# Tag Info

10

Suppose $x$ and $y$ are irrational while $x+y$ is rational. Assume on the contrary $x-y$ was rational. Then $x = \frac{(x+y)+(x-y)}{2}$ is rational. Contradiction.

10

Put $$a_n:=\sin\frac{n\pi}3\implies \begin{cases}a_{3n}=0\xrightarrow[n\to\infty]{}0\\{}\\a_{6n+1}=\frac{\sqrt3}2\xrightarrow[n\to\infty]{}\frac{\sqrt3}2\end{cases}$$ so we have two subsequences converging to different limits and thus the original sequence's limit cannot exist.

9

There is absolutely no need to use contradiction. Just prove the statement directly: there is no smallest positive real number. What this means is that if $r$ is a positive real number, then it isn't the smallest one. And indeed it's not because $\frac{r}{10}$ (or whatever) is smaller.

8

Let $m=\max\{|(x+y)_i|\}$. Then the maximium is achieved for some index, $\nu$ say. Then $$m=|(x+y)_\nu|\stackrel{(1)}=|x_\nu+y_\nu|\stackrel{(2)}\le |x_\nu|+|y_\nu|\stackrel{(3)}\le \max\{|x_j|\}+\max\{|y_k|\}$$ where $(1)$ follows from the definition of vector addition, $(2)$ is the triangle inequality, and $(3)$ follows from the definition of maximum.

7

Another way to show this is using the Archimedian property of natural numbers. That is, the natural numbers don't have an upper bound in the reals. Consider any small real number, $\epsilon>0$. Since natural numbers are unbounded, there exists some $n \in \mathbb{N}$ such that $n>\frac{1}{\epsilon}$. Rearranging gives that $\epsilon>\frac{1}{n}$. ...

