mersenneforum.org Standard crank division by zero thread
 Register FAQ Search Today's Posts Mark Forums Read

 2011-04-15, 14:46 #221 CRGreathouse     Aug 2006 3·1,993 Posts Actually that's not a problem, sm; take the contrapositive. His first mistake is somewhat later.
2011-04-16, 02:04   #222
schickel

"Frank <^>"
Dec 2004
CDP Janesville

1000010010102 Posts

Quote:
 Originally Posted by Condor BTW - is it possible to edit posts? I don't see an icon for it. I didn't realize I had cut-and-pasted formating when I pulled that one in from the editor I prefer, and I agree it looks ugly. (Oh - is it that only the last post can be edited? This one did get an "edit" icon.)
You have a time-limited window to edit a post (I believe it's an hour...)

 2011-04-17, 13:34 #223 Don Blazys     Feb 2011 163 Posts Poor "Condor", "science man" and "CRNuthouse"! They are so desperate, so frustrated, so obsessed and............. so stupid! After all their compussive and incessant postings, they still dont realize that any true equation, whether it be 2 + 3 = 5 or httр://donblazys.com/03.рdf is simply an actuality and that there is no lawyer-like argument to refute. Thus, the task of "refuting" this proof is utterly futile and truly Sisyphean ! (A fitting punishment for nincompoops!) Don.
2011-04-17, 15:00   #224
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Don Blazys Poor "Condor", "science man" and "CRNuthouse"! They are so desperate, so frustrated, so obsessed and............. so stupid! After all their compussive and incessant postings, they still dont realize that any true equation, whether it be 2 + 3 = 5 or httр://donblazys.com/03.рdf is simply an actuality and that there is no lawyer-like argument to refute. Thus, the task of "refuting" this proof is utterly futile and truly Sisyphean ! (A fitting punishment for nincompoops!) Don.
I almost fell for it:

Quote:
 $\frac{T}{T}C^z = T $$\frac{C}{T}$$^{\frac {ln( \frac{C^z}{T})}{ln(\frac{C}{T})}}=T $$\frac{C}{T}$$^{\frac {\frac{ln( \frac{C^z}{T})}{ln(T)}}{\frac{ln(\frac{C}{T})}{ln(T)}}} =T $$\frac{C}{T}$$^{\frac {\frac{ln( \frac{C^z}{T})}{ln(T)}-\frac{ln(T)}{ln(T)}}{\frac{ln(\frac{C}{T})}{ln(T)}-\frac{ln(T)}{ln(T)}}}=T $$\frac{C}{T}$$^{\frac {\frac{{z}*{ln(C)}}{ln(T)}-1}{\frac{ln(C)}{ln(T)}-1}}$
the last one has a different value for the exponent last I checked with T=6, C=3,and z=4, if these are within your bounds( haven't double checked) then it's done.

Last fiddled with by science_man_88 on 2011-04-17 at 15:55

 2011-04-17, 17:35 #225 tichy   Nov 2010 22×19 Posts I'm sorry - I could not resist any longer once I ran out of my popcorn: http://www.youtube.com/watch?v=qLrnkK2YEcE Can someone demonstrate any examples of widely used and commonly accepted proofs which would be rendered invalid when Don's reasonign is applied ?
2011-04-17, 19:07   #226
Don Blazys

Feb 2011

16310 Posts

Quoting science man:
Quote:
 The last one has a different value for the exponent last I checked with T=6, C=3,and z=4,
You are wrong. (They all result in 81.)
Now you need to check how you checked!

Don.

 2011-04-17, 20:34 #227 akruppa     "Nancy" Aug 2002 Alexandria 9A316 Posts Don Blazys, refrain from posting personal insults.
2011-04-17, 21:11   #228
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26·131 Posts

Quote:
 Originally Posted by Don Blazys Quoting science man: You are wrong. (They all result in 81.) Now you need to check how you checked! Don.
actually doing the $\frac{\frac{ln(c^z/t)}{ln(t)}}{\frac{ln(c/t)}{ln(t)}}$

as written in you thing is a different value than $\frac{ln(c^z/t)}{ln(c/t)}$ I did it out in pari:

Code:
(17:42)>c=3;z=4;t=6;print((log(c^z/t)/log(t))/(log(c/t)/log(t)))
-3.754887502163468544361216832
(17:42)>c=3;z=4;t=6;print(log(c^z/t)/ln(t)/log(c/t)/log(t))
***   obsolete function.

