- #36

webberfolds

- 65

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No, irrational numbersdoexist no matter whatnumericalsystem you use. The existence ofanynumbers is independent of how you wish to write them.

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- Thread starter Mu naught
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- #36

webberfolds

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No, irrational numbersdoexist no matter whatnumericalsystem you use. The existence ofanynumbers is independent of how you wish to write them.

- #37

funcalys

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- #38

Vorde

- 788

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If you are talking about a geometrical line with lengths and numbers, then of course rational numbers would exist.

I can't imagine how else you would imagine it, but if you decided to mean something else but a geometric line then the answer would lie in that field.

- #39

Robert1986

- 828

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Now, the actual question of whether irrational numbers are on the number line or "exist" in some sense aren't nearly as interesting as the intuition that the OP has. I haven't seen any posts mention this explicitly (that's not to say there aren't any, I might have missed some) and I think it is important to point out.

- #40

Studiot

- 5,441

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Secondly I assume you were referring to the second quote, from the originator, Robert.

Robert 1986

So, I don't want to beat a dead horse, but I think that the OP is, in some sense, on the exact right track. He talks about "moving" along the number line and "slowing" down at an irrational point. Of course, strictly speaking, this makes no sense; as many people have pointed out, numbers don't move. But his thought process is still valid. I mean, what he has essentially described is a sequence that converges to an irrational number, but this is one of the ways we define irrational numbers. We say "you know what, there are lots of rational sequences that converge to something that isn't rational so let's just 'create' numbers that fill these gaps."

Now, the actual question of whether irrational numbers are on the number line or "exist" in some sense aren't nearly as interesting as the intuition that the OP has. I haven't seen any posts mention this explicitly (that's not to say there aren't any, I might have missed some) and I think it is important to point out.

Mu naught

I still don't see how an irrational number can be represented by a definite point on a number line. Any point you choose, you can always move to the right by n x 10-j for whatever decimal you want to bring the number out to. This is what leads me to understand that an irrational number can only be represented by a point which is moving infinitesimally slowly towards some value but can't ever reach it, and if that is a valid way of thinking of things, then I don't believe you can ever represent an irrational number as a point on a number line.

I do believe in my post#32 I alluded to what you are saying, as did others in other ways.

A point is "that which has no part"

A line is an assembly of an infinite number of such points.

We can prove that the cardinality of this infinity is greater than the cardinality of the (also infinite) set of rational numbers.

In other words the set of real numbers has more points than the rationals.

Since they are not rational, we call these 'extra' points non rational or irrational.

It does not actualy matter which model we use to assemble the points into the real number line, the crucial fact is that there are more of them.

- #41

Natko

- 44

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- #42

Vorde

- 788

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I think I explained why I think this pretty clearly:

Any point you choose, you can always move to the right by n x 10-j for whatever decimal you want to bring the number out to. This is what leads me to understand that an irrational number can only be represented by a point which is moving infinitesimally slowly towards some value but can't ever reach it

But as has been said. You can easily construct a line segment of irrational length. One could claim that this is just an approximation done by the tools involved, but then you could make the same claim for rational numbers.

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