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 July 10th, 2015, 01:58 PM #1 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 All Fox are irrationals e^x, pi+/-e, etc... are all irrationals. I can prove that using the trick I've shown several times here to extract the square root with rising precision. I make the square of each unknown value, than I try to find the closest rational square root $\displaystyle (\pi+e)^2 =? K^2$ we know that: $\displaystyle (\pi+e)^2= 34.3381.....$ That is $\displaystyle 5^2 <34.3381... <6^2$ Since with my complicate modulus 2x-1 we have 34.3381...-1-3-5-7-11=9.3381... and 34.3381...-1-3-5-7-11-13=-4.3381 so since no integer solution we go rational using the rational complicate modulus: $\displaystyle Mnk = 2x/K-1/K^2$ We keep K=10 than we see that subtracting from 34.3381 Mnk calculated from x=0,1 to p (2 digit value)we have a value littlest than 34.3381 so a rest, while for p+1 we have a to big value. Than we rise K, f.ex 100 again we find a 3 digit value q littlest than 34.3381 while for q+1 we have a too big value. We are sure our value is irrational since we can prove that we feet it when K goes to infinity. This works for several unproved values.
July 10th, 2015, 05:04 PM   #2
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Quote:
 Originally Posted by complicatemodulus We are sure our value is irrational since we can prove that we feet it when K goes to infinity. This works for several unproved values.
So the part that actually requires proof is not addressed at all (other than an unsubstantiated claim that such a proof is possible).

And then you claim results for values that you don't specify.

How is anyone to take this seriously?

July 10th, 2015, 10:32 PM   #3
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 Originally Posted by v8archie So the part that actually requires proof is not addressed at all (other than an unsubstantiated claim that such a proof is possible). And then you claim results for values that you don't specify. How is anyone to take this seriously?
I already shown this procedure several times here. I'll be back on a PC Monday so if you're interested I'll rewrite it all. If you can't wait search: " ì " is irrational in my old post. Then you can square your unknown value and see if it has a zero in the rest for some K or require K to infinity. The condition is that the limit must exist and must be only one.

Sorry on android typing is very tricky. As example

$\displaystyle (\pi - e )^2 = \lim_{k\to\infty} \sum_{x=1/k}^{ p } ( 2x/k +1/k^2) = \int_ {0}^ {\pi - e } 2x dx$

Where p becomes $\displaystyle \pi - e$ just at the limit.

I hope it's not a round circle since we know for that fox that the limit exist and is only one.

Is the infinite descent I use for Fermat

Last edited by complicatemodulus; July 10th, 2015 at 11:08 PM.

 July 11th, 2015, 03:46 AM #4 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 Forgotten: e^x can be proved irrational just for known x.
 July 11th, 2015, 05:24 AM #5 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 What do you mean by "known x"? x= ln(2) is pretty "known" but for that x e^x is clearly not irrational. In fact, what do you mean to begin with by "e^x is irrational"? e^x is a continuous function which, for different x, takes on both rational and irrational numbers.
July 11th, 2015, 06:20 AM   #6
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Quote:
 Originally Posted by complicatemodulus closest rational square root
But for any real $x$ there are infinitely many rationals within $\varepsilon$.

July 11th, 2015, 09:23 AM   #7
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Quote:
 Originally Posted by Country Boy What do you mean by "known x"? x= ln(2) is pretty "known" but for that x e^x is clearly not irrational. In fact, what do you mean to begin with by "e^x is irrational"? e^x is a continuous function which, for different x, takes on both rational and irrational numbers.
Yes that's clear, my fault in the subtitle must by e^e or e^2.... and many others value.

July 11th, 2015, 09:50 AM   #8
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 Originally Posted by CRGreathouse But for any real $x$ there are infinitely many rationals within $\varepsilon$.
Yes and for so you,'ve to prove that rising k no result appear in Q till k goes to infinity at the limit so in R. You must start proving that the limit exist and is just one and is not in R. At the moment functions are out of the play.

There is probably a theorem on the rest that cut faster the story after some void turns, but I've to check.

