6

This is not an explanation of the steps in the book but I would like to point out that the proof in the book looks a bit silly to me. You don't require Rouche's Theorem. There is a very elementary 'High School' proof: suppose $f(z)=f(w)$. Then $z-w=a+a_2(w^{2}-z^{2})+a_3(w^{3}-z^{3})+\cdots$. Note that $|w^{n}-z^{n}| =|z-w|(|w^{n-1}+w^{n-2}z+\cdots+wz^{n-2}+...


5

Hint: Your squaring must be: $$(a-b)^2=a^2+b^2-2ab$$


4

Since the Legendre symbol is multiplicative we have: $$\left(\frac{-2}{p}\right)=\left(\frac{2}{p}\right)\left(\frac{-1}{p}\right)$$ The first one is equal to $1$ iff $p \equiv 1,7 \pmod{8}$ and the second iff $p \equiv 1 \pmod{4}$. So they are both one when $p \equiv 1 \pmod{8}$ and both $-1$ when $p \equiv 3 \pmod{8}$. In both cases their product is one ...


4

Writing $\cos2t=c,$ using $\cos2x=2\cos^2x-1,$ $$257+32(2c^2-1)-256\cdot\dfrac{1+c}2=64c^2-128c+97=64(c-1)^2+97-64\ge33$$ for real $c$


4

Note that\begin{align}32\cos(4t)&=32\bigl(2\cos^2(2t)-1\bigr)\\&=32\bigl(2\bigl(2\cos^2(t)-1\bigr)^2-1\bigr)\\&=32\bigl(8\cos^2(t)-8\cos(t)+1\bigr)\\&=256\cos^2(t)-256\cos(t)+32\\&\geqslant256\cos^2(t)-256+32\\&>256\cos^2(t)-257.\end{align}Your proof looks correct to me.


3

They are really unnecessary. Any morphism with both a left and a right inverse has an inverse, which is equal to both the given left and the right inverse. In fact it is more convenient in certain foundational systems like homotopy type theory to define an isomorphism as a morphism with both a left and a right inverse.


3

Your method is correct, but has some errors. You are considering $f(t) = 257 + 32 \cos 4t - 256\cos^2 t$ and want to prove $f(t_0)>0$, where $t_0$ is a critical point (for both max and min stationary points). So: $$f'(t)=\color{red}{-128}\sin4t\color{red}+256\sin 2t=0 \Rightarrow \sin 2t(\cos 2t-1)=0 \Rightarrow \sin 2t=0 \ \ \text{or} \ \ \cos2t=1 \...


3

Your proof looks OK to me on a quick viewing. For the other part of the proof (the "no derivatives"), you could write $$ \cos 4t = 2 \cos^2 2t - 1 $$ and then you can rewrite $$ \cos 2t = \cos^2 t - 1 $$ so that your original formula ends up as a quartic in $\cos t$...but one involving only even powers. So you can let $y = \cos^2 t$ and get a quadratic ...


3

Converting from base five to base ten by simply using the same digits and changing the base doesn't work. $31_5$ is $16_{10}$, not $31_{10}$. So the correct way of applying method 2 is $$ 31_5\cdot 2_5 = 16_{10}\cdot 2_{10} = 32_{10} = 112_5 $$


3

Your reasoning is correct. Substitute $Y = \dfrac{3x^2}{2y} X - \dfrac{x^2 - 2c}{2y}$ into $Y^2 = X^3 + c$ to obtain a cubic for $X$. Just don't brute force it. The key point is to observe that two of its roots are known, and equal to $x$ (do you see why?). Use a sum of roots theorem to get the third one. BTW, to use the sum of roots, yo don't need the ...


3

If all $5$ questions are chosen from good questions, we pass. Therefore, there are $\binom{35}{5}$ such choices. We also know that we can choose $5$ questions from $41$ questions with $\binom{41}{5}$, which is the total number of choices. Therefore, the probability of passing should be $$\frac{\dbinom{35}{5}}{\dbinom{41}{5}}$$ So your answer is correct.


3

If the triangles $T$ and $T'$ are similar with simialrity factor $k$, then ratio of their areas is $k^2$. So since the sides of the starting triangle are twice as median triangle the result is ${1\over 4}$. Have I missed something? Edit. You probably meant what is the area of a triangle formed by medians? In that case, note that area of $BDG$ where $D$ ...


3

Why does $f^{-1}(U)$ is closed mean $f$ is not continuous? (Yes, I know how it would go, but it is a rather long-winded way to do so.) It is better to say $f^{-1}(U)=\{0\}$ is not open, then it is just the definition of continuity to reach the desired conclusion.


2

It's tricky to speak definitively here, since we are missing some context, but I think there are some issues here. Specifically, you seem to be too reliant on citing results that say, more or less directly, what it is you're supposed to be proving. For example, $iii) \implies iv)$ Because column rank = row rank for any matrix, the row rank must be n ...


2

$(\sqrt{2}$ - $\sqrt{5})^2$ is not equal to 2 - 5; but rather, $2 - 2\sqrt{10} + 5$ = $7 - 2\sqrt{10}$


2

Using $\cos(2x)=2\cos^2(x)-1$ twice we have $257+32\cos(4t)-256\cos^2(t)=257+32(2(2\cos^2(t)-1)^2-1)-256\cos^2(t)$ Simplification leads to $257+256\cos^4(t)-512\cos^2(t)=257+256(\cos^2(t))^2-512\cos^2(t)=1+(16-16\cos^2(t))^2\geq 1$ The function is periodic and max value occurs at $t=n\pi$ thus we see that the inequality $257+32\cos(4t)-256\cos^2(t)>0$...


