The Math Thread

[civman2000]

A thread for talking about math and enjoying its wonderfulness.

Here's a fun little math challenge: Find uncountably many sets of integers such that the intersections of any two of them is finite.

[Frozzy]

Does 1 (1sf) + 1(1sf) = 2 (0dp) ?

[civman2000]

1sf? 0dp?

[SlowwHand]

Congrats. New(winning) thread.

[DRoseDARs]

Sorry, but "Does .9-repeating equal 1?" still retains its title as gayest math thread in all of existence.

[SlowwHand]

We need a vote, DR. Do you think?

[KrazyHorse]

Re: The Math Thread

Originally posted by civman2000
A thread for talking about math and enjoying its wonderfulness.

Here's a fun little math challenge: Find uncountably many sets of integers such that the intersections of any two of them is finite.

Oooo.

Tough.

[Petek]

Re: The Math Thread

Originally posted by civman2000
Here's a fun little math challenge: Find uncountably many sets of integers such that the intersections of any two of them is finite.

Does this work?

Consider

{a_1, a_2, a_3, ... : 0 <= a_i <= 9}

That is, the set of all sequences of integers whose elements are between 0 and 9.

These sequences obviously correspond to real numbers between 0 and 1 (by looking at the decimal representation .a_1 a_2 a_3 ...), and hence represent an uncountable collection. However, the intersection of any two such sequences is finite, because the elements of any two such sequences are contained in {0, 1, ... , 9}.

[DRoseDARs]

Originally posted by SlowwHand
We need a vote, DR. Do you think?

No, because I'm afraid of how they might try to tally the votes...

[Petek]

Assuming that my previous reply is correct, here's another challenge:

Given any integer N, show that some integer multiple of N consists entirely of the digits 0 and 1.

For example,

2 x 5 = 10
3 x 37 = 111
4 x 25 = 100

etc.

[KrazyHorse]

Got it.

For every real number R there exists a sequence of rational numbers qn such that lim(n->inf)qn = R.

Now, any two distinct real numbers (R1 and R2) are separated by a distance |R1-R2| = d > 0 and have corresponding sequences qn and vn (respectively). Since qn converges to R1 there exists M1 s.t |R1 - qn| < d/2 when n>M1. Similarly, there exists M2 s.t. |R2-vn| < d/2 when n>M2. Therefore, if qm = vn either m < M1 or n < M2. Therefore {qn} intersection {vn} contains less than M1+M2 elements (specifically, the set {q1,...,qM1} union {v1,...vM2}). Now, this is finite. Also, there are uncountably many of these sets (since there are uncountably many reals). The only problem is that these are sets over the rationals, not the integers. Oh, wait! Cantor's diagonal theorem to the rescue. There exists a bijection between the rationals and the integers. Call this mapping b: Q -> I. Represent the aforementioned converging sequence of rationals to a real R as {qRn}. Our uncountable set of subsets is the union over all R of {b(qRn}}

EDIT: formatting

[KrazyHorse]

Re: Re: The Math Thread

Originally posted by Petek


Does this work?

Consider

{a_1, a_2, a_3, ... : 0 <= a_i <= 9}

That is, the set of all sequences of integers whose elements are between 0 and 9.

These sequences obviously correspond to real numbers between 0 and 1 (by looking at the decimal representation .a_1 a_2 a_3 ...), and hence represent an uncountable collection. However, the intersection of any two such sequences is finite, because the elements of any two such sequences are contained in {0, 1, ... , 9}.

This answer makes no sense.

[KrazyHorse]

There are an infinite number of sequences of integers between 1 and 9. There are only a finite number of sets of integers between 1 and 9 (specifically 2^9)

[Flip McWho]

This thread stopped making sense for me after I read "Find uncountably many".

[KrazyHorse]

If it stopped making sense after that then you probably have a decent enough math background...