In many diciplines that utlizes mathematics, we often see the equation $$y=y(x)$$ where $y$ might be other replaced by whichever letter that makes the most sense in ...

Nov 30, 2016 · 0 This might be same argument as in previous answer: $$ (x^ {-1}\cdot yx)^2= (yx\cdot x^ {-1})^2 \,\,\,\, \Rightarrow \,\,\,\, x^ {-1}y^2x=y^2.$$

Aug 3, 2022 · I am showing that if for some group $G$, $xy=yx$ for every $x, y \in G$ then $$ (xy)^n=x^ny^n.$$ I claim this holds by induction on $n$. So base case if $n=1$, we ...

One thought I had is that "whenever $x^ {-1},y^ {-1}$ exist in $R$, then $xxyy=xyxy\implies xy=yx$".

Jul 3, 2024 · We know that $2^4 = 4^2$ and $(-2)^{-4} = (-4)^{-2}$. Is there another pair of integers $x, y$ ($x\\neq y$) which satisfies the equality $x^y = y^x$?

Jul 8, 2018 · Claim : $H$ and $K$ are two normal subgroups of group $G$ such that $H\cap K = \ {e\}$ then $ xy = yx $ for $x \in H$ and $y \in K$. Proof : Let $x \in H$ and $y\in K$ then I need to prove …

May 2, 2013 · 3 You could modify the proof of Sylvester's determinant theorem to show that $$\text {det} \left ( \lambda I_m - XY \right) = \text {det} \left ( \lambda I_n - YX \right)$$ for all $\lambda \neq 0$. …

May 15, 2020 · Show that if R is ring with identity, $xy$ and $yx$ have inverse and $xy=yx$ then y has an inverse. I said we have an $a=xy^{-1}=yx^{-1}$ I did some operations on $xy ...

When beginning to study multivariate calculus, you almost immediately come across the Schwarz-Clairaut Theorem which gives sufficient conditions to guarantee that mixed partials are equal. For the ...

May 12, 2017 · (A) $X+Y$ (B) $XY$ (C) $YX$ (D) $2 (X+Y)$ what I tried is $X^2 + Y^2 = XX+YY$ then putting $XY = Y$ and $YX = X$ in the equation I got $YX^2+XY^2$. how do I proceed from here