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a real pythagorean


7h
comment Show non-degenerate form of subspaces
Do you mean both $d(x,y)=d(y,x)$ and $d(x,y)=-d(y,x)$ are assumed to hold for all $x,y$? But then $d(y,x)=-d(y,x)\ \implies\ d(y,x)=0$ for all $x,y$ or $K$ has characteristic $2$.
1d
answered Prove that I can always write a number a, a>0 as any number c, c>0 to the power of some number (a=c^x)
1d
comment Freely homotopic but not homotopic
What do you mean by 'freely homotopic' here? You mean, the endpoints are not necessarily fixed?
1d
answered Prove that vector space and dual space have same dimension
1d
answered show that [T]β is a diagonal matrix
1d
comment A counterexample for an equation between arbitrary relations
No. Take any counterexample where $R_2\cap R_3=\emptyset$, then add, say $(s,s)$ to each of $R_1,R_2,R_3$ for an element $s$ that was not present in them.
1d
answered A counterexample for an equation between arbitrary relations
1d
comment equivalence relation problem - checking
@SMM: Ahh.. you are right. I forgot to take the absolute value sign into account..
1d
comment equivalence relation problem - checking
@Mario: It is a bad notation because $\aleph_{n+1}$ usually denotes the next cardinal after $\aleph_n$, and it is proven that the continuum hypothesis ($\aleph_1=c$) is independent from the axiom system ZFC.
1d
comment equivalence relation problem - checking
@SMM: No, Mario is right, as $A$ can be any subset of $\Bbb Z^-$, there are continuum many equivalence classes.
1d
answered Unique of represent object
1d
comment equivalence relation problem - checking
Did you mean $\aleph_1$ as continuum? You should rather use a $\mathrm c$-like letter for that, or $2^{\aleph_0}$.
Jan
29
revised Derivations on matrix algebra
added 69 characters in body
Jan
28
answered Derivations on matrix algebra
Jan
28
comment Intuition behind Hash Function
No, $h(k)=011010_2$ if $m=2^6=1000000_2$ (or $m=2^5$).
Jan
28
comment Prove that if $a$,$b$,$c$ are non-negative real numbers such that $a+b+c =3$, then $abc(a^2 + b^2 + c^2)\leq 3$
+1 that's a deeper 'why'
Jan
28
answered Quadratic to matrix form
Jan
28
answered f injective, g injective, $f\circ g(a) = a$ implies $f$ bijective
Jan
28
comment Quadratic to matrix form
Ok, it gets clearer. Then, $U$ would be the diagonal matrix of values $u_1,u_2,\dots$, wouldn't it?
Jan
28
comment Prove that if $a$,$b$,$c$ are non-negative real numbers such that $a+b+c =3$, then $abc(a^2 + b^2 + c^2)\leq 3$
It seems AM-GM for $a^2+b^2+c^2,\ ab+bc+ac,\ ab+bc+ac$.