## Real Analysis – Exercises

— 1. Words of Caution —

In my book of Quantum Mechanics by Sakurai at a given point in the preface it is said something like: “The student who has read the book but can’t do the exercises has learned nothing!”. And that is a true statement if I ever saw a true statement.

How many times I’ve heard people say: “I understand the theory, but I just can’t do the exercises…” This type of feeling is wrong and counter-productive.

Wrong because if the person had indeed understood the theory he/she would be able to solve more exercises than just the trivial ones, and counter-productive because hiding behind that sham will only make the student’s case worse and worse.

The current subject matter always builds from the previous subject matter and this is a snowball of I understand the theory, just can’t do the exercises that keeps on growing.

Instead, if people are grown up enough to understand that they don’t have to understand everything immediately, and most times have to work hard trying to understand what took many years of the best minds around to accomplish, positive results could be achieved.

Alas!, enough with the moral high ground and let’s get started with an integral part of this blog – Exercises!

Every now and then mid-subject exercises will appear and after the class notes are finished a batch of solved exams will be presented.

As a final thought I ask the reader not to immediately read my solution of the exercises but try to work through the exercises first and then go on to my solution and compare them both.

Even if you can’t do it by yourself that effort you put it into it at first will help you understand better what I’ve done.

Also bear in mind that the solutions that I post here are in no way unique or the best solutions, and in some instances they may even be wrong (though I hope this won’t happen too often).

— 2. Exercises —

 Exercise 1 Using Axiom I to Axiom V prove the following statements. ${ -0=0}$ ${0+(-0)=-0}$ by Axiom IV and ${ 0+(-0)=0}$ by Axiom V. Since the left-hand sides of both equalities are equal so must be their right-hand sides. Thus ${ -0=0}$. ${ 1^{-1}=1}$ ${ 1\cdot 1^{-1}=1^{-1}}$ by Axiom IV and ${ 1\cdot 1^{-1}=1}$ by Axiom V. Since the left-hand sides of both equalities are equal so must be their right-hand sides. Thus ${ 1^{-1}=1}$. ${ -(-x)=x}$ The symmetric of ${ (-x)}$ is just ${ -(-x)}$, and by Axiom V ${ -(-x)+(-x)=0}$. But we know that ${ x+(-x)=0}$ and so ${ x}$ is the symmetric of ${ -x}$. Since symmetric of ${ -x}$ is unique we must have ${ -(-x)=x}$. ${ (x^{-1})^{-1}=x}$ ${ (x^{-1})^{-1}\cdot x^{-1}=1}$. But we also have ${ x\cdot x^{-1}=1}$. The reciprocal for any real number (different of zero, of course) is also unique, thus both equalities imply ${ (x^{-1})^{-1}=x}$. ${ x\cdot y=x\cdot z \Rightarrow y=z}$ if ${ x\neq 0}$. $\displaystyle \begin{array}{rcl} x\cdot y &=& x\cdot z \\ x^{-1}\cdot (x\cdot y) &=& x^{-1}\cdot (x\cdot z) \\ (x^{-1}\cdot x)\cdot y &=& (x^{-1}\cdot x)\cdot z \\ 1\cdot y &=& 1\cdot z \\ y &=& z \end{array}$ ${ \dfrac{u}{v}\cdot \dfrac{x}{y}=\dfrac{u\cdot x}{v\cdot y}}$. $\displaystyle \begin{array}{rcl} \dfrac{u}{v}\cdot \dfrac{x}{y} &=& (u\cdot v^{-1})(x\cdot y^{-1}) \\ &=& u\cdot (v^{-1}x)\cdot y^{-1} \\ &=& u\cdot (x\cdot v^{-1})\cdot y^{-1} \\ &=& (u\cdot x)(v^{-1}\cdot y^{-1}) \\ &=& (u\cdot x)(vy)^{-1} \\ &=& \dfrac{u\cdot x}{v\cdot y} \end{array}$ For the more rigor inclined certainly a step in the previous demonstration seemed a bit fuzzy. Namely when we stated ${ v^{-1}y^{-1}=(vy)^{-1}}$. In the interest of completeness let us prove the previous equality. $\displaystyle \begin{array}{rcl} (vy)(v^{-1}y^{-1})&=& v(yv^{-1})y^{-1} \\ &=& v(v^{-1}y)y^{-1} \\ &=& (vv^{-1})(yy^{-1}) \\ &=& 1\times 1 \\ &=& 1 \end{array}$ ${ (vy)(vy)^{-1}=1}$ by the definition of reciprocal element. Thus ${ (vy)(v^{-1}y^{-1})=1}$ and ${ (vy)(vy)^{-1}=1}$. Hence ${ v^{-1}y^{-1}=(vy)^{-1}}$.

