%begin macros %\magnification=\magstephalf \magnification=\magstep1 \def\g{\bigtriangledown} \def\L{{\cal L}} %\baselineskip=14pt %\parskip=10pt \def\Tilde{\char126\relax} \font\eightrm=cmr8 \font\sixrm=cmr6 \font\eighttt=cmtt8 \def\P{{\cal P}} \def\Q{{\cal Q}} \parindent=0pt \overfullrule=0in \def\frac#1#2{{#1 \over #2}} \nopagenumbers %end macros {\bf MATH 437 Solutions to Exam II for Dr. Z.'s, Fall 2021, Dec. 6, 2021 Exam} \bigskip {\bf 1.} (10 pts.) Give {\bf two} proofs of the Pythagorean theorem. \bigskip {\bf Sol.} \bigskip {\bf First proof}: Consider an $(a+b) \times (a+b)$ square, and drawing four right-angled triangles with base $a$, height $b$ and hypotenuse $c$, in such a way that each side of the rectangle has one $a$ and one $b$. Then the emerging square at the middle is a $c \times c$ square, whose area is $c^2$, and hence the total area of the large square is $c^2+ 4 \cdot (ab/2)=c^2+2ab$. On the other hand, the area of the big square is $(a+b)^2=a^2+2ab+b^2$ so $$ a^2+2ab+b^2= c^2 + 2ab \quad . $$ Subtracting $2ab$ from both sides of the above equation, we get $a^2+b^2=c^2$. \bigskip {\bf Second Proof}: (Similar triangles). Let $ABC$ be a right-angled triangle with the angle at $C$ being the right-angle. Let $C'$ be the projection of $C$ on $AB$. Then the three triangles $\bullet$ $ACC'$ (with hypotenuse-size $|AC|=b$), $\bullet$ $CBC'$ (with hypotenuse-size $|CB|=a$), $\bullet$ $ABC$ (with hypotenuse-size $|AB|=c$), are {\bf similar} (they have the same angles), hence there is a non-zero constant $k$ (whose exact value does not matter) such that the area of triangle $ACC'$ is $kb^2$, the area of triangle $ABC'$ is $ka^2$ and the area of triangle $ABC$ is $kc^2$. But of course $$ Area (ACC') +Area (ABC') = Area(ABC) \quad , $$ so $$ ka^2+ kb^2=kc^2 \quad . $$ Dividing by $k$ gives $$ a^2+b^2 \, = \, c^2 \quad . $$ \bigskip {\bf 2.} (10 pts.) Prove that $\root{7}\of{3}$ is irrational. \bigskip {\bf Proof}: We need the obvious lemma that for every integer $N$ all the expoonents of $N^7$ in its prime decomposition are divisible by $7$. \bigskip Assume that $\root{7}\of{3}$ is rational. This means that there exist positive integers $m$ and $n$ such that $$ \root{7}\of{3} = \frac{m}{n} \quad . $$ Raise everything to the seventh-power $$ 3= \frac{m^7}{n^7} \quad . $$ Transposing: $$ m^7=3 n^7 \quad . $$ The left side as all the exponents of the primes, including the exponent of $3$, a multiple of $7$, but the exponent of $3$ on the right side has the form $7a+1$ (i.e. it gives reamiander $1$ when divided by $7$). By the uniqueness of the prime decomposition, this is a contradiction. QED. \bigskip {\bf 3.} (10 pts. total) (a) (5 points) Construct the Pascal triangle mod 2 Fractal using the first $7$ rows (i.e, row $0$ through row $7$). Highlight the middle $0$ section, and show that the remaining part consists of three identical triangles with $4$ rows, \bigskip {\bf Ans.} $$ 1 $$ $$ 1 1 $$ $$ 1 0 1 $$ $$ 1 1 1 1 $$ $$ 1 0 0 0 1 $$ $$ 1 1 0 0 1 1 $$ $$ 1 0 1 0 1 0 1 $$ $$ 1 1 1 1 1 1 1 $$ \bigskip (b) (5 points) Define the Feigenbaum constant. Explain everything! \bigskip {\bf Ans.} It turns out that that the mapping from $[0,1]$ to itself,$x \rightarrow rx(1-x)$ repeated many times, for very small $r$ ($r<1$) always go to $0$, no matter what the starting number, $x_0$ is . Later on, until $r=3$ it goes to some other fixed point, but only {\bf one}, but starting at $r_1=3$, for example, $r=3.1$, the limiting behavior vacillates between {\bf two} limiting points. Eventually, at $r_2$, the limiting behavior vacillates between {\bf four} limiting points. Soon later at $r_3$ , the limiting behavior vacillates between {\bf eight} limiting points. It is always a power of $2$, and the cut-off places in the parameter $r$ when it goes from a limiting orbit of $2^{n-1}$ to a limiting orbit of $2^n$ are called $r_n$. Michell Feigenbaum proved that $$ \lim_{ n \rightarrow \infty} \frac{r_n-r_{n-1}}{r_{n+1}-r_n} \quad , $$ tends to a fixed constant, about $4.66\dots$ and it is called the {\bf Feigenbaum constant}. \vfill \eject {\bf 4.