This holds for if $x$ is natural nunbers greater than $1$ so i think it would work for every real nunber $x$ greater than $1$.
69355
For $\alpha\in\mathbb{N}$ we can use the Binomial and get: $$\left(a+b\right)^{n}=\sum_{k=0}^{n}{n \choose k}a^{k}b^{n-k}=\sum_{k=1}^{n-1}{n \choose k}a^{k}b^{n-k}+a^{n}+b^{n}&gt;a^{n}+b^{n} $$  But what about rational and irrational powers?  Respectively, can we also say that $\left(a+b\right)^{\alpha}&lt;a^{\alpha}+b^{\alpha}$ for all $\alpha&lt;1$?
30665
Suppose that $a,b,m,n$ are natural numbers, prove the following statement:  If $a({m \over n})=b$ and $(a,n)=1$, then ${m \over n} \in \mathbb{N}$
80733
Problem :   Let $n$ be a positive integer, for a real number $x$, let $[x]$ denote the largest integer not exceeding x and $\{x\} =x -[x]$ Then $$\int^{n+1}_1 \frac{(\{x\})^{[x]}}{[x]}dx$$ is equal to   (a) $\log_e(n)$   (b)  $\frac{1}{n+1}$   (c)  $\frac{n}{n+1}$   (d)  $1+\frac{1}{2}+\cdots +\frac{1}{n}$   Please suggest how to proceed in such cases not getting any clue how to proceed in cases of fractional part and greatest integral cases. please guide will be of great help thanks.
7937
You can fool some of the people all of the time, and you can fool all of the people some of the time, but you can't fool all of the people all of the time.    My attempt is   $$((\exists(x),A(x)) \land(\forall(x),S(x))) \land \neg (\forall(x), A(x))$$  Such that  $A(x)=$ You can fool all the time  $S(x)=$ You can fool some of the time  Just starting out basic logic and I'm looking to see where I"m making mistakes. Thanks
108585
Let $n$ be a natural number. Is it possible to write $$(x+y)^n \leq C(x^n + y^n)$$ for some constant $C$??  It is obvious for $n=2$ (using Young's inequality) but not obvious to me for other $n$.  Let $x$ and $y$ be positive reals.
40316
Exercise 3.5.12. Let $ f: \mathbb N \times \mathbb N \to \mathbb N$ be a function, and let c be a natural number. Show that there exists a function $a:\mathbb N\to \mathbb N$ such that $$a(0)=c$$ and $$a(n++)=f(n,a(n))\;for\;all\;n\in \mathbb N,$$ and furthermore that this function is unique. Tao gives the hint: first show inductively, by a modification of the proof of Lemma 3.5.12, that for every natural number $N\in \mathbb N$, there exists a unique function $a_N:\{ n\in \mathbb N:n\lt N\}\to \mathbb N $ such that $a_N(0)=c$ and $ a_N(n++)=f(n,a(n))$ for all $n\in \mathbb N$ such that $n\lt N$.  In my opinion, we could solve it by proving that $\forall n\in \mathbb N,a(n)\in \mathbb N$ according to mathematcial induction.  The property P(n) is:$$\forall n\in \mathbb N, a(n)\;is\;unique$$ P(0) is true. And if P(n) holds, P(n++) is true because different natural numbers must have different successors. $$a(n++)=f(n,a(n))$$ $$a(m++)=f(m,a(m))$$ $$n++=m++\Rightarrow n=m\Rightarrow a(n++)=a(m++)$$ It's similar to the proof of  Proposition 2.1.16 (Recursive definitions): Suppose for each natural number $n$, we have some function $f_{n} : \mathbb{N} \rightarrow \mathbb{N}$ from the natural numbers to the natural numbers. Let $c$ be a natural number. Then we can assign a unique natural number $a_{n}$ to each natural number $n$, such that $a_{0} = c$ and $a_{n{++}} = f_{n} (a_{n})$ for each natural number $n$.  This proof indicates both the existence and uniqueness of the function $a$. I don't know why Tao says it's informal.  But regarding the hint, how do we get the function f even if we've got the conclusion from the hint. Also it seems like the function a(n) can be diffentent considering different N. I don't know if the hint is correct.
