I have already proved $ x &lt; y =&gt; x^n &lt; y^n $ I am currently trying to prove the converse statement, but I am getting nowhere. The textbook suggests trying to set up a contradiction, but I can't seem to succeed with this.
17436
Here, we suppose that $x,y\in\mathbb{R}$ and that $x^n=y^n$, where $n$ is odd. I want to prove that $x=y$.  Maybe we can use that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1})$  So, it suffices to show that if $x^n=y^n$ then $x^{n-1}+x^{n-2}y+...+xy^{n-2}+y^{n-1}\neq 0$  Any hint?
22556
How can the following inequality be proven? $$x^y+y^x\gt1$$ for $(x\gt0,y\lt1)$
54284
For all $n \in \mathbb{N}$, $n &gt; 1$ and $x,y \in \mathbb{Z}$ prove that      \begin{equation}   (x+y) \text{ mod } n   = ((x \text{ mod } n)+(y \text{ mod } n))\text{ mod } n. \end{equation}   Would this be a proof if I say:  Let $x = x' \text{ mod } n$ and let $y=y' \text{ mod } n$. Then  \begin{equation}   x+y = x'+y' \text{ mod } n. \end{equation}  Now, if we look at the current thing we want prove, this will follow:  \begin{equation}   x+y \text { mod } n = (x \text{ mod } n + y \text{ mod } n) \text{ mod } n. \end{equation}    I'm not sure if this is allowed or will count as a "proof" at all. Maybe you can show me some other ways of doing this, more precise ones?
14021
I would like to prove the statement in the title.  Proof: We prove that if $f$ is not strictly decreasing, then it must be strictly increasing. So suppose $x &lt; y$.  And that's pretty much how far I got.  Help will be appreciated.
522
Can't quite finish this proof:     Prove that if $x$ and $y$ are not both $0$ then $ x^2 +xy +y^2&gt; 0$   $ x^2 +xy +y^2 +xy -xy&gt; 0$  $ (x +y)^2 -xy&gt; 0$ Without loss of generality define $x\geq y$  Firstly the case where $y&lt;0$ and $x&gt;0$ is trivial as both $ (x +y)^2$ and $-(xy) &gt;0$.   I want to say by transitivity  as each are >0 that there sum must be ${}&gt; 0$ can I do that?  The second trivial case is $y=0$ as we simply have $ x^2 &gt;0 $ where $x&gt;0$.  Lastly we have define $c= x+y$  since both $x$ and $y$ must have the same sign we know $|c|&gt;x$ since $x\geq y$ we know that $ x^2 \geq xy$ finally since  $ |c|^2 = c^2$ $ \forall c \in \mathbb{R} $  We have $ c^2 &gt;x^2 \geq xy$ hence $c^2 -xy &gt;0$  I feel like I have all the pieces but it doesn't feel finished how do I fix it?  Edit: I am really glad i asked this question love the variety of different good answers no idea what a discriminant is but im going to go look that up. I wanted to give the accepted answer to Edward as he actually answered my question but Anurag s proof was just so much better then my attempt. ^^
76995
This is exercise $1.5$A from Engelking's book, page $47$.  Verify that $X$ is $T_{0}$ if and only if $\overline{\{x\}} \neq \overline{\{y\}}$ for every pair of distinct points $x,y$.  My try:  Suppose first that $X$ is $T_{0}$ and let $x,y$ be two distinct points of $X$. Then since $X$ is $T_{0}$ wlog we can find an open set $U$ such that $x \in U$ and $y \not \in U$. Let us show that $\overline{\{x\}} \neq \overline{\{y\}}$. Suppose otherwise, then $x \in \overline{\{x\}} \subseteq \overline{\{y\}}$. This in turn implies that $U \cap \{y\} \neq \emptyset$ so that $y \in U$, a contradiction.  For the other direction, let $x,y$ be distinct points of $X$. Then either $\overline{\{x\}}$ is not contained in $\overline{\{y\}}$ or $\overline{\{y\}}$ is not contained in $\overline{\{x\}}$. Assume the former, then $x \not \in \overline{\{y\}}$ so we can find an open set $U$ such that $U$ contains $x$ and $U$ does not intersects $\{y\}$. If the other case happens we can find an open set $V$ such that $V$ contains $y$ and $x \not \in V$. Therefore $X$ is $T_{0}$.   Is this OK?
62597
I know that this is valid because by thinking about it, you can see it's true, but I'm just not sure how I would go about actually proving it mathematically. I've kind of tried proof by contradiction, since that's what I use most of the time, but it doesn't seem to work.    In my mind, A ∪ B ⊆ C kind of just implies that A ⊆ C and B ⊆ C.