} (10 points altogether) (a) (2 points) Define a {\bf Platonic soild} (b) (2 points) Let $a$ be the number of edges meeting each vertex, and let $b$ be the number of edges surrounding each face. Express $V$ (the number of vertices) and $F$ (the number of faces) in terms of $E$ (the number of edges), and $a$ and $b$. (c) (2 points) Find an expressions for $F$, in terms of $a$ and $b$. (d) (4 points) Obviously both $a$ and $b$ must be at least $3$, and $F$ (and hence $V$ and $E$) must be positive. It is easy to see (you don't have to do it) that $a,b$ must be both between $3$ and $5$, leaving $9$ potential scenarios. Find those values of $a$ and $b$ that make sense, and thereby prove that there are exactly $5$ Platonic solids. For each of them, find $F$ (the number of faces) and give the name of the corresponding Platonic solid. {\bf Sol.} (a) A platonic solid is a {\it polyhedron} (a solid body with finitely many faces) where all the faces are identical, and every vertex has the same number of edges coming out of it. (b) Each of the $V$ vertices have $a$ edges meeting it, hence there are altogether $aV$ edges. But every edge has exactly $2$ vertices, so $aV$ is a {\bf double count} hence $$ a\,V=2\,E \quad . $$ Also very face as $b$ edges around it, so the total number of edges is $bF$. Once again, every edge belongs to exactly two faces, so the above is a double-count. Hence $$ b\,F=2\,E \quad . $$ Hence, using algebra $$ V= \frac{2E}{a} \quad, \quad F= \frac{2E}{b} \quad . $$ (c) Using Euler's famous $V-E+F=2$, we get $$ \frac{2E}{a}-E+\frac{2E}{b}=2 \quad. $$ Hence $$ E= \frac{2}{\frac{2}{a} - 1 + \frac{2}{b}} \, =\, \frac{2ab}{2a+2b-ab} \quad. $$ We also have (from the above expression, $F=\frac{2E}{b}$) $$ F\, =\, \frac{4a}{2a+2b-ab} \quad. $$ Hence we have an expression for $E$ in terms of $a$ and $b$. Of course $a \geq 3$ and $b \geq 3$. So one has to try out the $9$ cases $a \in \{3,4,5\}$, $a \in \{3,4,5\}$. Only $(a,b)=(3,3)$,$(a,b)=(3,4)$,$(a,b)=(4,3)$,$(a,b)=(3,5)$,$(a,b)=(5,3)$ give values of $E$ that are positive inetgers. $\bullet$ $a=3$, $b=3$ $$ F=\frac{4(3)}{2(3)+2(3)-(3)(3)} \, = \, \frac{12}{3} \, = \,4 \quad $$ This gives the {\bf tetrahedron} ({\bf four} faces). $\bullet$ $a=3$, $b=4$ $$ F=\frac{4(3)}{2(3)+2(4)-(3)(4)} \, = \, \frac{12}{2} \, = \,6 \quad $$ This gives the {\bf hexahedron} ({\bf six} faces), more commontly called the {\bf cube}. $\bullet$ $a=3$, $b=5$ $$ F=\frac{4(3)}{2(3)+2(5)-(3)(5)} \, = \, \frac{12}{1} \, = \,12 \quad $$ This gives the {\bf dodacahedron} ({\bf twelve} faces). $\bullet$ $a=4$, $b=3$ $$ F=\frac{4(4)}{2(4)+2(3)-(4)(3)} \, = \, \frac{16}{2} \, = \,8 \quad $$ This gives the {\bf octahedron} ({\bf eight} faces). $\bullet$ $a=5$, $b=3$ $$ F=\frac{4(5)}{2(5)+2(3)-(5)(3)} \, = \, \frac{20}{1} \, = \,20 \quad $$ This gives the {\bf icasahedron} ({\bf twenty} faces). \bigskip {\bf 5.} (10 points) \bigskip Prove Lagrange's theorem that if $H$ is any subgroup of a group $G$, and $|H|$ and $|G|$ are their number of elements, respectively, then $|G|/|H|$ is always an integer. \bigskip {\bf Sol.} Let the number of elements of $G$ be $n$ and the number of elements of $H$ be $m$. Let's write $$ H=\{ h_1, \dots, h_m \} \quad, $$ where, of course, $h_1,h_2, \dots, h_m$ are all {\bf different}. If $G=H$ then we are done and $n/m=1$. Otherwise, there exist a $g_1 \in G$ {\bf not} in $H$. Consider the set (called ``coset'') $$ g_1 H=\{g_1 h_1, \dots, g_1 h_m\} \quad, $$ The elements of $g_1H$ are all {\bf different}, since if not there would have been for some $1 \leq i