anonymous
  • anonymous
A piecewise function is defined as follows: for all rational numbers, f(x) = 1, for all irrational numbers, f(x) = 0. will the resulting graph be a line on f(x) = 0 with discontinuities? a line on f(x) = 1 with discontinuities, or a scattering of individual points?
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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dumbcow
  • dumbcow
i tend to agree with line on y=0 with discontinuities, there are far more irrational numbers than rational when looking at all real numbers
anonymous
  • anonymous
is that because they are uncountable?
dumbcow
  • dumbcow
kind of , rational numbers are uncountable too but for every gap between 2 rational numbers there could be an infinite number of irrational numbers

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anonymous
  • anonymous
well you will "see" two lines , y = 1 and y = 0
anonymous
  • anonymous
it doesnt matter there are more irrationals than rational, so what
anonymous
  • anonymous
Are there more irrational numbers than rational is really the question
anonymous
  • anonymous
as far as you are concerned, between any two irrationals is a rational, between any two rationals is an irrational, etc
anonymous
  • anonymous
or more generally, pick two points, between them is an irrational and a rational
anonymous
  • anonymous
pick 2 distinct points
anonymous
  • anonymous
so it is a scattering of individual points, neither line exists
dumbcow
  • dumbcow
very true, imagine we had a microscope ;)
anonymous
  • anonymous
right, its not a line ,. but at the macro level it "looks" like a line
anonymous
  • anonymous
So the argument I have heard that because irrational numbers are uncountable, it is a "larger" infinity than the countable set of rational numbers, is false - or that there is only one more
anonymous
  • anonymous
that is a true statement, but that doesnt mean that you can order them , rational, irrational, rational, irrational, etc
anonymous
  • anonymous
on the number line. so ...
anonymous
  • anonymous
it gets even weirder. between any 2 rationals is an irrational. between any two irrationals is a rational. so where exactly are all the "extra" irrationals ?
anonymous
  • anonymous
between any 2 rational numbers there exists an irrational number. between any 2 irrational numbers there exists a rational number. so you might think, the number of rational and irrational should be about the same, since you can keep going with this statement
anonymous
  • anonymous
okay, you said, "that is a true statement, but that doesnt mean that you can order them , rational, irrational, rational, irrational, etc" does that mean that there can and cannot be two irrational numbers without a rational between them?
anonymous
  • anonymous
there cannot be two distinct irrationals without a rational between them, correct
anonymous
  • anonymous
so its kind of paradoxical
anonymous
  • anonymous
Do you have a good source where I could read more? Wiki gets too techie too quickly
anonymous
  • anonymous
yeah, thats true. well you can google cantor's infinity
anonymous
  • anonymous
but this question is not addressed specifically, its something that my math professor told me
anonymous
  • anonymous
there are many layman easy websites
anonymous
  • anonymous
Yeah, I want something between the two levels
anonymous
  • anonymous
yeah i know the feeling
anonymous
  • anonymous
Also, I think it's neat to see hebrew letters come into math for this question!
anonymous
  • anonymous
well college libraries have some gradual immersion books
anonymous
  • anonymous
yeah, thats cool
anonymous
  • anonymous
the big question in advanced math, does there exist a cardinality between aleph null ( natural numbers) and the cardinality of the real numbers
anonymous
  • anonymous
it can be shown that aleph null is the "least" infinity , as weird as that sounds
anonymous
  • anonymous
and we can make a hierarchy of infinite cardinalities. use the natural number cardinality, and take the cardinality of the power set of the natural numbers
anonymous
  • anonymous
so we start with | N | , where N = { 1,2,3,...}
anonymous
  • anonymous
ok
anonymous
  • anonymous
where |N | means cardinality , that means the size of the set
anonymous
  • anonymous
right
anonymous
  • anonymous
now we know for any set , |P (S)| > |S|
anonymous
  • anonymous
P(S) is the power set of S
anonymous
  • anonymous
only for finite sets?
anonymous
  • anonymous
for any set
anonymous
  • anonymous
ok
anonymous
  • anonymous
it comes in handy for infinite sets
anonymous
  • anonymous
what a power set ?
anonymous
  • anonymous
2^S, right?
anonymous
  • anonymous
the number of permutations within the set
anonymous
  • anonymous
right, thats the shorthand for it
anonymous
  • anonymous
well, its the set of all subsets of the set
anonymous
  • anonymous
i dont know about permutations, hmmm
anonymous
  • anonymous
I thought infinity was defined as the set that has the same cardinality as any of its proper subsets
anonymous
  • anonymous
yes thats a definition of an infinite set
anonymous
  • anonymous
but im talking about power set, what is the power set operation
anonymous
  • anonymous
ok say you have { 1, 2 } , the power set is { {}, {1}, {2} , {1,2} }
anonymous
  • anonymous
S= { 1, 2 } , then P(S)= { {}, {1}, {2} , {1,2} }
anonymous
  • anonymous
right
anonymous
  • anonymous
now cantor proved a delightful theorem , that | P ( S) | > | S| for any set S
anonymous
  • anonymous
so the powerset of integers is larger I mean has a higher cardinality than the set of integers
anonymous
  • anonymous
exactly
anonymous
  • anonymous
ok so
anonymous
  • anonymous
| P ( N ) | > | N | , correct ?
anonymous
  • anonymous
I sense a trick coming up...
anonymous
  • anonymous
but, yes
anonymous
  • anonymous
http://en.wikipedia.org/wiki/Cardinality_of_the_continuum
anonymous
  • anonymous
ok it turns out that |R| which we use shorthand c , it turns out |R| = | P ( N ) | ,
anonymous
  • anonymous
now the question is, is there an infinity between |N| and | P ( N ) |
anonymous
  • anonymous
is the real numbers the smallest infinity after the natural numbers
anonymous
  • anonymous
Wouldn't rational and irrational be smaller than real?
anonymous
  • anonymous
we have a hierarchy of infinities, |N| < |P(N)| < |P (P(N))| < ...
anonymous
  • anonymous
oh boy, rational is the same cardinality as natural numbers
anonymous
  • anonymous
the irrationals are uncountable
anonymous
  • anonymous
So cantor's work doesn't cover uncountable sets?
anonymous
  • anonymous
I find it hard to assign the same cardinality to rationals as naturals, which means I only have a working understanding of cardinality and not a real grasp on it.
anonymous
  • anonymous
help

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