# Tag Info

8

Yes, your solution is correct. Of course you could have multiplied both sides of your inequality by $$\frac {1}{ab}$$ to get the result in one shot.

7

The idea is fine, but look again at what you wrote; it is just a sequence of assertions, with no links between them. I suggest that you type it as follows:\begin{align}a<b&\implies-a>-b\\&\implies(-a)^2>(-b)^2\text{ (since $-a>-b\geqslant0$)}\\&\iff a^2>b^2.\end{align} Another possibility consists in noting that$$a^2-b^2=\overbrace{... 5 This is true, but messier than necessary and lacking justification. Here's an easier way: Note that a^2 is positive, so we can multiply both sides of the inequality a^3>a by it without changing the direction of the >, getting that a^5>a^3>a, using the assumption for the last >. On the concept of obviousness: on one hand, sure, it ... 4 d|n^2+7-(n+5)(n-5), where the right hand side is 32. 4 Note that Bézout's identity is used to prove an existence of an equation, but the existence of such an equation doesn't necessarily prove the reverse in terms of the factors or the gcd (e.g., there exists x = -5 and y = 5 such that 3x + 4y = 5, but \gcd(3,4) = 1 rather than \gcd(3,4) = 5). However, as shown in Bill Dubuque's and steven gregory's ... 3$$a+b+c+d=(a+b+c+d)\sum_{cyc}\frac{a}{b+c+d}==\sum_{cyc}\frac{a^2+a(b+c+d)}{b+c+d}=a+b+c+d+\sum_{cyc}\frac{a^2}{b+c+d},$$which says$$\sum_{cyc}\frac{a^2}{b+c+d}=0.$$In the full writing the last equality it's:$$\frac{a^2}{b+c+d}+\frac{b^2}{c+d+a}+\frac{c^2}{d+a+b}+\frac{d^2}{a+b+c}=0.$$3 Solution without cases: Because$$a^5-a=(a^2+1)(a^3-a).$$3 Let p_u,f_u be the planted/fallow acreage in the U.S. and p_s,f_s be the planted/fallow acreage in the Soviet Union. Also, let y_u be the yield per planted acre in the U.S. and y_s be the yield per planted acre in the Soviet Union. The information given says that$$\frac{y_s}{y_u} = 0.68 \; \iff \; y_s = 0.68y_u \tag{1}\label{eq1}$$The total yield ... 3 The proof is correct, there's no need to improve anything unless you want to be very formal. You can omit the quantifier \forall, because if something is true for all x, then it's particularily true for the one x that you're given. 2 A quick solution... We note that d|(n+5)\implies d|(n+5)^2\implies d|(n^2+10n+25), but d|(n^2+7) and hence we get,$$d|(n^2+10n+25-(n^2+7))\implies d|(10n+18)$$Also, d|(n+5)\implies d|(10n+50), but d|(10n+18) and hence we get,$$d|(10n+50-(10n+18))\implies d|32$$This completes the proof... 2 Here's an interesting solution. For sake of contradiction, assume that there exists two non equal real numbers x_1 and x_2, for which f(x_1)=f(x_2). Since, f is a linear function, its graph will be strictly monotonic, and this forces x_1=x_2 which contradicts our assumption. 2 a) true (but what you are saying is not a proof of it). Let s,s'\in S with f(s)=f(s'). Then also (g\circ f)(s)=g(f(s))=g(f(s'))=(g\circ f)(s') so that we are allowed to conclude that s=s' because g\circ f is injective. b) false in general. Let U=\{u\} and S\neq\varnothing so that automatically g\circ f is surjective. If however T ... 2 Suppose that you have proven your third case. What this would mean to me is that you have shown that if you have an arbitrary collection of open sets U_i, some of which are non-empty, then their union is open. If some of them are non-empty (and you don't know which), and you've proven that their union is open, then you could happily assume that they're ... 2 You approach is correct (and two answers posted are also correct, and they use some "shortcuts"). If one is to stay closer to your approach, one may provide some more details, For example, Case 1 could be explained further using that: If a>1 and b>0 then ab>b. Indeed ab-b=(a-1)b>0. Using the above, with b=a^3 and then with b=a^4 ... 2 Yes, you can use Bezout: \ 1 = \gcd(n,k) = an\!+\!bk = (a\!\color{#c00}{+\!b)n - b(n\!}-\!k)\,\Rightarrow\,\gcd(n,n\!-\!k)=1 Or   if \ d\mid n\  then \ d\mid k\!\iff\! d\mid n\!-\!k,\, so \,n,k\, and \,n,n\!-\!k\, have the same set S of common divisors d, so they have the same greatest common divisor (= \max S) Or \!\bmod d\!:\:  if \,... 2 Let c\leq d. Thus,$$ac-bd=a(c-d)+d(a-b)<0,$$which is a contradiction. 1 This is a proof by contraposition. Omitting the justifications of the arithmetic manipulations, it could be written like this: Prop. Let f:\mathbb{R}\to\mathbb{R}, f:x\mapsto7-3x. Then \forall x_1,x_2\in\mathbb{R}, x_1 \ne x_2 \implies f(x_1) \ne f(x_2). Proof. We will prove the following equivalent statement:$$\forall x_1,x_2\in\mathbb{R}, f(x_1)=f(...