42667
Is there a clever way to show that  $$|x| - |y| \leq |x-y|$$  I believe I can think of a way to solve this using cases, but the book I'm working out of said that "A very short proof is possible if you write things in the right way" which has motivated me to figure out this quick way, but I've failed thus far and any help would be appreciated
61024
My proof: By the way of contradiction, suppose that we found $(x,y) \in \mathbb N^*$ satisfies:$$\frac{x+y}{xy}=5$$ So : $$x+y=5xy \\ \iff x=5xy-y \iff x=y(5x-1)$$ Since $x\in \mathbb N^* $ so the minimum value of $5x$ is $5&gt;1 \implies y\mid x$ By the same method we have: $$y=x(5y-1) \implies x\mid y$$ $x\mid y, y\mid x \implies x=y$, Now we can write the original equation as : $$\frac{2x}{x^2}=5 \iff 2=5x \implies x\notin \mathbb N^* $$ Therefore there is no solution to this equation: $$\frac{x+y}{xy} =5$$ Does this considered a proof?
116030
I know how to prove this using an element and subset proof but that's not what I'm being asked to do. I'm suppose to prove it using the distributive law and the associative law but I'm just not seeing how to do this. Can you give me some advice on how to start this. I figure the first thing would be $$((A \cup B) \cap A) \cup ((A \cup B) \cap B^c))$$ But I'm not sure where to go from there.
48127
I was trying to prove it this way. Let $a,b\in\mathbb{R}$ with $a&lt;b$. Let $ I=(a,b)$. As for all $x\in I$,$x&lt;b$, we get $b$ is an upper bound. let $u=\sup I$, then $u\leq b$. If $u=b$ it's done. Otherwise, let's suppose $u&lt;b$. Furthermore, for all $x\in I$, $a&lt;x$ and $x\leq u$, so $a&lt;u$, then $u\in I$...  Can I get somewhere this way?
14779
How would one prove that  $$(n+1)^{n-1}&lt;n^n \ \forall n&gt;1$$  I have tried several methods such as induction.
3350
I am trying to prove a result, for which I have got one part, but I am not able to get the converse part.  Theorem. Let $R$ be a commutative ring with $1$. Then $f(X)=a_{0}+a_{1}X+a_{2}X^{2} + \cdots + a_{n}X^{n}$ is a unit in $R[X]$ if and only if $a_{0}$ is a unit in $R$ and $a_{1},a_{2},\dots,a_{n}$ are all nilpotent in $R$.  Proof. Suppose $f(X)=a_{0}+a_{1}X+\cdots +a_{n}X^{n}$ is such that $a_{0}$ is a unit in $R$ and $a_{1},a_{2}, \dots,a_{r}$ are all nilpotent in $R$. Since $R$ is commutative, we get that $a_{1}X,a_{2}X^{2},\cdots,a_{n}X^{n}$ are all nilpotent and hence also their sum is nilpotent. Let $z = \sum a_{i}X^{i}$ then $a_{0}^{-1}z$ is nilpotent and so $1+a_{0}^{-1}z$ is a unit. Thus $f(X)=a_{0}+z=a_{0} \cdot (1+a_{0}^{-1}z)$ is a unit since product of two units in $R[X]$ is a unit.   I have not been able to get the converse part and would like to see the proof for the converse part.
47722
I need to prove the following statement: if $xy=0$   then  $x=0$ $or$ $y=0$.  My first idea was to prove the negation of the original problem if $x\neq0$ and $y\neq0$ then $xy\neq0$ but I still don't know how to prove it rigorously.
109622
I'm trying to do my Maths assignment but I can't get this done. I can do other questions but this one is different. This is a picture of the question.   (source: )    My usual first step is to proof it when n = 1. But the problem is, that I don't know what x is. It says x is not equal to 1, but I still don't know what to do next.  Thank you.
60388
$x$ and $y$ are real numbers.  Given that $1&lt;x^2-xy+y^2&lt;2$, how can I show that $\frac 29&lt;x^4+y^4&lt;8$ ?  Then can I use that to prove that for any natural number $n&gt;3$  $$x^{2^n}+y^{2^n}&gt;\frac 2{3^{2^n}} \text{?}$$
30704
I have a "proof" that has an error in it and my goal is to figure out what this error is. The proof:  If $x = y$, then  $$ \begin{eqnarray} x^2 &amp;=&amp; xy \nonumber \\ x^2 - y^2 &amp;=&amp; xy - y^2 \nonumber \\ (x + y)(x - y) &amp;=&amp; y(x-y) \nonumber \\ x + y &amp;=&amp; y \nonumber \\ 2y &amp;=&amp; y \nonumber \\ 2 &amp;=&amp; 1 \end{eqnarray} $$    My best guess is that the error starts with the line $2y = y$. If we accept that $x + y = y$ is true, then   $$ \begin{eqnarray} x + y &amp;=&amp; y \\ x &amp;=&amp; y - y \\ x &amp;=&amp; y = 0 \end{eqnarray} $$  Did I find the error? If not, am I close?