Math ASCII Notation Demo

Mathematical content on Apronus.com is presented in Math ASCII Notation which can be properly displayed by all Web browsers because it uses only the basic set of characters found on all keyboards and in all fonts.

The purpose of these pages is to demonstrate the power of the Math ASCII Notation. In principle, it can be used to write mathematical content of any complexity. In practice, its limits can be seen when trying to write complicated formulas (containing, for example, variables with many indexes or multiple integrals).

Despite its limitations the Math ASCII Notation has much expressive power, as can be seen from browsing through these pages.

Let A[1],A[2],A[3],... be a decreasing sequence of sets.
Can we conclude that {A[n]\A[n+1]}(n:-|N) is a disjoint collection?
YES
page 97 in 2nd measure
Let A[1],A[2],A[3],... be a sequence of sets, having empty intersection.
Decide if we can conclude that
\/k:-|N {A[n]\A[n+1]}_(n=k to n=oo) is disjoint.
NO
(1) Let A[2n]={1}, A[2n+1]=O.
//Let A[2n] = {2n, 2n+1, 2n+2, 2n+3, ...}. Let A[2n+1] = {2n}.
//Then //\\(n:-|N) A[n] = O. And for every even n:-|N,
//(A[n]\A[n+1]) n (A[n+2]\A[n+3]) = {n+1,n+2,n+3,n+4,...} n {n+3,n+4,...} != O
Let B[1],B[2],B[3],... be an increasing sequence of sets.
Decide if we can conclude that
\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],
where B[0] = O.
YES
page 98 in 2nd measure
Let B[1],B[2],B[3],... be a sequence of sets.
Decide if we can conclude that
\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1],
where B[0] = O.
YES
The right-hand set is easily seen to be contained in the left- hand set.
Take any x belonging to the left-hand set. We have (1) \/n:-|N x:-B[n].
Let's argue by contradiction. Suppose that (2) /\n:-|N ( x:- B[n] => x:-B[n-1] ).
Let B[1],B[2],B[3],... be a sequence of sets.
Decide if we can conclude that
\\//(n=1 to n=oo) B[n] = \\*//(n=1 to n=oo) B[n]\B[n-1],
where B[0] = O.
NO
We can indeed conclude that
\\//(n=1 to n=oo) B[n] = \\//(n=1 to n=oo) B[n]\B[n-1]
but the right-hand union does not have to be disjoint.
Let B[1],B[2],B[3],... be an increasing sequence of sets.
Decide if we can conclude that
/\n:-|N B[n] = \\*//(k=1 to k=n) B[k]\B[k-1],
where B[0] = O.
YES
page 42 in 1st measure
Let B[1],B[2],B[3],... be a sequence of sets.
Decide if we can conclude that
/\n:-|N B[n] = \\//(k=1 to k=n) B[k]\B[k-1],
where B[0] = O.
NO
Let B[1] = {7}, B[2] = O.
Then B[2] != B[1] u B[2]\B[1].
We can only conclude:
/\n:-|N B[n] c \\//(k=1 to k=n) B[k]\B[k-1],
Let /\n:-|N 0<=a[n]<=oo and 0<=b[n]<=oo.
Suppose that both sequences a[n] and b[n] are increasing.
What can we conclude?
There exist a,b:-[0,oo] such that
(1) lim a[n] = a
(2) lim b[n] = b
(3) lim a[n]*b[n] = a*b
page 93 in 2nd measure
Let /\n:-|N 0<=a[n]<=oo and 0<=b[n]<=oo.
Let a,b:-[0,oo].
(1) lim a[n] = a,
(2) lim b[n] = b.
Decide if we can conclude that
(3) lim a[n]*b[n] = a*b.
NO
1) a[n] = 1/n; b[n] = n
2) a[n] = 1/n; b[n] = n^2
Let F be a s-algebra in X.
Let f : X -> |R* be measurable.
What can we conclude about the measurability of 1/f ?
It is measurable, no matter how we define: 1/0, 1/oo, 1/-oo.
page 93 in 2nd measure