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10

(assuming $p>2$) odd squares are all $1$ mod($4$), your expression is $3$ mod($4$).


4

Suppose $2p+1=k^2$. Clearly $p=2$ fails so we may assume $p$ is odd. We may rewrite it as $2p=(k-1)(k+1)$. Now since the LHS is even, so must the RHS. Note now that $k+1$ and $k-1$ must always have the same parity. Thus $k$ must be odd, so $k+1$ and $k-1$ are even. But this means that $(k+1)(k-1)$ must be divisible by $4$, and the LHS is clearly not ...


4

Hint: if $a=\cos\alpha$, $\sqrt{1-a^2}=\sin\alpha$, $b=\cos\beta$ and $\sqrt{1-b^2}=\sin\beta$, then you need to show that $$ \lvert\cos\alpha\sin \beta+\cos \beta\sin\alpha-\sqrt3 \sin\alpha\sin\beta+\sqrt3 \cos\alpha\cos\beta| =\lvert\sin(\alpha+\beta)-\sqrt3\cos(\alpha+\beta)|\le2, $$ that is, $$ ...


4

Actually, you need both to hold; in classical logic, $P \vee Q \to R$ is equivalent to $(P \to R) \wedge (Q \to R)$. The intuition is as follows: assuming $P \vee Q$, you know at least one of $P$ or $Q$ is true, but you don't know which. Thus, to deduce $R$ from the assumption $P \vee Q$, you need to be able to deduce $R$ from whichever one is true—since ...


4

In any field (well, at least in math and physics, the two fields I am interested in) one of the skills acquired over time by a serious researcher is a feel for what problems are open, what problems are interesting enough to be published, who is currently working on what, and so forth. In general, one doesn't have this fell until at least the second year of ...


2

Let $a_1=a_2=a_3=a_4=\frac{1}{16}$. Then $A=\frac{1}{16}$. Note that $B=(1/4)^{1/4}\gt A$. Perhaps $B$ is supposed to be $(a_1\cdots a_n)^{1/n}$.


2

You're overthinking. The dimension formula $$ \dim(V_1\cap V_2)=\dim V_1+\dim V_2-\dim(V_1+V_2) $$ already tells you that $\dim(V_1\cap V_2)>0$, because $$ \dim V_1+\dim V_2>\dim V\ge\dim(V_1+V_2) $$ A subspace with dimension greater than $0$ contains a nonzero vector.


2

It's fine.   The predicate is a material implication.   This is evaluated as true if either its antecedant is false or the consequent is true. So for any two arbitary integers, we claim either they are not both odd or their product is odd. Thus, in the case where at least one of our arbitrary integers is not odd, then our claim is clearly true. ...


2

you don't need that result(actually usually that result is proved USING the fact that maximal planer graphs have only triangular regions!) To prove your result, observe that it isn't true for $n=1$ and $n=2$ so let $n\geq 3$. Assume to the contrary that there is a maximal planar graph $G=(V,E)$ embedded in the plane with a region that is not a triangle. ...


2

A homeomorphism by definition is a bicontinous BIJECTION. Meaning they must have the same cardinalities. $\mathbb{R}$ has cardinality $2^{\aleph_0}$ while $\mathbb{Q}$ has cardinality $\aleph_0<2^{\aleph_0}$.


2

You don't need induction, since it is enough to show $$n(n+2)<(n+1)^2$$


2

Lemma: Lagrange identity Let $x=(x_1,...,x_n),y=(y_1,...,y_n)\in\mathbb{R}^n$. Then $\|x\|^2 \|y\|^2-\langle x,y\rangle^2=\sum_{i<j}(x_iy_j-x_jy_i)^2$. Before prove this formula, note that it means $$\sum_{i=1}^nx_i^2\sum_{j=1}^ny_i^2=\sum_{i<j}(x_iy_j-x_jy_i)^2+\left(\sum_{i=1}^nx_iy_i\right)^2.$$ Thus, if we take $y_i=1$, then ...


2

Yeah, that's fine. Remember that you're allowed to use words in proofs; feel free to use as many as necessary in order to be clear. Maybe the following would be a good example: Consider the binomial expansion of $(\sqrt{n} + 1)^8$, where the first term is $n^4$. So take $(n^4 + \ldots + 1)$ and (...your next steps here)


2

Hint: If $M_C$ is the midpoint of $AB$, then $A M_CS$ and $M_CBS$ have equal bases and share the same height. Can you proceed for the first part? For the second part, a similar approach, except consider the triangles $CM_CB$ and $CM_CA$. Are these areas equal? Express each of the areas in terms of $X, Y$ and $Z$. Can you finish up for the second part?


