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December 30th, 2012, 01:53 AM   #1
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Sum of rational numbers is irrational?

There must be some aspect of Taylor's Theorem or perhaps arithmetic which I'm not recalling and would appreciate someone pointing out what it is.

The sum of two rational numbers is rational is always true. It can be proven that the exponential function is equal to the sum of its Maclaurin series and in particular when ,



Since the product of natural numbers are natural numbers meaning factorials are natural numbers and since their reciprocals are ratios of natural numbers (rational), and it can be shown through induction that the partial sum of the first terms of the series, no matter how large we choose is rational, then what happens along the way that makes the infinite sum converge to , an irrational (and transcendental) number?

Thank you for the input.
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December 30th, 2012, 05:05 AM   #2
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Re: Sum of rational numbers is irrational?

Quote:
Originally Posted by policebox
what happens along the way that makes the infinite sum converge to e, an irrational (and transcendental) number?
That's because we are summing up infinite number of rational numbers here. Sum of finite number of rationals is always a rational but the "infinite" part makes everything unpredictable.

.
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December 30th, 2012, 08:44 AM   #3
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Re: Sum of rational numbers is irrational?

Quote:
Originally Posted by policebox
Since the product of natural numbers are natural numbers meaning factorials are natural numbers and since their reciprocals are ratios of natural numbers (rational), and it can be shown through induction that the partial sum of the first terms of the series, no matter how large we choose is rational, then what happens along the way that makes the infinite sum converge to , an irrational (and transcendental) number?
Another way to look at this is that a limit of an infinite sequence of rationals can be irrational. For example the numbers 3, 3.1, 3.14, 3.141, ... are all rational; yet their limit is pi, an irrational.
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January 1st, 2013, 01:05 AM   #4
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Re: Sum of rational numbers is irrational?

Quote:
Originally Posted by mathbalarka
Quote:
Originally Posted by policebox
what happens along the way that makes the infinite sum converge to e, an irrational (and transcendental) number?
That's because we are summing up infinite number of rational numbers here. Sum of finite number of rationals is always a rational but the "infinite" part makes everything unpredictable.

.
Can you direct me to a proof of this? I mean, obviously I'm convinced that this is a true statement because and are both examples, but I'm unsettled as to why one can perform an operation an unlimited number of times (while still considering the number of times to be finite) but get a different result than if you performed the same operation ad infinitum. Further examples aren't going to convince me, because I am not in doubt as to the validity of the statement. I'm curious about the "why".
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January 1st, 2013, 02:51 AM   #5
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Re: Sum of rational numbers is irrational?

No proof is needed to show that a statement is false. All that you need to show a statement is false is a counterexample. Here is a simple counterexample:

Let x1=.1, x2 = .001, x3 = .000001, x4 = 0000000001, ...

These are all terminating decimals, thus rational numbers. But their sum is the irrational number .101001000100001...
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January 1st, 2013, 02:54 AM   #6
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Re: Sum of rational numbers is irrational?

I don't understand, how is this a counterexample to my statement?
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January 3rd, 2013, 07:19 AM   #7
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Re: Sum of rational numbers is irrational?

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I don't understand, how is this a counterexample to my statement?
It's NOT. It is a counterexample to the statement policebox was implying- that a sequence of rational numbers must have a rational limit.
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January 3rd, 2013, 10:46 AM   #8
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Re: Sum of rational numbers is irrational?

let's take a detour:

claim: between every two (unequal) real numbers (rational OR irrational) there lies a rational number.

now this is obvious for two rational numbers r and s, because we have x = r + (s-r)/2 that fits the bill (assuming r < s). but this doesn't give us much help in the case(s) where r,s or both are irrational.

i will do "the hard case" (where both r and s are irrational), you may find it entertaining to try and adapt this to the case where only ONE of r and s is irrational.

since r is irrational it lies between two integers, say k and k+1.

if s < k, then k will do. if s > k+1, then k+1 will do. so the only difficulty in finding some rational x with r < x < s (or s < x < r) is if k < r,s < k+1.

since we can always add -k to both sides, we may as well assume r,s are in (0,1).

since r ? s, we have |r - s| = ? < 1. suppose we pick a positive integer N so that N ? 1/? < N+1.

now we divide [0,1] into N-ths. there will be some positive integer M between 0 and N with: M/N < r < (M+1)/N.

note that s is ? ? 1/N away from r, so s in not in the open interval (M/N,(M+1)/N). one of the endpoints is thus a rational number between r and s.

ok, what THIS means, is given any irrational number r, we have r in the open interval (r - (1/n), r + (1/n)), for any integer n.

this means we can find two rational numbers x_n and x_n' with x_n in (r - (1/n),r) and x_n' in (r,r + (1/n)).

how far apart are these two rational numbers? |x_n - x_n'| ? |r - (1/n) - (r + (1/n)| = |-2/n| = 2/n, so they are at most 2/n apart.

so as n gets very large, the rational numbers x_n and x_n' get very close to each other.

so it's strange, but two sequences of rational numbers can both "converge" to an irrational number.

this is similar to how, for an integer n, 1/n can get "very close" to 0, but it never actually "equals" 0.

the conclusion is: there must be convergent sequences of rational numbers that are irrational. in fact, this is a profitable way to think of "what" irrational numbers ARE. they are numbers that aren't rational, but can be APPROXIMATED by rationals. most irrational numbers are impossible to list "in their FULL form" (all the decimal digits, for example). the best we can do is get "close enough" (and this approximation, which will not be PERFECT, will be rational).
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