 July 12th, 2015, 11:38 AM #9 Banned Camp   Joined: Dec 2012 Posts: 1,028 Thanks: 24 The procedure to check unknown irrational NUMBERS is: Knowing that the number you are looking for is not an integer: For example $\displaystyle (\pi + e ) = 5,85987....$ make the square: $\displaystyle (\pi + e )^2 = 34.33812...$ Knowing the rules for integer square we can make the recorsive difference that gives the square root of each square with rest zero, or a rest if it is not an integer (equal to the sum in the other direction): $\displaystyle 34.33812...=? \sum_{x=1/k}^{ p } ( 2x +1) =$ 34.33812 - (1+3+5+7+9) = 34.33812.... -25 = 9,33812... too little 34.33812 - (1+3+5+7+9) = 34.33812.... -36 = -1,66187... too big So it's a number between 5^2 and 6^2 We start to look if 34.33812.. has a rational root with 1 digit precision or if: $\displaystyle (\pi + e )^2 = 34.33812.. =? \sum_{x=1/k}^{ p } ( 2x/k +1/k^2)$ With k=10 so we tabulate: x 2x/10-1/100 Sum 0.1 0.01 0.01 0.2 0.03 0.04 0.3 0.05 0.09 0.4 0.07 0.16 0.5 0.09 0.25 0.6 0.11 0.36 0.7 0.13 0.49 0.8 0.15 0.64 0.9 0.17 0.81 1 0.19 1 1.1 0.21 1.21 1.2 0.23 1.44 1.3 0.25 1.69 1.4 0.27 1.96 1.5 0.29 2.25 1.6 0.31 2.56 1.7 0.33 2.89 1.8 0.35 3.24 1.9 0.37 3.61 2 0.39 4 2.1 0.41 4.41 2.2 0.43 4.84 2.3 0.45 5.29 2.4 0.47 5.76 2.5 0.49 6.25 2.6 0.51 6.76 2.7 0.53 7.29 2.8 0.55 7.84 2.9 0.57 8.41 3 0.59 9 3.1 0.61 9.61 3.2 0.63 10.24 3.3 0.65 10.89 3.4 0.67 11.56 3.5 0.69 12.25 3.6 0.71 12.96 3.7 0.73 13.69 3.8 0.75 14.44 3.9 0.77 15.21 4 0.79 16 4.1 0.81 16.81 4.2 0.83 17.64 4.3 0.85 18.49 4.4 0.87 19.36 4.5 0.89 20.25 4.6 0.91 21.16 4.7 0.93 22.09 4.8 0.95 23.04 4.9 0.97 24.01 5 0.99 25 5.1 1.01 26.01 5.2 1.03 27.04 5.3 1.05 28.09 5.4 1.07 29.16 5.5 1.09 30.25 5.6 1.11 31.36 5.7 1.13 32.49 5.8 1.15 33.64 too little 5.9 1.17 34.81 too big... Now we try k=100 x 2x/100-1/10000 Sum 0.01 0.0001 0.0001 0.02 0.0003 0.0004 0.03 0.0005 0.0009 0.04 0.0007 0.0016 0.05 0.0009 0.0025 0.06 0.0011 0.0036 ... 5.82 0.1163 33.8724 5.83 0.1165 33.9889 5.84 0.1167 34.1056 5.85 0.1169 34.2225 too little.. 5.86 0.1171 34.3396 too big ! So we can continue till K-> infinity where: $\displaystyle (\pi + e )^2 = \lim_{k\to\infty} \sum_{x=1/k}^{ p } ( 2x/k +1/k^2) = \int_ {0}^ {\pi + e } (2x) dx$ Where the STOPPING POINT becomes exactly $\displaystyle \pi + e$ just at the limit, so $\displaystyle (\pi + e) =$ irrational root of $\displaystyle (\pi + e)^2$ Why we are sure it is irrational ? Since the limit, if exist (and we prove that exist and is the root $\displaystyle \pi + e$) and is only one. I hope, ...as Dedekind states. In this way we can check many (all?) unknown suspected irrtationals NUMBERS. Waiting for good math opinions. Thanks Ciao Stefano
 July 12th, 2015, 04:18 PM #10 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,138 Thanks: 872 Math Focus: Wibbly wobbly timey-wimey stuff. What's p? When you defined your first sum it went from 1/k to p, which is a little weird in itself. But near the bottom of the post you are evidently setting $\displaystyle p = \pi + e$, in which case if k is a rational number: you can't sum 1/k (a rational number) to p (which would then have to also be a rational number.) Your procedure seems to be circular reasoning. -Dan

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