Let's look at the number line. You can see -5,-4,-3,-2,-1,0,1,2,3,4 and so on. If you look between 0 and 1 there is going to be a rational number 1/2 exactly in the middle. There is going to be a rational number 1/4 exactly in the middle of 0 and 1/2. You can do this for any two numbers and make the distance between them smaller and smaller.
So you can fill up the whole number line by cutting it in halves. Therefore, the whole number line is made up of rational numbers and irrational numbers cannot exist.
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Anonymous2007-01-23 22:19
This is a pretty weak troll. Rationals are dense; no matter how close two of them are, there are always an infinite number of other rationals between them.
Go calculate pi for us. Come back when you're done.
What are you, stupid? If you say that Irrational numbers do not exist, then how do you explain the use of pi, e, and i in advanced algebra and calculus?
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Anonymous2007-01-23 23:53
>>3
That never stops the .999... denialists. All of higher education in math is wrong, apparently.
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Anonymous2007-01-23 23:55
>>1
All you've proven is that (a) irrational numbers aren't rational numbers with power-of-two denominators (WOW!), and (b) you're a fucking idiot.
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Anonymous2007-01-24 6:57
>>5
No I just proved that there cannot be any holes in the number line, so irrational numbers can't exist because there is no space for irrational numbers to exist in. Therefore pi, e, and so on are made up numbers and they cannot exist the number line.
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Anonymous2007-01-24 8:32
>>6
first prove that you haven't left any rationals out. i.e that you can reach to any rational a/b. by the process of bisection along two points. ok. let's try a specific example. starting from the intergers ...,-1,0,1,... and doing your bisection thingy, show that you can reach to 1/17.
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Anonymous2007-01-24 8:35
>>7
It doesn't have to be just bisection. You can replace division by two with division by any number there you go. Irrationals still do not exist.
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Anonymous2007-01-24 12:12
>>8
Just because you won't reach them rationally doesn't mean they don't exist. What, are you trying to find a finite decimal expansion for a number that does not have a finite decimal expansion? Look up proofs that sqrt(2) is irrational. You see, those proofs actually use maths in their arguments, whereas you don't.
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Anonymous2007-01-24 12:37
>>6
You didn't prove that there are no holes. You merely proved there are an infinite number of rational numbers. However, infinity is not enough to fill up even the number line between 0 and 1. Cantor proved that there are an uncountable (for all intents and purposes, "uncountable" is a larger quantity than "infinity") number of real numbers. He also proved that there are a countable (but infinite) number of rational numbers. The gap between infinite and uncountable is filled in by transcendental numbers (irrational numbers that are non-algebraic).
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Anonymous2007-01-24 15:09
>>3
i is not a real number
to the OP, there will always be space between two numbers
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Anonymous2007-01-24 15:23
I r nots teh math genius, but maybe irrational numbers are irrational because they indicate the presence of higher dimensions. Assuming you can't just do something simple like change the base number system to rationalize them.
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Anonymous2007-01-24 17:28
>>12
Next time, just stop after the word "genius."
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Anonymous2007-01-25 18:39
>>10
If there are an infinite number of rational numbers between 1 and 2, then that means that there are no holes. How can there be an infinite number of rational numbers between two numbers and it to have holes? It is impossible and proves that irrational numbers cannot exist.
>>15
Whut? He talks about 'uncountable' numbers, like there is something higher than infinity, which is ridiculous. If you have an infinite number of things in a finite amount of space, then there cannot be any holes at all, that is all.
Here's your homework: Find a ratio of two integers that will form a decimal expansion where there is eventually a repeating sequence that when squared equals two.
Want to cheat? You won't find it because you CAN'T find it. There are simple proofs out there that the square root of two is an irrational number.
Irrational numbers exist.
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Anonymous2007-01-25 21:34
>>16
Something isn't wrong just because you are too stupid to understand it.
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Anonymous2007-01-25 22:06
>>16
Look at the set of rational numbers other than 1/2 in [0,1]. There are infinitely many of them, and only "finite space"! I guess 1/2 doesn't exist either. But wait! What about 1/3! OH SHI-
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Anonymous2007-01-26 12:28
If there are an infinite amount of rational numbers in finite space, then irrational numbers are beyond infinite and outside of finite space. But that doesn't mean they don't exist.
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Anonymous2007-01-26 12:43
the correct reply to end this troll is
Where did you get the flawed assumption that numbers "exist" if they lie upon some imaginary line.
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Anonymous2007-01-26 15:15
Assume Sqrt(2) is rational. Then, by definition, there exists two relatively prime integers a and b such that a/b = sqrt(2). Implying, a^2/b^2 = 2, and a^2 = 2b^2.
Thus, a must be even, since only the square of an even number is even, and there exists an integer k such that 2k = a. Thus, (2k)^2 = 2b^2 and 2k^2 = b. Therefore, b is also even. But if a and b are both even then they can't be relatively prime integers.
Contradiction.
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Anonymous2007-01-26 15:37
>>16
Also, yes there are higher orders of infinity. You fail; Troll harder.