For full compatibility with GP 1.39.15, type "default(compatible,3)", or set "compatible = 3" in your GPRC file.

New syntax: ln(x) ===> log(x)

log(x): natural logarithm of x.

(17:46)>c=3;z=4;t=6;print(log(c^z/t)/log(t)/log(c/t)/log(t))
-1.169600413701046825003995111
(17:46)>c=3;z=4;t=6;print(log(c^z/t)/log(c/t))
-3.754887502163468544361216832

Last fiddled with by science_man_88 on 2011-04-17 at 21:31

2011-04-18, 12:30   #229
Condor

Apr 2011

31 Posts

Quote:
 Originally Posted by science_man_88 I almost fell for it: the last one has a different value for the exponent last I checked with T=6, C=3,and z=4, if these are within your bounds( haven't double checked) then it's done.
You should double check - the "identity" is indeed valid, as it is merely an expansion and re-arrangement of the terms. You need to add some parentheses to your code.

But the problem is the part inside the [] below:
$\frac{T}{T}C^z = T $$\frac{C}{T}$$^{\frac {ln( \frac{C^z}{T})}{$ln(\frac{C}{T})$}}$
Anything to the right of this in Don's derivation is no longer part of a "true equation" if T=C. Actually, the term becomes indeterminate; but Don will deny that.

It seems sad that a person who obviously has the capability to imagine combinations in new and interesting ways can so delude himself about what they mean. And about how he applies a double standard (and is using "lawyer-like" arguments) when he insists on letting C=T because of the "truth" inherent in his "identity," yet insists the same argument doesn't apply to numbers that would allow him to see that it is indeterminate.

2011-04-18, 14:19   #230
science_man_88

"Forget I exist"
Jul 2009
Dumbassville

26×131 Posts

Quote:
 Originally Posted by Condor You should double check - the "identity" is indeed valid, as it is merely an expansion and re-arrangement of the terms. You need to add some parentheses to your code. But the problem is the part inside the [] below: $\frac{T}{T}C^z = T $$\frac{C}{T}$$^{\frac {ln( \frac{C^z}{T})}{$ln(\frac{C}{T})$}}$Anything to the right of this in Don's derivation is no longer part of a "true equation" if T=C. Actually, the term becomes indeterminate; but Don will deny that. It seems sad that a person who obviously has the capability to imagine combinations in new and interesting ways can so delude himself about what they mean. And about how he applies a double standard (and is using "lawyer-like" arguments) when he insists on letting C=T because of the "truth" inherent in his "identity," yet insists the same argument doesn't apply to numbers that would allow him to see that it is indeterminate.
what I was pointing out is without more parentheses that second exponent is clearly different that the first. an example to go on:

3/4/2/4 or (3/4)/(2/4) first one is 3/(4*2*4) = 3/32 the second is (3/$\strike {4}$)/(2/$\strike {4}$) =3/2 = 1.5

Last fiddled with by science_man_88 on 2011-04-18 at 14:28

2011-04-19, 09:54   #231
Don Blazys

Feb 2011

163 Posts

Quoting Condor:
Quote:
 Anything to the right of this in Don's derivation is no longer part of a "true equation" if T = c.
That's not true. It all depends on the value of z. Please follow this carefully.

If z=1,

then (T/T)*c^z = T*(c/T)^((z*ln(c)/(ln(T))-1)/(ln(c)/(ln(T))-1))

becomes (T/T)*c^1 = T*(c/T)^1

where clearly, we can let T = c because doing so gives us the "true equation"

(c/c)*c^1 = c*(c/c)^1

Quoting Condor:
Quote:
 Actually, the term becomes indeterminate; but Don will deny that.
As I just demonstrated, indeterminate forms are not an issue in my proof.

They are "removable singularities" that are easily avoided and don't even exist if we
do the algebra correctly and evaluate the exponents at z = 1 before we let T = c.

Don.

 Similar Threads Thread Thread Starter Forum Replies Last Post Uncwilly Miscellaneous Math 85 2017-12-10 16:03 req Math 4 2011-12-06 04:17 Mini-Geek Forum Feedback 21 2007-03-06 19:21 amateurII Miscellaneous Math 40 2005-12-21 09:42 jinydu Puzzles 5 2004-01-10 02:12

All times are UTC. The time now is 04:19.

Thu Oct 21 04:19:31 UTC 2021 up 89 days, 22:48, 1 user, load averages: 3.26, 2.87, 2.65