I already have the demonstration that $x &gt; y$ implies that $x^{n} &gt; y^{n}$ for $x$ and $y$ positives and I read that to prove the inverse way I could use the contrapositive form but this would give me that $x^{n} \le y^n \Rightarrow x\le y$ not $x^n &lt; y^n\Rightarrow x&lt;y$. So, how can I prove this?
120645
I tried proving it but I'm quite skeptical about it:  $(\Rightarrow )$ Assume $x&lt;y$. Clearly, the base case for the induction is done. Assume $x^k&lt;y^k$ for some $k\in \mathbb{N}$.  Now, $x^{k+1}=x^k\cdot x&lt;y^k\cdot y=y^{k+1}$.   $(\Leftarrow )$ Assume for contradiction that $x\not&lt; y$. Then $y\le x$. Now, suppose $y\ne x$. Then $y&lt;x$. It follows that $y^n&lt;x^n$ for every $n\in \mathbb{N}$ from $(\Rightarrow )$. This contradicts our hypothesis. Also, if $y\not&lt; x$ then $x=y$. Then $y^n=x^n$ for every $n\in \mathbb{N}$ which also contradicts our hypothesis.  Is this proof correct?
122936
I know I have to prove both sides here so: $\implies$: Suppose $|a| &lt; b$. Since $|a|$ is a positive number, we know $b$ is a positive number greater than $|a|$:  If $a$ is positive, $a &lt; b$ If $a$ is negative, $a &lt; b$  But for $|a| &lt; b$, it has to be that $a &gt; -b$. $a &lt; b$ and $a &gt; -b$, therefore $–b &lt; a &lt; b$ is true. $\impliedby$: Suppose $-b &lt; a &lt; b$.  If $a$ is positive, then $b$ is a greater positive number, and $-b$ would be less than $a$ If $a$ is negative, then $b$ can only be a greater positive number for $-b &lt; a$ to hold  So, $b$ is a greater positive integer than $a$. Therefore, it must be that $|a| &lt; b$. This was what I came up with but it seems convoluted and kind of forced with the cases. Any suggestions?
98111
Prove that $x^2+y^2=z^n$ has a solution $(x, y, z)$ in $\mathbb{N}$ for all $n\in\mathbb{N}$  I tried to prove this by induction, but couldn't. ( This was probably because the solution for some $n$ isn't necessarily related to the solution for $n+1$)  I can't seem to see any other way other than induction for proving the statement. Any help/hints on solving this problem, or any alternative approach will be highly appreciated :)
9583
So lets say I have $$ n^{2} \le n! $$  For what positive integers is this not true? $n=2$ and $ 3$  Base case? $$n=4 \implies  16 \le 24 $$  What is the inductive hypothesis and how do I show the inductive proof? Thanks
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?
67814
Prove $$\left(x^2 - y^2\right)\left(\frac1y - \frac1x\right) \ge 0$$ where $x$ and $y$ are positive real numbers.   Can I simplify $\frac{(x^2 - y^2)(x-y)}{xy} \ge 0$ and then $(x^2 - y^2)(x-y) \ge 0$ cancelling out the $xy$?  Is this valid because then  \begin{align} (x+y)(x-y)(x-y)&amp;\ge  0\cdot(x-y)^2\ge 0 \end{align}  which is always true and then prove backwards from here?
50770
We have the sets $X,Y$ and $f:X\to Y, \ g: Y\to X $      Prove if $f$ has a right inverse function: $f\circ g=id_Y$ $\iff$ $f$ is onto $Y$ (surjective).   $\Leftarrow \forall y\in Y : \exists x\in X$ then $f(y)=x$ I need to show that $f\circ g(x)=x$ but I don't know how.  I have no clue on to start in the other direction.   BTW, I found below similar question but they're either not quite the same as this one or I didn't understand the answers:
122407
I want to show that $\forall x$,$y\in G$, $(xy)^2=x^2y^2 \iff G $ is an abelian group.  Proof of the backward direction seems easy enough:  Assume $G$ is an abelian group. Then consider $(xy)^2=(xy)(xy)$. By associativity, $(xy)(xy)=xyxy$. Since $G$ is abelian, $xyxy=xxyy=x^2y^2$.   Proof of the forward direction I attempt by contrapositive:  That is, assume $G$ is not an abelian group. Then, consider $(xy)^2=xyxy$ as above. Yet, $xyxy\neq xxyy=x^2y^2$ since we no longer have commutativity.  Is this proof by contrapositive completely correct?
56780
I'm doing exercise on discrete mathematics and I'm stuck with question:     If $f:Y\to Z$ is an invertible function, and $g:X\to Y$ is an invertible function, then the inverse of the composition $(f \circ\ g)$ is given by $(f \circ\ g) ^{-1} = g^{-1} \circ\ f^{-1}$.   I've no idea how to prove this, please help me by give me some reference or hint to its solution.