 Exercise 2 ${ x > 0 \Leftrightarrow x^{-1} > 0}$. Let us prove this proposition by the method of proof by contradiction. That is we’ll prove that ${ x > 0 \land x^{-1} 0 \land x < 0}$ also leads to a contradiction. First we prove the necessary condition: ${ x > 0}$, and ${ x^{-1} < 0}$. Multiplying both sides of the first inequality by ${ x^{-1}}$ we have by Axiom V and Theorem VII ${ x\cdot x^{-1} < 0\cdot x^{-1} \Leftrightarrow 1 < 0}$ which is a false statement. The proof for the sufficient condition is exactly the same and so validity of ${ x > 0 \Leftrightarrow x^{-1} > 0}$ is established. Notice that it is an equivalence relationship we intend to prove, so either we prove it by equivalence symbols or we prove it, as we did, the implications in either way. In this instance I think most people will understand what I mean if I express it symbolically. Imagine that we wanted to prove that ${ A \Leftrightarrow B}$ is a true proposition. One way to do that would be to prove ${ A \Leftrightarrow A' \Leftrightarrow A'' \Leftrightarrow \cdots \Leftrightarrow A^{(n)} \Leftrightarrow \cdots \Leftrightarrow B}$. This way we would go through an ${ n}$ number of equivalent propositions between ${ A}$ and ${ B}$ and since the equivalence relationship is transitive we get that ${ A \Leftrightarrow B}$. Another way to do that is to prove that we can get from ${ A}$ to ${ B}$ and from ${ B}$ to ${ A}$. Symbolically ${ A \Rightarrow P_1 \Rightarrow P_2 \Rightarrow \cdots \Rightarrow P_n \Rightarrow \cdots \Rightarrow B}$ and ${ B \Rightarrow Q_1 \Rightarrow Q_2 \Rightarrow \cdots \Rightarrow Q_n \Rightarrow \cdots \Rightarrow A}$. I’ll give the best example I was given in order to understand what a necessary and sufficient condition means. Lets us suppose you’re doing an exam whose range of grades goes from ${ 0\,}$ to ${ 20}$, ${ 0\,}$ being ${ 10}$ the minimal grade to pass the exam. For you to pass the exam it is sufficient for you to have a grade greater than ${ 15}$, but thank goodness it isn’t necessary! And if you passed, your grade necessarily had to be greater than ${ 5}$, but unfortunately having a grade greater than ${ 5}$ isn’t sufficient for you to pass. ${ x > 1 \Leftrightarrow x^{-1}\in \rbrack 0,1\lbrack}$ If ${ x > 1}$ then ${ x > 0}$ also and by the previous exercise ${ x^{-1} > 0}$. Thus $\displaystyle \begin{array}{rcl} x &>& 1 \\ x\cdot x^{-1} &>& 1\cdot x^{-1} \\ 1 &>& x^{-1} \end{array}$ So ${ x^{-1} > 0}$ and ${ x^{-1} <1 }$. Taking both inequalities into account we write ${ 0 < x^{-1} < 1 \Leftrightarrow x^{-1} \in \rbrack 0,1 \lbrack}$