2

How is $\Pr(A_3) = \Pr(A_1 ∩ A_2 ∩ A_3)$? If GS 1, 2, 3, and 4 are in different groups, then GS 1, 2, and 3 are in different groups; and GS 1 and 2 are in different groups. Therefore, if $A_3$, then $A_2$ and $A_1$. In set-theoretic terms, $A_3 \subseteq A_2 \subseteq A_1$. The intersection of a set with any one of its subsets is the subset. ...


2

Hint: For the second portion, try a proof by contradiction: For non-zero divisors $a, b \in R$, suppose $(ab)c=0$ for nonzero $c$, then by associativity $a(bc)=0$...


1

An alternative solution: The number of ways to arrange the $16$ students in a line and then split it into $4$ equal segments: $$16!=20922789888000$$ The number of ways to do it such that each segment contains a graduate student: ...


1

The specifics of your actual questions (about how $A_3=A_3\cap A_2\cap A_1$ and such) seem to be adequately answered above. Here instead, I provide an alternative solution to the stated problem which may be more intuitive than the book's approach. Let us approach this via direct counting. Temporarily assume that the four groups are considered distinct ...


1

Let $S$ be the reference set. One can indeed consider $X$, $Y$, $U$ and $V$ as elements of the Boolean ring $\mathcal{P}(S)$ of subsets of $S$, where the addition is the symmetric difference and the product is intersection. Then the union of two subsets $A$ and $B$ is $A + B + AB$, the complement of $A$ is $1 + A$, the set difference $A \ B$ is $A(1+B) = A ...


1

If you have already on hand a theorem that uniform convergence implies convergence of the integral, like Baby Rudin 7.16, then the proof you've been given is perfectly rigorous. The triangle comment establishes that the value of the integral is independent of $n$. In more detail, the integral from $0$ to $1/n$ is ${1\over 2}(1/n)n={1\over 2}$ because ...


1

If the point $(r(t)x, t)$ is to lie on $S^{n+1}$, then $r(t)^2 + t^2 = 1$, so, since $r(t)$ is non-negative, one must have $r(t) = \sqrt{1 - t^2}$. But this function does not satisfy $r(0) = 0$. The way to fix this is consider $I = [-1,1]$ instead, and carry the proof as you did, just changing $0$ and $1$ to $-1$ and $1$, respectively. Of course, the ...


1

If $a=0$, consider $(b\epsilon)^2=0$. If $a \neq 0$, consider $(a+b\epsilon)(a-b\epsilon)=a^2 \in \mathbb R^*$.


1

Hint: By Taylor expansion about zero, $$ \mu(\theta_n)=\mu(0)+\mu'(0)(\theta_n-0)+o(\theta_n-0)\\ \iff\sqrt{n}(\mu(\theta_n)-\mu(0))=\sqrt{n}\mu'(0)\theta_n+\sqrt{n}o(\theta_n)\\ \iff \sqrt{n}(\mu(\theta_n)-\mu(0))=\sqrt{n}\theta_n\mu'(0)+\sqrt{n}\theta_n o(1)\\ $$ Furthermore, $\frac{\mu'(0)}{\sigma(0)}=o(1)$ is just a constant. Given the other stuff you ...


1

Take a region which is not a triangle. What is your definition of region? Can you add an edge? Hope that helps,


1

The condition $(d,a)=1$ is part of the hypothesis. So the first two sentences of the proof are superfluous, and should be deleted. Once that is done, the proof is good. The "By Bezout" part is too informal. You should say that by Bezout, there are integers $s$ and $t$ such that $\dots$. The proof is, as you observed, essentially the same as the proof of ...


1

If $2p+1$ is a square then $2p=n^2-1=(n-1)(n+1)$ for some odd number $n\ge 3$. Then $n-1$ and $n+1$ are even. Now write $$p=\frac{n-1}2(n+1)$$ But then $p$ is an even number greater than $2$. Contradiction.


1

The formal formula is $$ \forall x \forall y (x \in \mathbb Z \land y \in \mathbb Z \land x \text{ odd}\land y \text{ odd} \implies xy \text{ odd}) $$ Since $P \implies Q$ is true when $P$ is false, one generally omits the cases when $P$ is false, because there is nothing to prove in these cases. In our context, $P$ is false when $x \not\in \mathbb Z \lor ...


1

Hint : if $B \in \text{Aut}(H)$, then you can consider $ \beta = \phi^{-1} B \phi \in \text{Aut}(G)$.


1

You need the the assumption that the series $\sum_n a_n$ is convergent: otherwise, the quantity $\sum_{k=n+1}^\infty a_k$ cannot be manipulated (it is not defined). Furthermore, the inequality that actually holds is in the other direction. See below. Assuming the series $\sum_n a_n$ is convergent, you can write: $$ \sum_{k=n+1}^\infty a_k = ...



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