.

Substituting the deﬁnition of z and subtracting |y| from both sides gives the result.

Problem 1.17: Prove that if x, y ∈ R

, then

kx + yk

+ kx − yk

=2kxk

+2kyk

Interpret this result geometrically in terms of parallelograms.

Solution: Using the deﬁnition of the norm in terms of scalar products, we have:

kx + yk

+ kx − yk

= hx + y, x + yi + hx − y, x − yi

= hx, xi + hx, yi + hy, xi + hy, yi

+ hx, xi−hx, yi−hy, xi+ hy, yi

=2hx, xi+2hy, yi =2kxk

+2kyk

The interpretation is that 0,x,y,x+ y are the vertices of a parallelogram, and

that kx + yk and kx −yk are the lengths of its diagonals while kxk and kyk are each

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求志达道

2018-03-22

真是感觉不错啊！
u010829203

2017-01-21

答案很全，很好
sjtujgxxaq

2011-10-24

这份答案很全，在其他地方找的答案一般只包括了前5章，这个全部有了很好。
palandsword

2014-06-06

习题不是很全，部分题目有答案。不过是免费资源，还是很感谢youwubing上传
jzhang_ioa

2014-07-14

很好，答案比较全了。

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