Dror Bar-Natan: Classes: 2002-03: Math 157 - Analysis I:

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Solution of Term Exam 1

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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 $r^3=2$.

Solution.

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

   $$(b-a)(b+a)>0.$$ (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 $g(x)=x$.

Solution.

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

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

(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 $$\lim_{x\to a}f(x)=l$$ and give examples to show that the following definitions of $$\lim_{x\to a}f(x)=l$$ 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, $$\lim_{x\to a}c=c$$ 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 $$\lim_{x\to 0}f(x)=0$$ 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 $$\lim_{x\to 0}f(x)=0$$ is satisfied.

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