Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I: (69) Next: Homework Assignment 7
Previous: Term Exam 1

Solution of Term Exam 1

this document in PDF: Solution.pdf

Problem 1.

  1. Prove directly from the postulates for the real numbers and from the relevant definitions that if $ a,b\geq 0$ and $ a^2<b^2$, then $ a<b$. If you plan to use a formula such as $ b^2-a^2=(b-a)(b+a)$ you don't need to prove it, but of course you have to be very clear about how it is used.
  2. Use induction to prove that any integer $ n$ can be written in exactly one of the following two forms: $ n=2k$ or $ n=2k+1$, where $ k$ is also an integer.
  3. Prove that there is no rational number $ r$ such that $ r^3=2$.


  1. If $ a=0$ then $ a^2<b^2$ means that $ 0<b^2$ and therefore $ b\neq 0$. Along with $ a\geq 0$, $ b\geq 0$ and P11, it follows that $ b+a>0$. If $ a>0$ then again along with $ b\geq 0$ and P11 we get that $ b+a>0$, so in either case $ b+a>0$ is assured. Now $ a^2<b^2$ is by definition the same as $ b^2-a^2>0$, and therefore

    $\displaystyle (b-a)(b+a)>0.$ (1)

    Had $ (b-a)$ been negative, then $ -(b-a)$ would have been positive and by $ b+a>0$ and P12 we'd have that $ -(b-a)(b+a)>0$, contradicting Equation (1). Hence $ (b-a)$ is positive, and this by definition means that $ b>a$.

  2. First we show using induction that every natural number (positive integer) $ n$ can be written in at least one of the forms $ n=2k$ or $ n=2k+1$ for an integer $ k$. Indeed, for $ n=1$ we write $ 1=2\cdot 0+1$ as required. Now if $ n$ is of the form $ 2k$ for an integer $ k$, then $ n+1=2k+1$ is of the second allowed form, and if $ n$ is of the form $ 2k+1$ for some integer $ k$ then $ n+1=2k+1+1=2(k+1)$ is of the first allowed form, for $ k+1$ is also an integer. Therefore if $ n$ can be written in either of the required forms then so is $ n+1$, and the inductive proof is completed.

    Let's deal with 0 and with the negative integers now. First, $ 0=2\cdot 0$ so 0 is of the form $ 2k$. Next, if $ n<0$ is an integer, then $ (-n)$ is a positive integer and therefore $ (-n)=2k$ or $ (-n)=2k+1$. In the former case, $ n=2(-k)$ and we are done. In the latter case, $ n=2(-k)-1=2(-k-1)+1$ and again we are done.

    Finally, an integer cannot be of both forms at the same time, for if we could write $ n=2k_1$ and $ n=2k_2+1$ with integer $ k_1$ and $ k_2$, then we'd have that $ 2k_1=2k_2+1$, which is $ 2(k_1-k_2)=1$. But $ k_1-k_2$ is an integer and it is easy to show that $ 1$ is not twice an integer.

    Having said all that, we can call the integers of the form $ 2k$ ``even'' and the integers of the form $ 2k+1$ ``odd'', and then every integer is either even or odd but never both.

  3. First, $ (2k+1)^3=8k^3+12k^2+6k+1=2(4k^3+6k^2+3k)+1$ and hence if $ n$ is odd then so is $ n^3$. Therefore if $ n^3$ is even then so is $ n$. Now assume by contradiction that there is some rational number $ r$ with $ r^3=2$ and write $ r=p/q$ where $ p$ and $ q$ are integers and $ q$ is the least positive integer for which it is possible to present $ r$ in this form. Now $ r^3=p^3/q^3=2$ hence $ p^3=2q^3$ is even hence $ p$ is even hence we can find an integer $ k$ with $ p=2k$. But then $ p^3=2q^3$ becomes $ 8k^3=2q^3$ and hence $ q^3=2(2k^3)$ so $ q^3$ is even and hence so is $ q$, so $ q=2l$ for some positive $ l$ (which of course is smaller than $ q$ itself). But now $ r=p/q=2k/2l=k/l$ contradicting the minimality of $ q$. Therefore there is no rational number $ r$ with $ r^3=2$.

Problem 2.

  1. Suppose $ f(x)=x+1$. Are there any functions $ g$ such that $ f\circ g=g\circ f$?
  2. Suppose that $ f$ is a constant function. For which functions $ g$ does $ f\circ g=g\circ f$?
  3. Suppose that $ f\circ g=g\circ f$ for all functions $ f$. Show that $ g$ is the identity function $ g(x)=x$.