1

We know that: $$d | n + 5, \quad d | n^2 + 7,\quad \Rightarrow d | (n+5)^2 - (n^2+7) = 10(n+5) - 32$$ This means that: $$d | 10(n+5) - 32,\quad \Rightarrow d | 32$$ beacause $d | n+5$.

1

a) is true, take $f(a)=f(b)$ then $g(f(a)) = g(f(b))$ and since $g\circ f$ is injective we get $a=b$ so $f$ is injective. b) is not true, take $f(x)=\arctan x$ and $g(x)=\tan x$ Then $g(f(x))=x$ so it is surjective while $f$ is not. The following is wrong: (but it is usefull to see why, look at the comment): c) is not true: Take $f(x)=e^x$ and $g(x)=\... 1 If two continuous real-valued functions$f$and$g$do not intersect on the interval$(a, b)$then the difference$f-g$has the same sign on the interval. That is an immediate consequence of the “intermediate value theorem”: If$f(x_1) - g(x_1) > 0$and$f(x_2) - g(x_2) < 0$then$f(x) -g(x) = 0$for some$x$between$x_1$and$x_2$, contrary to the ... 1 Let $$h(x)=f(x)-g(x)$$ Note that$h(x) $is continuous The intermediate theorem state that if there is a change in sign of a continuous function$h(x)$there must be at least one$c$such that$h(c) =0$That provides a point of intersection for$f$and$g$1 We can solve this system by the following way. Let$x=a^6$,$y=b^6$and$z=c^6$, where$a$,$b$and$care positives. Thus, we have $$\frac{a^3}{2b^3}+\frac{a^2c^2}{3b^4}=\frac{a^2b^2}{3c^4}$$ and $$1+\frac{b^3}{2a^3}+\frac{b^2c^2}{3a^4}=\frac{a^2b^2}{3c^4}.$$ The first gives $$\frac{a}{2b^3}+\frac{c^2}{3b^4}=\frac{b^2}{3c^4}$$ and $$a=\frac{2b^5}{3c^4}... 1 There are two ways to prove an implication P\to Q: you prove that the consequent Q is true, or you prove that the antecedent P is false. In other words, P\to Q is equivalent to \lnot P\lor Q. Therefore, your formula says as much as that there exists some number x such that x\neq 1 or such that for all y we have x+y=y. So informally there ... 1 Yes, correct. Alternative way that makes use of the following characterization of "interior": The interior of set C is the union of all open subsets of C. (If this characterization is not yet familiar to you then I advice you to make it familiar to you) You could also say: \mathsf{int}(C) is the largest open subset of C. This in the sense that ... 1 Sketch of a proof If we start with the axioms for the reals we know that they are a field that means that there are two different elements 0 and 1. Thus, assuming 0=1 we have a contradiction with the previous axiom and we can use the tautology : \lnot P \to (P \to Q) to derive directly : (0+y=y) Then we discharge the assumption to get : 0=1 ... 1 \gcd(x,y)=g if and only if g is the smallest positive integer that can be expressed in the form g=ax+by where a,b \in \mathbb Z. Since \gcd(k,n)=1 is given, then you know that there exists integers a and b such that ak+bn = 1$$ So \begin{align} 1 &= ak + bn \\ &= -a(-k) -an + an + bn \\ &= -a(n-k) + (a+b)n \\ ... 1 To prove the given implication you need to show that if the antecedent, \forall_{x} \exists_{y} [y \Delta y = x]\ $, is true then the consequent,$\ \forall_{x} \exists_{a} \exists_{b} \left[ (a \Delta a) \Delta b = x \right]\ $, must also be true—that is, given any$\ x\ $you can find$\ a\ $and$\ b\ $such that$\ (a \Delta a) \Delta b = x\ $. The ... 1 Since$ac \ge bd$, we can write: $$ac \ge bd \Longrightarrow 1 < \frac{b}{a} \le \frac{c}{d} \Longrightarrow 1 < \frac{c}{d} \Longrightarrow c > d$$ 1 You only have that$adq\geq bd,$not$>.$It’s still true that$q>1,$but in either case it is not clear exactly how you know that$q >1.$But you could have extended your chain of inequalities like this: $$0 < ad < bd \leq ac,$$ and from this you get$ad < ac.$Then use the fact that$a>0.$1$\|g(x)\|$attains its minimum value (by compactness of$K$and continuity). Let$m$be this minimum value. Choose$N$such that$\|g_n(x)-g(x)\| <m/2$for$n \geq N$. Then$|g_n(x)| \geq |g(x)| -|g_n(x)-g(x)|>\frac m 2$so$n \geq N$which implies$g_n(x) \neq 0$for every$x$. This is what is meant by saying that$\frac {f_n} {g_n}\$ is everywhere ...

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