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From what I could gather from some google searches (1, 2, 3-pdf, and Krantz has a good book on writing mathematical prose) the most common reasons for using we instead of I are: To emphasize participation by the reader and ensure that he or she is included. To not sound egotistical. As in, to stress the mathematics and reduce the role of the author in ...

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You have an implication $p\to q$, where $p$ is the sum of a certain set of $N$ natural numbers is less than $N+2$, and $q$ is each of the $N$ numbers is less than $3$. To prove such an implication by contradiction, you assume that $p$ is true and $q$ is false, and you try to derive a contradiction from that assumption. Assuming that $p$ is true ...

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Begin by defining $u[{x\atop t}]$ when $u$ is a variable or a constant. The first clause will be $x[{x\atop t}]=t$; there will be two more clauses, one for the case where $u$ is a variable other than $x$, and one for the case where $u$ is a constant. After that comes the recursion clause, where you'll define $f(t_1,\dots,t_n)[{x\atop t}]$ in terms of the ...

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You could also use gcd and lcm in regards to prime decomposition. Let $a=\prod_{k=1}^{m}p_k^{i_k}, b=\prod_{k=1}^{m}p_k^{j_k}$ where $p_k$ is the $k$'th prime number. Then $$d=\gcd(a,b)=\prod_{k=1}^{m}p_k^{\min\{i_k,j_k\}}$$ $$m=\text{lcm}(a,b)=\prod_{k=1}^{m}p_k^{\max\{i_k,j_k\}}$$ Can you see it from here? $d|m \Rightarrow m=dx, x \in \mathbb{Z} ... 3 If you’re going to try to prove directly that$A=B$, you don’t want to start with some$x\in(A\cup B)^c$: you want to start with some$x\in A$and show that it’s in$B$, and vice versa. (Incidentally,$x\in(A\cup B)^c$means that$x\notin A\cup B$, which means that$x\notin A$and$x\notin B$, not that$x\notin A$or$x\notin B$.) Suppose that$x\in A$. ... 2$\forall x \in B$.$x \in B, A\cap B = \emptyset \Rightarrow x \not\in A$.$x \in B,A \cup B \subseteq C\cup D \Rightarrow x \in C \cup D$.$x \not\in A, C\subseteq A \Rightarrow x \not\in C$.$ x \in C \cup D, x \not\in C \Rightarrow x \in D$. Therefore,$B \subseteq D$. 2 Can only one of them be odd? Next assume they are both odd. Write$m = 2s + 1$and$n = 2t + 1$for some integers$s, t$(this can be done since we assumed$m, n$to be odd). Try inserting this into$m^2 + n^2$, expand the resulting parentheses, and see if there are any choices for$s$and$t$which makes the result divisible by$4$. If you get the answer ... 2 No, you cannot, at least under the definition of big-$\Omega$used in computer science. For example:$f(n) = n^2$if$n$is of the form$\displaystyle 2^{4^k}$and$f(n) = f(n-1)$otherwise;$g(n) = n$. Then$f(n)$is as large as$n^2$infinitely often, so$f(n)$is not$O(g(n))$. Also,$f(n)$is smaller than$2\sqrt{n}$infinitely often (for instance, ... 2 The question is to be answered with no; WLOG assuming$(a,b) = (0,1)$. @HughDenoncourt pointed out, that$\frac34 = 0.\overline{20}_3$will never become part of any$A_i$by construction of$A_i$and thus$\frac34 \notin A$. ($A_i$is the union of the closed sets up to the$i$-th iteration,$A = \bigcup_{i\in\mathbb N} A_i$is the set in question) More ... 2 It may help to think of$\epsilon$as the error you're willing to tolerate between$f$and$L$. Then$\delta$is the deviation which is allowed for$x$from$x_0$to get the error less than$\epsilon$. The definition is then simply saying that no matter how small you wish your error to be, there is some distance around$x_0$, the deviation, which will allow ... 2 If you stick strictly with a direct proof (denoting two consecutive integers by$n$and$n+1$, summing them to get$2n + 1$, therefore odd), you'll be fine. For one thing, your assumed contradiction, the negation of "the sum of two consecutive numbers is always odd" is not correctly stated; its negation needs to be "it is not the case that that the sum of ... 2 Your proof of injectivity is perfect. I would approach surjectivity slightly differently. Take some$m\in\Bbb R,$and then let$k=(b-a)m+a.$Then$k\in\Bbb R,$and you can show that$f(k)=m.$"Let$m=\frac{k-a}{b-a}$" basically assumes what you're trying to prove, though you'll certainly want to start there with your calculations. As an alternate approach, ... 2 (This is not an evaluation of your proof, but an alternative one. For clarity I'll be using a slightly different notation:$\;f^{-1}[Y]\;$instead of$\;f^{-1}(Y)\;$.) Let's start at the most complex side, here the right hand side, and see if we can use the definitions to find out which elements$\;x\;are in this set: \begin{align} & x \in f^{-1}[G] ... 1 You have to prove thatA\cap B$is a subset of$C$. At this level, things do not require immense bursts of creativity, or a stream of great ideas. Usually just verifying the definitions is fairly straightforward and short. So you need to show that if$x\in A\cap B$then$x\in C$. If$x\in A\cap B$then$x\in A$and$x\in B$. Now comes the point to use the ... 1 The relevant fact is that the sum of the angles in a triangle equals$\pi.$So, summing all the angles, you get$\pi$times the number of triangles. But you ALSO get sum over the vertices of$-k(v) 2\pi$(I am assuming the surface is closed for simplicity, the general case is the same). So, you get$\pi T = -\sum k(v) + 2\pi V.$Now, recall that in a ... 1 Remember that any odd number can be written in the form$2k+1$with$k$integer. Now remember that the square of an odd number is odd, and the square of an even number is even, and the sum of two numbers one of which is even and the other is odd is always odd (and therefore cannot be divisible by$4$). The only two possible alternatives are thus "both ... 1 I'm a little confused by how you're going about your proof. You prove that such an$\Omega$is not possible since it not connected, but then you modify it regardless? How do you know that$A$is an open subset of$\mathbb C$? A more straightforward proof might be thus: first, prove an open, connected subset of$\mathbb C$is path connected (which you seem ... 1 I'm not sure to understand what you are doing. But your point$1.$is essentially sufficient for the proof of the following stronger fact: Let$\Omega$be any subset of$\mathbb{C}$, containing at least one point in$\mathbb{H}$and one point in$\bar{\mathbb{H}}$. Then if$\Omega$is connected,$\Omega \cap \mathbb{R}$is not empty. (Remark that I don't ... 1 Spivak wants to show that zero is the unique additive identity on$\mathbb{R}$. That is, he want to prove that if we have$a+x=a$then$x$must identical to zero. He assumes P1, P2 and P3 to prove this. In particular, he uses P2 in the last step. If$0+x=0$then using P2 we can conclude that$x=0$without P2 we can not conclude this. 1 Here is how I would think about how to solve this problem: Groups are an abstraction of the nice properties of a ring under the operation of addition. That is, when you add two integers, you get another integer, addition is associative, you have an identity (0), and every element has an inverse ($(x)+(-x)=0\$). (Warning: Commutativity is not (necessarily) a ...

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