
Re: Chapt8: welldefined Reals in light of 10^603 #524 Correcting Math 3rd ed
Posted:
Aug 8, 2011 6:47 PM


In article <LowSp.33495$m22.15361@newsfe05.ams2>, "Androcles" <Headmaster@Hogwarts.physics.June.2011> wrote:
> "Earle Jones" <earle.jones@comcast.net> wrote in message > news:earle.jones4E6487.21540410072011@news.giganews.com... >  In article >  <f8639f7e74394d608fd8855ae0db23a6@v8g2000yqb.googlegroups.com>, >  Archimedes Plutonium <plutonium.archimedes@gmail.com> wrote: >  >  [...] >  >  > So what is the square root of 2? If Irrationals are just imprecise >  > notions? Well, you take the square root of 2 out >  > to 603 decimal places to the right of the decimal point. It is still a >  > rational number and it is what the square root of >  > 2 really is.... >  >  * >  Archie: If sqrt(2) is rational (as you say above), it can be written >  as >  >  sqrt(2) = a/b >  >  where a and b are integers (definition of a rational number.) >  >  Please tell us what integers a and b are. >  >  Thanks, >  >  earle >  * > Oh dear... now you've left yourself open to accepting the burden of proof > instead of leaving it to Archie. > a = 54608393 > b = 38613965 > > > Of course if you'll accept four figure accuracy then > a = 8119 > b = 5741 > > is good enough, and for two figure accuracy > a = 99, b = 70 > > There exists two integers a and b such that a/b = sqrt(2) to any desired > accuracy. > Provide a counterexample (or other disproof).
* The proof that sqrt(2) is irrational is a freshman algebra exercise. What that means is that sqrt(2) cannot be expressed as the ratio of two integers (definition of irrational.) It can certainly be 'approximated' to whatever degree of accuracy one wants.
earle *