2

Yes, you are right. On the other hand, note that the first words of the that textbook are “This book is designed to introduce the reader to the theory of semisimple Lie algebras over an algebraically closed field of characteristic $0$”. So, there is no reason for the author to mention what occurs in characteristic $2$.


2

Comment:An experimental approach; following conditions cover a set of these types of primes: For $8k+1$: Let $k=k_1+2$ ⇒ $p=8(k_1+2)+1=8k_1+8+8+1=2[4(k_1+1)]+3^2$ If $k_1+1=c^2$ then we have: $p=3^2+2\times (2c)^2$ Examples; $k_1=3$ ⇒ $p=41=3^2+2 (2\times 2)^2=3^2+2\times 4^2$ $k_1=15$ ⇒ $c^2=15+1=4^2$ ⇒ $p=137=3^2+2\times 8^2$ For $p=8k+3$ it can ...


2

It's not quite as simple as you make it in your proof because the sequence may approach the $\lim \sup$ from above. The following proof works, though: Let $M_1= \lim \sup \frac{a_n}{b_n}$. Then $\exists N~ n \gt N \Rightarrow \frac{a_n}{b_n} \lt M_1+1.$ Let $M_2 = \max \{ \frac{a_n}{b_n} \vert ~n \leq N \}.$ Let $M = \max \{M_1+1, M_2+1 \}$. Then $\...


2

In general, $A\subseteq A\cup S$ for any set $S$, since if $a\in A$ then it is true that $a$ is either in $A$ or in $S$. Thus, to prove (ii), it remains to show that in each case $A\cup S\subseteq A$ (since if $A\subseteq B$ and $B\subseteq A$, then $A=B$). Now if $b\in A\cup A$, then by definition of the union either $b\in A$ or $b\in A$, which means that ...


2

In group theoretical terms, $G^{op}$ is the same as $G$ but with multiplication $x\cdot_{op} y=yx$. As you see, not even at this level the identity is a homomorphism, since $f(x)\cdot_{op}f(y)=x\cdot_{op} y=yx\neq f(xy)=xy$. An isomorphism at a group theoretical level that you can translate into the categorical language is $f:x\mapsto x^{-1}$. This way, $f(...


2

If $\frak{a}$ is a left ideal and you want $(\frak{a}:\frak{b})$ to be a left ideal, or if you are working in a commutative ring, this is fine. In a non-commutative ring, if you want $(\frak{a}:\frak{b})$ to be a two-sided ideal, you need $\frak{a}$ and $\frak{b}$ to be left ideals so that you can handle both $rx$ and $xr$ when $x\in(\frak{a}:\frak{b})$. ...


2

Basically you are right and maybe simply overanalysing a simple situation. The 'amount to' notion means that if you have ordered sets $A,B$ and you view them as categories as you describe, then when you look into what it means to have a functor $A\to B$ you find essentially the same information as when you look into what it means to have an order preserving ...


2

We can solve for any integer $A$, not just for primorial divided by something, and proof is clearly easier. Here, we should state that, for any positive number $m$, there is a number $x$ in $[m, m+A)$ which $gcd(x(x+2), A) = 1$. The necessary/sufficient condition for $gcd(x(x+2), A) = 1$ is, $gcd(x, A)=1$ and $gcd(x+2, A) = 1$. It is obvious that ...


2

Your proof is correct. Anyway note that the function $f$ must meet the line $y=x$ in atleast one point, and that point is the fixed point. To see this geometrical idea into a formal proof, set $g(x)=f(x)-x$ and apply IVT to $g$.


1

No, your proof has errors. A set belongs to $\mathcal A_1$ if it has the form $A\cap E$ for some $A \in \mathcal A$. $\emptyset=\emptyset\cap E$ and $\emptyset \in \mathcal A$, so $\emptyset \in \mathcal A_1$. $E=X\cap E$ and $X \in \mathcal A$, so $E \in \mathcal A_1$. If $A_n \in \mathcal A_1$ for all $n$ then there exist sets $B_n \in \mathcal A$ such ...


1

Seems good to me, however there is a proof that doesn't invoke Poincare Duality and also is quite a bit less tedious, in my opinion. Recall that if a compact manifold has no subgroup of $\pi_1(M)$ with index 2, then the manifold is orientable. Clearly, $H_1(M,\mathbb{Z}_2)$ has no subgroup of index 2. By the universal coefficient theorem, $H_1(M,\mathbb{Z}...


1

Kevin essentially already said this in an answer, but let me rephrase it (i.e. spell out more explicitly). Let $f: A \to B$ be any arrow, then if $f$ has a left inverse $h: B \to A$ (i.e. $hf = Id_A$) and a right inverse $g: B \to A$ (i.e. $fg = Id_B$) we have that $f$ has an inverse on both sides and this inverse is given by $h=g$ (so they are the same ...


1

You reached the correct answer but missed a few key facts. 1) $P[X\geq x]=1-P[X<x]$ So you have to put strict inequality instead of inequality in your 2nd step. 2) Use the fact that $P[|X-1|=2]=0$, it basically impies that $P[X=3]=0$ and $P[X=-1] =0$. So you have to use this to bring back the equality in the strict inequality as your second last step ...


1

This is correct. In other words $T_e O(n)$ is the sub vector space of skew-symmetric matrix. The fact that $d(-)^t(A)=A^t$ can be written $d_e(-)^t=(-)^t$. This is true because $(-)^t$ is a linear map, hence it is equal to its differential.


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