Recent content by poutsos.A

The only proof i can thing of ,is the following: \frac{1}{b}\frac{1}{d} = \frac{1}{bd}\Longleftrightarrow (bd)\frac{1}{b}\frac{1}{d} = (bd)\frac{1}{bd}\Longleftrightarrow d(b\frac{1}{b})\frac{1}{d} = 1\Longleftrightarrow d\frac{1}{d} = 1. But then again this is not a rigorous proof ,is there??

commutativity and associativity

\frac{a}{b}*\frac{c}{d}= a\frac{1}{b}*c\frac{1}{d}=ac*\frac{1}{bd}=\frac{ac}{bd} That is how far i can go.But then this is not a rigorous proof, is there??

Give a rigorous proof using the appropriate axioms and the definition ,\frac{a}{b}=a\frac{1}{b} of the following: \frac{a}{b}*\frac{c}{d}=\frac{ac}{bd} \frac{a}{b}:\frac{c}{d} =\frac{ad}{bc}

Yes, you right thank you. But if we define a set to be closed if its complement is open, how then we prove its closure to be a closed set??

Given that f is a function from R(=real Nos) to R continuous on R AND ,A any subset of R,IS THE closure of f(A) ,a closed set??

There could be not a better description for the state of affairs that socialistic civilization has brought up the whole planet to

WHAT is p{X(n)=m} ? apart from other things

GET THE book : VECTOR ANALYSIS and an introduction to TENSOR ANALYSIS by MURRAY R. SPIEGEL in SCHAUM'S OUTLINE SERIES IT is very good in this kind of vector identities

||x||=||x-y+y||\leq||x-y||+||y|| ======> ||x||-||y||\leq||x-y||............1 ||y||= ||y-x+x||\leq||x-y||+||x||======> ||y||-||x||\leq||x-y|| =======> ||x||-||y||\geq-||x-y||............2 from (1) and (2) we get: -||x-y||\leq ||x||-||y||\leq ||x-y||...

the whole sentence

yes A= { x: (x\in A\Longrightarrow y\in A)\wedge x\neq y} sorry

expanding set builder notation is: x\in A\Longleftrightarrow[(x\in A\Longrightarrow y\in A)\wedge x\neq y] Now how from the above you get : \forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right).

. LETS take it line by line: 1st line you have written: \forall x\;\; x\in A\Longleftrightarrow \left(x\in A\Rightarrow\forall y\neq x\;\; y\in A\right). DO you actually mean:x\in A\Longleftrightarrow(x\in A\Longrightarrow\forall y(y\neq x\wedge y\in A)). If yes, how did you get that?

A Is a set of x elements in a such way that if x belongs to A THEN ANY y different from x belongs to A. Doesn't that make sense??