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- functions - What does y=y (x) mean? - Mathematics Stack Exchange
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
- If $H$ and $K$ are normal subgroups with $H\cap K=\ {e\}$, then $xy=yx . . .
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 that $xy =yx$
- The relationship between the eigenvalues of matrices $XY$ and $YX$
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 $\lambda \neq 0$ This shows equivalence for all nonzero eigenvalues For the zero eigenvalues, an application of the fundamental theorem of algebra is sufficient
- If $xy=yx$ in group $G$, then $ (xy)^n=x^ny^n$ [duplicate]
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
- $x^y = y^x$ for integers $x$ and $y$ - Mathematics Stack Exchange
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 $x^y = y^x$?
- Let $G$ be a group. Prove that $o(xy) = o(yx)$ for all $x,y \\in G$
i guess you use morphisms, which i haven't learned yet was trying to make a proof with some basic properties
- abstract algebra - Verify my proof of $G$ is nilpotent iff $xy=yx . . .
Verify my proof of $G$ is nilpotent iff $xy=yx \forall x,y\in G$ such that $ (o (x),o (y))=1$ Ask Question Asked 10 years, 10 months ago Modified 10 years, 10 months ago
- calculus - Finding $y$ by implicit differentiation if $x^y=y^x . . .
Both your expressions are the same :) It is just multiplying by (-1) in the numerator and denominator
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