6

Let $g(x)=f(x)(x-a)(b-x)$ Then $$\int_a^b f(x)^2 (x-a)(b-x) dx =0$$ Note that $$f(x)^2 (x-a)(b-x)\geq 0,\ (x-a)(b-x) > 0\ (x\in (a,b))$$ If for some $x\in [a,b]$, $f(x)\neq 0$ then since $f$ is continuous, there exists a closed set $x\in [s,t]$ : $$f(x)^2\geq c > 0\ {\rm on}\ [s,t] \subseteq [a,b]$$ Hence $$\int_a^b f(x)^2 (x-a)(b-x) dx \geq ... 5 Here is a slightly different way to organize the proof. What we will do is split it into two parts: For every positive real number there is another positive real number less than it. Proof: Let x>0. Then since 0<\frac{1}{2}<1, we have x>\frac{1}{2}x>0, and so \frac{1}{2}x is such a number. There is no smallest positive real ... 5 If n\ge 4, then n! is divisible by 4, but x^2+1 cannot be. For n=3, 6 is not of the form x^2+1. n=0, 1, 2 were addressed in a comment above. 5 Here is a formal proof. Fix 1\leq i \leq n, then it is always the case that$$ |x_i|\leq \max\{|x_j|\}. $$In particular, and using the triangle inequality, we have$$ |x_i+y_i|\leq |x_i|+|y_i|\leq \max\{|x_j|\} + \max\{|y_k|\}. $$Since we pick an arbitrary i, the previous inequality is true for all 1\leq i \leq n, hence also true for the i giving ... 5 It looks right. Another possible approach is the following one - since:$$ f(x)=x^2+3x+2 = \left(x+\frac{3}{2}\right)^2-\frac{1}{4} $$we have that f(x) attains its minimum in x=-\frac{3}{2} (the abscissa of the vertex) and for every r>-\frac{1}{4} the equation$$ f(x)=r $$has two solutions, symmetric with respect to x=-\frac{3}{2}. It follows ... 5 Suppose$$n\lt \left(\frac32\right)^n$$then$$\left(\frac32\right)^{n+1}=\left(\frac32\right)\left(\frac32\right)^n\gt\frac32n\gt n+1\,\,\,\,\,\forall n\ge 3$$Therefore verify your assertion for n=1, 2 separately. Then Mathematical Induction will leads to your answer. 4 The proof given looks fine to me, modulo my comment (see above). Here's another way to see it, sans calculus: Choose x_1, x_2 \in [-1, 1] with x_1 > x_2. Then f(x_1) - f(x_2) = x_1^2 + 3x_1 + 2 - x_2^2 - 3x_2 - 2 = x_1^2 - x_2^2 + 3(x_1 - x_2) = (x_1 + x_2)(x_1 - x_2) + 3(x_1 - x_2) = (x_1 + x_2 + 3)(x_1 - x_2); \tag{1} note that x_1 - x_2 ... 4 Here is a sketch of a contraposition argument. Suppose that 3 is not a factor of m or of n. Then each is either one greater or one less than a multiple of 3, i.e., there exist integers j and k such that m=3j\pm1 and n=3k\pm1 (with independent sign choices). Multiply these together and you will have 3(\text{stuff})\pm 1, so mn is not a ... 4 The way to express your ideas formally is to use modular arithmetic. To say that the possible remainders for a^2 are 0,1,2, and 4 is to say that$$ a^2 \equiv 0,1,2,4 \pmod 7. $$The same is true for b^2, and 7|a^2 + b^2 if and only if a^2 + b^2 \equiv 0 \pmod 7. But notice that if i,j\in \{0,1,2,4\}, then i+j \equiv 0 \pmod 7 if and only if ... 4 Yes it does. Suppose x \in B^c, we want to show that x \in A^c. In other words, we want to show that x \notin A . Suppose not. Then x \in A . By hypothesis, A \subset B , therefore x \in B  which is a contradiction since we assumed x \in B^c ( x \notin B ). Therefore, x \in A^c. We have shown that B^c \subset A^c  as desired. 4 It is obvious that ceiling of a real number is at least as much as that number, therefore, Ceiling(x)Ceiling(y)\geq xy. As Ceiling(xy) is defined to be such minimum integer, we have Ceiling(x)Ceiling(y)\geq Ceiling(xy). 4 Following my hint in the comments, one of the numbers \{n,n+2, n+4\} must always be a multiple of 3. However the hypotheses of the problem are that all three are prime. Hence one of them must be the specific prime 3, as that is the only prime number that is also a multiple of 3. So there are three cases: n=3. Excluded, since p>3 forbids ... 4 If you want to prove that every polynomial map \mathbb R^n\to\mathbb R is continuous, it is not a good idea to consider a general polynomial and give an \varepsilon-\delta-proof. Instead, you should start with the building blocks, i.e. the projections x\mapsto x_i. Use an \varepsilon-\delta-proof to show that all the projections are continiuous. ... 4 Definitions are just shorthand. For example "f is a function from A to B" is shorthand, in set theory, for$$f\subseteq A\times B \land \left(\forall x(x\in A\implies\exists_1 y((x,y)\in f))\right)$$And here, even \exists_1 is a shorthand. Definitions are a way to avoid writing the same thing over and over again. Another example: Saying "p,q ... 4 You lacked to mention how addition and multiplication on cuts are defined. Addition is simple: A+B:=\{\,a+b\mid a\in A\land b\in B\,\}. One readily shows that this is indead a cut if A,B are cuts. Thankfully, this definition is so simple that commutativity of + follows immediately from commutativity of + in \mathbb Q and from commutativity of ... 4 Hint:$$n!>n^2(n+1)!= (n+1)\cdot n! > (n+1)\cdot n^2 > (n+1)^2$$3$$A+A\cdot C = A\cdot 1 + A\cdot C = A\cdot(1+C) $$since A\cdot 1=A, and you're free to use identities "in reverse". 3 Elaborating on N. S.'s comment, if f is non zero at some x_0 \in [a,b], then it is non-zero in some region [x_0- \delta , x_0 + \delta ]. WLOG take f > 0 in this region. Then find a g s.t. g(a) = g(b) = 0 but g(x) > 0  \forall x \in [x_0-\delta,x_0+\delta] (think of defining g piecewise). Further, make g non zero only in this ... 3 Simpler counter-example: F(x) = 1, 0 = G(x) :) However it's true that: assuming F(x_0) \le G(x_0) and F'(x) \le G'(x) for x \ge x_0 we have F(x) \le G(x) for all x \ge x_0. 3 Using continuity of f at x = x_0, you can find an open interval (x_0-\delta,x_0+\delta) such that |f(x)-f(x_0)| < \dfrac{f(x_0)}{2} \to f(x) > \dfrac{f(x_0)}{2} \to \displaystyle \int_{a}^b f(x)dx \geq \displaystyle \int_{x_0-\delta}^{x_0+\delta} f(x)dx > \displaystyle \int_{x_0-\delta}^{x_0+\delta} \dfrac{f(x_0)}{2}dx = \delta\cdot f(x_0) ... 3 If x>y, then \max (x,y)=x and \min (x,y)=y. So,$$\max (x,y)+\min (x,y)=x+y$$If x\le y, then \max (x,y)=y and \min (x,y)=x. And we have,$$\max (x,y)+\min (x,y)=y+x=x+y

3

Hint: First, prove that $f(1)=0$ (writing $1=\frac11$). Then, prove that $f(-1)=0$ (writing $-1=\frac{-1}1$ and $-1=\frac1{-1}$). Then, using this prove that $f(-x)=f(x)$ (writing $x-=\frac{x}{-1}$) for any $x\neq0$.

3

What definitions are in the most primitive sense (mathematical and linguistic): Definition, n. $\qquad$ The settings of bounds or limits; limitation, restriction. The above is the primary meaning of definition from the authoritative Oxford English Dictionary. Of course, you are more interested in the role(s) of definitions in mathematics. The following ...

3

Hint: $y = \frac{(x+y) - (x-y)}{2}$

3

Name the vertices with $X=\{x_i\mid i=1,\dots,n\}$ and $Y=\{y_i\mid i=1,\dots,n\}$ such that $M=\{\{x_i,y_i\}\mid i=1,\dots,n\}$ is a perfect matching. Assume, towards contradiction, that we have more than $\binom{n}{2}$ edges which is not in any perfect matching. Partition $E(G)\setminus M$ to $E_{i,j}=\{\{x_i,y_j\},\{x_j,y_i\}\}\cap E(G)$ for $i<j$ and ...

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