  1. Yes. For example, the identity function $ g(x)=x$ has this property.

  2. Suppose $ f(x)=c$ for all $ x$. Then $ f\circ g=g\circ f$ if and only if $ \forall x  (f\circ g)(x)=(g\circ f)(x)$ iff $ \forall x  f(g(x))=g(f(x))$ iff $ \forall x  c=g(c)$ iff $ c=g(c)$. So $ g$ satisfies $ f\circ g=g\circ f$ iff $ g(c)=c$.

  3. If $ f\circ g=g\circ f$ for all functions $ f$ then in particular $ f\circ g=g\circ f$ for all constant functions $ f(x)=c$. But then by the previous part for all constant $ c$ we have $ g(c)=c$ and this precisely means that $ g$ is the identity function $ g(x)=x$.

Problem 3. Sketch, to the best of your understanding, the graph of the function

$\displaystyle f(x)=x^2-\frac{1}{x^2}. $

(What happens for $ x$ near 0? For large $ x$? Where does the graph lie relative to the graph of the function $ y=x^2$?)

Solution. The first graph below shows $ x^2$ (above the $ x$ axis) and $ -1/x^2$ (below the $ x$ axis. The second shows the sum of the two, the desired function $ x^2-1/x^2$:


For $ x$ near 0 our function goes to $ -\infty$, for large $ x$ it goes to $ +\infty$. It is always below $ x^2$ but for large $ x$ it is very near $ x^2$.

Problem 4. Write the definition of $ \displaystyle\lim_{x\to a}f(x)=l$ and give examples to show that the following definitions of $ \displaystyle\lim_{x\to a}f(x)=l$ do not agree with the standard one:

  1. For all $ \delta>0$ there is an $ \epsilon>0$ such that if $ 0<\vert x-a\vert<\delta$, then $ \vert f(x)-l\vert<\epsilon$.
  2. For all $ \epsilon>0$ there is a $ \delta>0$ such that if $ \vert f(x)-l\vert<\epsilon$, then $ 0<\vert x-a\vert<\delta$.

Solution. The definition is: For every $ \epsilon>0$ there is a $ \delta>0$ so that whenever $ 0<\vert x-a\vert<\delta$ we have that $ \vert f(x)-l\vert<\epsilon$. The required examples:

  1. This is satisfied whenever $ \vert f\vert$ is bounded and regardless of its limit. Indeed, choose $ \epsilon$ bigger than $ \vert l\vert+M$ where $ M$ is a bound on $ \vert f\vert$, and $ \vert f(x)-l\vert<\epsilon$ is always true.
  2. According to this definition, for example, $ \displaystyle\lim_{x\to
a}c=c$ is false, and hence it cannot be equivalent to the standard definition. Indeed, in this case $ \vert f(x)-l\vert<\epsilon$ means $ 0=\vert c-c\vert<\epsilon$. This imposes no condition on $ x$, so $ \vert x-a\vert$ need not be smaller than $ \delta$.

Problem 5. Suppose that $ g$ is continuous at 0 and $ g(0)=0$ and that $ \vert f(x)\vert\leq\sqrt{\vert g(x)\vert}$ for all $ x$. Show that $ f$ is continuous at 0.

Solution. At $ x=0$ the inequality $ \vert f(x)\vert\leq\sqrt{\vert g(x)\vert}$ reads $ \vert f(0)\vert\leq\sqrt{\vert g(0)\vert}=\sqrt{\vert\vert}=0$, and hence $ f(0)=0$. We claim that $ \displaystyle\lim_{x\to 0}f(x)=0$ and hence that $ f$ is continuous at 0. Indeed let $ \epsilon>0$ be given. Then $ \epsilon^2>0$ and by the continuity of $ g$ at 0 we can find a $ \delta>0$ so that whenever $ \vert x\vert<\delta$ we have that $ \vert g(x)\vert<\epsilon^2$. But then if $ \vert x\vert<\delta$ then $ \vert f(x)\vert\leq\sqrt{\vert g(x)\vert}=\sqrt{\epsilon^2}=\epsilon$ and the definition of $ \displaystyle\lim_{x\to 0}f(x)=0$ is satisfied.

The generation of this document was assisted by LATEX2HTML.

Dror Bar-Natan 2002-10-23