55879
The following is an exercise (Exercise #3 (a), Chapter 3, page 28) from Richard Stanley's Algebraic Combinatorics.  Let $P(x)$ be a nonzero polynomial with real coefficients.  Show that the following two conditions are equivalent:  There exists a nonzero polynomial $Q(x)$ with real coefficients such that all coefficients of $P(x) Q(x)$ are nonnegative.  There does not exist a real number $a &gt; 0$ such that $P(a) = 0$.    That the first item implies the second is straightforward (since if $a$ is a positive real root of $P(x)$ then it is a positive real root of $P(x)Q(x)$, but since $P(x)Q(x)$ is nonzero with all coefficients nonnegative, this is impossible).  However, I can't seem to find a way to prove that the second item implies the first.  I would very much like to see a proof if someone can provide it.
104310
I need to prove that for every natural n that happens and the prove needs to be by induction, I wont prove for n=1 because its obvious,so lets move to n and n+1 $$ \frac{(2n!)}{(n!)^2} ≥ \frac{(4^n)}{(2n+1)} $$ the assumption for n+1 $$ \frac{(2n+2)!}{(n+1)!^2} ≥ \frac{(4^{n+1})}{2n+3} $$ now I dont now how to progress,I tried this for n+1 $$ \frac{(2n+2)(2n+1)(2n!)}{(n+1)!^2} ≥ \frac{(4^{n}4^1)}{(2n+1)+2)} $$ and had a suggestion to use $ 2n! = (n!) \prod _{n=n}^{2n}\:n\cdot )! $ but cant see how it helps,ty.
78540
I am trying to prove that for a continuous $f\colon[1,\infty)\rightarrow\mathbb R$ and $f(x)\to a$ as $x\to\infty$ it could be approximated by $g(x)=p(1/x)$ where $p$ is a polynomial.
5557
The mapping is supposed to be from $\mathbb{N}$ to $\mathbb{N}$.  I'm still trying to understand if this is possible, I mean if it was from $\mathbb{R}$ to $\mathbb{N}$, I guess $x^2$ would work.
10362
Does $1.0000000000\cdots 1$ (with  an infinite number of  $0$ in it) exist?
10408
Prove that for any natural number $n&gt;1$, we have $(2n)!&lt;(n(n+1))^n$    I tried using induction, but I failed in that approach, I rather found it was untrue, but on several case testing, I found it to be true. I am out of ideas here, please help. Thank you.
105641
This is a problem from Rudin, but I wanted to add my own intuition to it. It uses Rudin's definition of Archimedean property. I'd just like to know if my version holds  If $x \in \mathbb R$, $y\in \mathbb R$ and $x&lt;y$, then $\exists p \in \mathbb Q$ such that $x &lt; p &lt; y$  Since $x &lt; y$, then $y-x&gt;0$. Applying Archimedean property, we see $\exists n \in \mathbb Z^+$ such that $n(y-x)&gt;1$. Allow $A=\{i\in \mathbb Z\mid i &gt; nx\}$. Obviously $A$ is bounded by $nx$, so take $m = \inf A$. So we have $m &gt; nx$ and $m-1 \leq nx$ by set and $\inf$ definition. This gives us $m-1 \leq nx &lt; m$. Combining and rearranging with $n(y-x) &gt; 1$ gives us $nx &lt; m \leq nx +1 &lt; ny \implies nx &lt; m &lt; ny$. Thus $x &lt; \frac{m}{n} &lt; y$
45171
Show that there is at least one real value of $x$ for which      $$x^{1/3} + x^{1/2} = 1$$   I did draw the graphs of $x^{1/3}$ and $1-x^{1/2}$ and showed that they met at a point, but I don't think it's a good algebraic proof.  How should I proceed after substituting $x$ for $z^{1/6}$ and getting a polynomial in terms of $z$
95138
I'm trying to construct a continuous surjection from $(0,1]$ onto $[0,1]$, but I'm not getting anywhere. I don't immediately see a contradiction which falsifies the existence of such a function, so my intuition tells me one exists. I feel like an absolute value function would work, but I'm not sure how to arrive at it in the proper way. Thanks for any help.
55328
Show given any $x\in\mathbb{R}$ show there exists a unique $n\in\mathbb{Z}$ such that $n-1\leq x &lt;n$. I already know that there exists a $n\in\mathbb{N}$ such that $x&lt;n$ by the archimedean property. But to prove that $n-1\leq x$ part I'm not sure. I think I have to create some set such that $n-1$ is a lower bound. But I'm not sure.
120688
I'm thinking of $f$ as a function either from the reals to the reals or from the naturals to the naturals.   Edit: Okay, what if $f$ is from the naturals to the naturals?