Differentproof there are more irrational numbers than rational numbers

[japplepie]

you can list and match up all rational numbers with irrational numbers this way..

lets say i have an irrational number 'c'.
Rational->Irrational
r1->cr1
r2->cr2
.
.
.
rn->crn

There exists an irrational number that is not on this matching, (not equal to any of the crx's)

this irrational number can be made by multiplying c to another irrational number 'b'

and I can prove that this is not on the list because cb never equals crx because b is irrational and rx is rational

is this a valid proof?

 

[pbuk]

You also need to prove that the product of an irrational number and a rational number is irrational, otherwise it is possible that some of the numbers on the right hand side of the list are rational, and also that it is possible to select b such that bc is irrational, otherwise the fact that bc is not on the right hand side of the list means nothing.

 

[japplepie]

well, if we assume crx is rational then c must be rational which is not the case (proof by contradiction)

 

[mathman]

you can list and match up all rational numbers with irrational numbers this way..

lets say i have an irrational number 'c'.
Rational->Irrational
r1->cr1
r2->cr2
.
.
.
rn->crn

There exists an irrational number that is not on this matching, (not equal to any of the crx's)

this irrational number can be made by multiplying c to another irrational number 'b'

and I can prove that this is not on the list because cb never equals crx because b is irrational and rx is rational

is this a valid proof?

no!

Try the same approach substituting "algebraic" for "irrational". The "proof" looks the same. However "algebraic" numbers form a countable set.

 

[pbuk]

well, if we assume crx is rational then c must be rational

Why?

This also does not address how to choose b.

 

[pbuk]

Actually Mathman is right - the whole approach is flawed. Just because one mapping from members of set A to set B does not span all the elements of set B does not prove that set B has a higher cardinality than set A.

Take for instance the mapping of integers $k$ to $2k$. There are integers not on the right hand side, but this does not mean that there are more integers than there are integers!

 

[japplepie]

Actually Mathman is right - the whole approach is flawed. Just because one mapping from members of set A to set B does not span all the elements of set B does not prove that set B has a higher cardinality than set A.

Take for instance the mapping of integers $k$ to $2k$. There are integers not on the right hand side, but this does not mean that there are more integers than there are integers!

well, how did the diagonalization argument side-step this problem?

if I am not mistaken, you're saying that i need show there exists no bijection in any kind of listing?

 

[micromass]

well, how did the diagonalization argument side-step this problem?

if I am not mistaken, you're saying that i need show there exists no bijection in any kind of listing?

The diagonalization argument works because it considers every kind of listing. That is: you take an arbitrary listing, then diagonalization says it can't work. If the diagonalization argument only considered one specific listing, then it wouldn't work.

 

[japplepie]

that's absolutely right

how can i show that without diagonalization?

 

[micromass]

that's absolutely right

how can i show that without diagonalization?

Show what?

 

[japplepie]

Show what?

show something that can be generalized to every possible bijection (without the logic behind diagonalization).

and also, can i show that one infinite set is bigger than the other if there is at least one matching that is infinite to one (for all elements).

 

[Bacle2]

you can list and match up all rational numbers with irrational numbers this way..

lets say i have an irrational number 'c'.
Rational->Irrational
r1->cr1
r2->cr2
.
.
.
rn->crn

There exists an irrational number that is not on this matching, (not equal to any of the crx's)

this irrational number can be made by multiplying c to another irrational number 'b'

and I can prove that this is not on the list because cb never equals crx because b is irrational and rx is rational

is this a valid proof?

Well, you can use your map to inject the rationals into the irrationals, but , if it were also possible to inject the irrationals into the rationals, you would have a bijection between the two--by Schroeder-Bernstein.

 

[japplepie]

can't I really do this? cause it really looks like i can

i understand that everything that everyone above is true and I am grateful for such smart replies

but my intuition tells me that this is "kind of" valid

for every rational, i can construct an irrational which gives a bijection
there is NO WAY of creating another rational that is not accounted for
and there are infinitely many ways of creating another irrational that is not yet accounted for

it seems valid in the perspective of constructivism

 

[pbuk]

can't I really do this?

No you can't.

for every rational, i can construct an irrational which gives a bijection

No, that's an injection. A bijection goes both ways.

for every rational, i can construct an irrational which gives a bijection
there is NO WAY of creating another rational that is not accounted for
and there are infinitely many ways of creating another irrational that is not yet accounted for

So what? $\aleph_0 + \aleph_0 = \aleph_0$.

 

[japplepie]

No, that's an injection. A bijection goes both ways.

its a bijection f(x)=cx;
unique rational <-maps-> unique irrational

 

[micromass]

its a bijection f(x)=cx;
unique rational <-maps-> unique irrational

It's not a surjection. Not every irrational is mapped to.

 

[japplepie]

It's not a surjection. Not every irrational is mapped to.

its a bijection only by that step

 

[micromass]

its a bijection only by that step

Prove it. Take any irrational $a$, what is the rational number $x$ such that $f(x) = a$?