75449
Just would like to see if I'm right.  Proof: Let $G$ be generated by $x$ (Let $G=\langle x \rangle$ for some $x$ in $G$.) We wish to show that there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$. First, we will show that for all $x^n$ in $G$, $f(x^n)=f^n(x)$. We proceed by induction.   Base case: $n=1$. $f(x*e_G)=f(x)f(e_G)=f(x)$. Now, assume that $f(x^k)=f^k(x)$ for some arbitrary positive integer k. We show that $f(x^{k+1})=f^{k+1}(x)$. Observe that $f(x^{k+1})=f(x^kx)=f(x^k)f(x)=f^k(x)f(x)=f^{k+1}(x).$   Next we will show that for all $x^{-n}$ in $G$, $f(x^{-n})=f^{-n}(x)$. We proceed by induction. Base case: $n=-1$. $f(x^{-1}*e_G)=f(x^{-1})f(e_G)=f(x)$. Now, assume that $f(x^{-k})=f^{-k}(x)$ for some arbitrary positive integer k. We show that $f(x^{-k-1})=f^{-k-1}(x)$. Observe that $f(x^{-k-1})=f(x^{-k}x)=f(x^{-k})f(x)=f^{-k}(x)f(x)=f^{-k-1}(x).$ Because $f$ is onto, $Im(f)=G'$ and therefore $Im(f)=(...,f^{-2}(x),f^{-1}(x),e_G,f(x),f^2(x),...)=G'$.  Thus, there exists an $x'$ in $G'$ such that $G'=\langle x' \rangle$ and the proof is complete by the principle of mathematical induction. Q.E.D.  I have a feeling the proof shouldn't be this long. Any suggestions for a shorter, cleaner proof?
107714
I am trying to understand the proof above, Proposition 7, given by the author, but I got stuck at some points.   My explanation:  Now, suppose for contradiction there is an $\epsilon &gt; 0$, for all $x &gt; 0$ such that  $NOT (x^2 &lt; 2 &lt; (x+ \epsilon)^2)$,  Since $NOT (x^2 &lt; 2 &lt; (x+ \epsilon)^2) \ \ iff \ \ (x^2 \geq 2 \ \ OR  \ \ (x+ \epsilon)^2 \leq 2)$,  We must show that for some $\epsilon&gt;0$ and for all $x &gt; 0$, the statement $x^2 \geq 2 \ \ OR  \ \ (x+ \epsilon)^2 \leq 2$ leads to a contradiction.  I think in the proof above, author tries to get a contradiction using right hand side of $OR$ clause, namely $(x+ \epsilon)^2 \leq 2$. But no contradicition is achieved by him for the left hand side side of $OR$ clause namely, $x^2 \geq 2$.  First Question is, am I correct at this point, i.e. the need for a contradiction for the left hand side side of $OR$ clause?  Second question is, I don't understand how he derived the following sentence in his proof above,     Since $0^2 &lt; 2$, we thus have $\epsilon^2 &lt; 2$, which then implies $(2 \epsilon)^2 &lt; 2$ and indeed $(n \epsilon)^2 &lt; 2$   Why does he feel a need to multiply $\epsilon$ constantly by $n$
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.
69535
Let $f(x) = |x|$ for $x \in [-1,1]$ and extend $f$ to all of $\mathbb R$ by requiring $f(x+2)=f(x)$. Prove that $|f(x)-f(y)|\le|x-y|$ for all $x$ and $y$.   This is easy to observe on a graph but my proof is long and a bit tedious. Just curious if there is a quick/elegant way?
116157
If $n\geq m$ show that: $(x^m+y^m)^{1/m} \ge (x^n+y^n)^{1/n}$, all numbers being positive real.   Obviously since $n\geq m$, for every $a\in \Bbb Z^+$: $a^{1/m}\geq a^{1/n}$, but does it help?
38566
Prove that $\forall x,y&gt;0, x^x+y^y \geq x^y + y^x$   A friend of mine told me none of the teachers in my school have succeeded in proving this seemingly simple inequality (it was asked at an oral exam last year). I tried it myself, but I made no significant progress.
19432
Let $x,y \in L_p$   such that  $\|x\|_p=\|y\|_p=1$ , $1&lt; p&lt;\infty$ and $x\neq y.$ Why is $\|x+y\|_p&lt;2$  ?  I'm not sure how to start the proof.. I don't know how to handle integral of $(x+y)^p$ and it seems that using the binomial theorem won't be a great success.
102547
Prove if $x, y ∈ R$ and $xy &gt; 0$ then either $x &gt; 0$ and $y &gt; 0$, or, $x &lt; 0$ and $y &lt; 0$ using only the field axioms.  These include the Field axioms for addition, multiplication, distribution, order axioms (trichotomy and transitivity), ordered field axioms. I'm stuck on this and not sure how to approach it. Would proof by contrapositive be a good approach in this case?