Since Rudin frequently skips steps either for compactness, or to make the student work out the answers themselves, I thought I’d post my workings online so that I don’t forget them later on. I like to have everything spelled out for me, so I’ll be as detailed as possible. Here are the workings for Theorem 1.21 where I feel Rudin achieves results without telling us how he got there.

**Statement:**

$latex \text{If}\;t=\frac{x}{(1+x)}\;\text{then}\;0\le{t}<1$

**Proof:**

$latex \text{Since}\;0<1\;,\text{we have}\;x<x+1\\\\

\text{and}\;0<x<x+1,\;\text{since}\;x>0\;\text{by definition}$

By Proposition 1.18(e), we have

$latex 0<\frac{1}{x+1}<\frac{1}{x}\\\\

\text{Since} \;\frac{1}{x+1}>0\;\text{and}\;x>0,\;\text{By Definition 1.17(ii), we have}\\\\

x.\frac{1}{x+1}=\frac{x}{1+x}=t>0$

So the first part of the inequality is satisfied.

$latex \text{Again, since}\;x+1>x,\;\text{and}\;\frac{1}{x+1}>0\;\text{we have}\;\frac{1}{x+1}.(x+1)>\frac{1}{x+1}.x\;\text{by proposition 1.18(b)}\\\\

\text{Therefore}\;\frac{x}{x+1}<1\;\text{Which is the second half of the inequality. So we have}\linebreak0<t<1$

**Statement:**

$latex t^n\le t<x$

**Proof:**

From the above result:

[latex] 0<t<1\;\therefore\;t.t<t.1\implies t^2<t.\;\text{Also}\;t.0<t.t\implies t^2>0[/latex] (Again from Proposition 1.18(b))

And so,

[latex]0<t^2<t<1[/latex]

Similarly, we can obtain:

[latex]0<t^n\le t<1\;\text{Since n can be equal to 1}[/latex] So we have the first half of the inequality.

Now from Proposition 1.18(d), we have [latex]x^2>0\;\text{Since}\;x>0\implies x\neq0[/latex]

From Definition 1.17(i),

$latex x+x^2>x+0\\\\\implies x(1+x)>x

\\\\\text{Since}\;\frac{1}{x+1}>0,\;\frac{1}{x+1}.x(1+x)>\frac{x}{1+x}&s=1\\\\

\implies x>\frac{x}{1+x}\\\\

\implies x>t$

Which is the second part of the inequality. So we obtain:

$latex t^n\le t<x$

Later, Rudin introduces this identity:

$latex b^n-a^n=(b-a)(b^{n-1}+b^{n-2}a+….+a^{n-1})\\\\

\text{when }0<a<b$

This is not proved directly, but we can prove it from right to left by expanding the right hand term like so:

[latex](b-a)(b^{n-1}+b^{n-2}a+….+a^{n-1}) = b(b^{n-1}+b^{n-2}a+….+a^{n-1})-a(b^{n-1}+b^{n-2}a+….+a^{n-1})=(b^{n}+b^{n-1}a+….+ba^{n-1}) – (b^{n-1}a+b^{n-2}a^2+….+a^{n})=b^n-a^n[/latex]

(Thank you to Axiomer from the physicsforums.com for supplying me with this). Don’t ask me how someone first came up with this formula!

**Statement:**

[latex]b^n-a^n<(b-a)nb^{n-1}[/latex]

**Proof:**

$latex \text{Since }0<a<b,\;0<a^n<b^n\\\\

\implies (b^{n-1}.a^n).b>(b^{n-1}.a^n).a\\\\

\implies b^n.a^n>b^{n-1}.a^{n+1}\\\\

\text{So every successive term of }(b^{n-1}+b^{n-2}a+….+a^{n-1})\text{ is lesser than }b^{n-1}\\\\

\text{since each term reduces the power of “b”by 1 and increases the power of “a” by 1.}\\\\

\implies nb^{n-1}>(b^{n-1}+b^{n-2}a+….+a^{n-1})\\\\

\text{Since }b>a,\;(b-a)nb^{n-1}>(b-a)(b^{n-1}+b^{n-2}a+….+a^{n-1})\\\\

\implies b^n-a^n<(b-a)nb^{n-1}$

**Statement:**

$latex 0<k<y$

Rudin defines *k*:

$latex k=\dfrac{y^n-x}{ny^{n-1}},$

But never explains why $latex 0<k<y$. Here is the reason.

**Proof:**

$latex \text{Since } 1\le n,\\\\

1-n\le 0\text{. Also, }x>0\text{ and }y>0\therefore y^n>0\\\\

\implies \dfrac{x}{y^n}>0\ge 1-n\\\\

\implies \dfrac{x}{y^n}>1-n\\\\

\implies x>y^n(1-n)

\implies -x<y^n(n-1)\text{ By proposition 1.18(c) with x=-1}\\\\

\implies -x<ny^n-y^n\\\\

\implies y^n-x<ny^n\\\\

\implies \dfrac{y^n-x}{ny^n}<1\\\\

\implies \dfrac{y^n-x}{ny^{n-1}.y}<1\\\\

\implies \dfrac{y^n-x}{ny^{n-1}}<y\text{ (}y>1\implies \dfrac{1}{y}>1\text{ by proposition 1.18(d)}\text{)}\\\\

\implies k<y\\\\

\text{Also, } 0<x<y^n\text { (by assumption) and so }y^n-x>0\text{ and }ny^{n-1}>0\\\\

\implies \dfrac{y^n-x}{ny^{n-1}}>0\implies k>0\\\\

\text{And so }0<k<y$

**Statement:**

$latex y^n-t^n\le y^n-(y-k)^n<kny^{n-1}=y^n-x\text{ if }t\ge y-k$

**Proof:**

First we’ll show that

$latex \text{If }a>b,\text{ then }a^n>b^n\text{ when } 0<b<a$

$latex a.a>b.a\text{ from proposition 1.18(b)}\\\\

\text{and }b.a>b.b\\\\

\implies a^2>ba>b^2\text{. Similarly, }a^n>b^n$

$latex \text{So if }t\ge y-k,\\\\

t^n\ge (y-k)^n\\\\

\implies -t^n\le -(y-k)^n\text{ Again by proposition 1.18(c) with x=-1}\\\\

\implies y^n-t^n\le y^n-(y-k)^n\text{ And so the first part of the inequality is proved}\\\\

\text{ Next, we’ve already seen that }b^n-a^n<(b-a)nb^{n-1}\\\\

\text{Putting }b=y\text{ and }a=y-k\\\\

y^n-(y-k)^n<kny^{n-1}=y^n-x\text{ from the definition of k. And so}\\\\

y^n-t^n\le y^n-(y-k)^n<kny^{n-1}=y^n-x$

To me, these are the main missing steps in Theorem 1.21. If I’ve left anything out, let me know!

Instead of these details, it is better to use a more conceptual proof:

Consider the function f(y)=y^n on the interval [0,infty): its a continuous function so its range must be

a connected subset of [0,infty), i.e. an interval. Let us call this interval I.

Obviously, 0=f(0) is in the range of f. So, 0 is the lower limit of the interval I.

Also the range of f(y)=y^n cannot be bounded above, so the upper limit of the interval I must be infty.

Hence proved: range of f is [0,infty)

So for any positive number x in [0,infty) there is some y in [0,infty) such that f(y)=y^n=x.

In reply toAbhishekAt the stage of this book, the idea of functions hasn’t been introduced and neither have the concepts of infinity or intervals, nor continuity.

This is proof from the field axioms alone and nothing more.

This was incredibly helpful, thank you.

In reply toAlexeiGlad you found it useful!