 Exercise 3 The factorial of a natural number ${ n}$ can be defined by recursion by the following relationships: ${ 0!=1}$ and ${ (n+1)!=n!\times(n+1)}$. ${ \dbinom{n}{k}=\displaystyle\dfrac{n!}{k!(n-k)!}}$ denotes the binomial coefficient which is the number of ways you can choose ${ k}$ elements from an ${ n}$ element set. Prove that ${ \dbinom{n}{k}+\dbinom{n}{k+1}=\dbinom{n+1}{k+1}}$ $\displaystyle \begin{array}{rcl} \dbinom{n}{k}+\dbinom{n}{k+1} &=& \dfrac{n!}{k!(n-k)!}+\dfrac{n!}{(k+1)!(n-k+1)!} \\ &=& \dfrac{k!(n-k-1)!n!(k+1+n-k)!}{k!(n-k-1)!(n-k)!(k+1)!} \\ &=& \dfrac{n!(n+1)}{(k+1)!(n-k)!} \\ &=& \dbinom{n+1}{k+1} \end{array}$ Prove that ${ (x+y)^n=\displaystyle\sum_{k=0}^n \binom{n}{k} x^k y^{n-k}}$ by mathematical induction. For ${ n=0}$ it is $\displaystyle \begin{array}{rcl} (x+y)^0 &=& \displaystyle \sum_{k=0}^0 \binom{0}{k} x^k y^{0-k} \\ 1 &=& 1 \end{array}$ which is a true statement. Let us now suppose that we have for some ${ n \in \mathbb{N}}$ ${ (x+y)^n= \displaystyle \sum_{k=0}^n \binom{n}{k} x^k y^{n-k}}$ $\displaystyle \begin{array}{rcl} (x+y)^{n+1} &=& (x+y)^n (x+y) \\ &=& x(x+y)^n+y(x+y)^n \\ &=& x\displaystyle\sum_{k=0}^n \dbinom{n}{k} x^k y^{n-k}+y\displaystyle\sum_{l=0}^n \dbinom{n}{l} x^l y^{n-l} \\ &=& \displaystyle\sum_{k=0}^n \dbinom{n}{k} x^{k+1} y^{n-k}+ \displaystyle\sum_{l=0}^n \dbinom{n}{l} x^l y^{n+1-l} \end{array}$ Making the change of variables ${ l=k+1}$ in the first summation symbol: ${ \displaystyle \sum_{l=1}^{n+1} \dbinom{n}{l-1}x^l y^{n-l+1}=\displaystyle \sum_{l=1}^{n} \dbinom{n}{l-1}x^l y^{n-l+1}+x^{n+1} }$ And noting that ${\displaystyle\sum_{l=0}^n \dbinom{n}{l} x^l y^{n+1-l}=y^{n+1}+\displaystyle\sum_{l=1}^n \dbinom{n}{l} x^l y^{n+1-l}}$ we can write $\displaystyle \begin{array}{rcl} (x+y)^{n+1} &=& \displaystyle\sum_{k=0}^n \dbinom{n}{k} x^{k+1} y^{n-k}+ \displaystyle\sum_{l=0}^n \dbinom{n}{l} x^l y^{n+1-l} \\ &=& \displaystyle \sum_{l=1}^{n} \dbinom{n+1}{l}x^l y^{n-l+1}+x^{n+1}+y^{n+1} \end{array}$ Now ${ x^{n+1}}$ is ${ \dbinom{n+1}{l}x^l y^{n-l+1}}$ with ${ l=n+1}$ and ${ y^{n+1}}$ is ${ \dbinom{n+1}{l}x^l y^{n-l+1}}$ with ${ l=0}$. Hence we can write ${(x+y)^{n+1}= \displaystyle\sum_{l=0}^{n+1} \binom{n+1}{l}x^l y^{n-l+1}}$ which finishes our proof.

 Exercise 4 Prove by induction ${ \displaystyle\sum_{n=1}^{n}\dfrac{1}{r^2}\leq 2-\dfrac{1}{n}}$. For ${ n=1}$ it is ${ \displaystyle\sum_{n=1}^{1}\dfrac{1}{r^2}\leq 2-\dfrac{1}{1} \Leftrightarrow 1\leq 1}$ which is a valid statement. Now we want to prove that the validity ${ \displaystyle\sum_{n=1}^{n+1}\dfrac{1}{r^2}\leq 2-\dfrac{1}{n+1}}$ follows from the validity of ${ \displaystyle\sum_{n=1}^{n}\dfrac{1}{r^2}\leq 2-\dfrac{1}{n}}$ for some ${ n \in \mathbb{N}}$. $\displaystyle \begin{array}{rcl} \displaystyle\sum_{r=1}^{n+1}\dfrac{1}{r^2} &=& \sum_{r=1}^{n}\dfrac{1}{r^2}+\dfrac{1}{(n+1)^2} \\ &\leq& 2-\dfrac{1}{n}+\dfrac{1}{(n+1)^2} \\ &=& 2-\dfrac{1}{n}+\dfrac{1}{(n+1)^2}+\dfrac{1}{n+1}-\dfrac{1}{n+1} \\ &=& 2-\dfrac{1}{n+1}+\dfrac{-(n+1)^2+n+n(n+1)}{n(n+1)^2} \\ &=& 2-\dfrac{1}{n+1}+\dfrac{-n^2-2n-1+n+n^2+n}{n(n+1)^2} \\ &=& 2-\dfrac{1}{n+1}-\dfrac{1}{n(n+1)^2}\leq 2-\dfrac{1}{n+1} \end{array}$ In conclusion ${ \displaystyle\sum_{n=1}^{n}\dfrac{1}{r^2}\leq 2-\dfrac{1}{n}}$ ${ \forall n \in \mathbb{N}: n > 1}$

 Exercise 5 Prove by induction on ${ n}$ that ${ \displaystyle\sum_{r=1}^{n}r(r+1)\cdots(r+k-1)=\dfrac{n(n+1)\cdots(n+k)}{k+1}}$ For ${ n=1}$ it is $\displaystyle \begin{array}{rcl} \displaystyle \sum_{r=1}^1 r(r+1)(r+2)\cdots(r+k-1) &=& \dfrac{1.2.3\cdots (1+k)}{k+1} \\ \displaystyle 1\times 2\times 3\times\cdots\times k &=& \dfrac{(k+1)!}{k+1} \\ k! &=& k! \end{array}$ Now our inductive hypothesis is ${\displaystyle\sum_{r=1}^{n}r(r+1)\cdots(r+k-1)=\displaystyle\dfrac{n(n+1)\cdots(n+k)}{k+1}}$ and we want to prove ${ \displaystyle\sum_{r=1}^{n+1}r(r+1)\cdots(r+k-1)=\displaystyle\dfrac{(n+1)(n+2)\cdots(n+1+k)}{k+1}}$ It is $\displaystyle \begin{array}{rcl} \displaystyle \sum_{r=1}^{n+1}r(r+1)(r+2)\cdots(r+k-1) \\ \displaystyle \sum_{r=1}^{n}r(r+1)\cdots (r+k-1)+(n+1)(n+2)\cdots(n+k) \\ \dfrac{n(n+1)(n+2)\cdots(n+k)}{k+1}+(n+1)(n+2)\cdots(n+k) \\ (n+1)(n+2)\cdots(n+k)\left\lbrack \dfrac{n}{k+1}+1 \right\rbrack \\ (n+1)(n+2)\cdots(n+k)\left\lbrack \dfrac{n+k+1}{k+1}\right\rbrack \\ \dfrac{(n+1)(n+2)\cdots(n+k)(n+k+1)}{k+1} \end{array}$

### 6 Responses to “Real Analysis – Exercises”

1. you lost me where it said 1. Using AI to AV ….good luck!

2. Hi wesley. When I said: “Using AI to AV” it was short for: “Using Axiom I to Axiom V”. And in the other bit where I mention TVII it is Theorem VII. But I just didn’t want to type all of that.

Any more comments are more than welcomed so that any other point that wasn’t made explicit enough can be explained.

3. hey, can we pause the snow kinda distracted me when i tried to understand that stuff tnx,
for the statement above you should tell what the abbreviation stands for the first time you use it so if some one does not know they could refer to that(you lost me there too lol)
so far no luck getting this stuff ill try some more tomorrow.

4. Ok. The post is updated. No more shady acronyms for now on.

5. […] was proven in here we can write . Writting out the terms we […]

6. he as got a pc program that makes any thing realated with maths