Irrational Square Roots
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From: Bart Goddard on 4 Aug 2010 22:47 bill <b92057(a)yahoo.com> wrote in news:d698bbe6-93d4-4b13-bf88- d81debce69ea(a)u38g2000prh.googlegroups.com: >> Again, you had better give a precise definition of "finite decimal." > > I'll rely on any of the several published definitions > of "finite decimal" or "terminating decimal" or > "regular number". > So pick one, and then use it to tell us whether 3.99999..... is a finite decimal. -- Cheerfully resisting change since 1959.
From: Gerry Myerson on 4 Aug 2010 23:39 In article <d698bbe6-93d4-4b13-bf88-d81debce69ea(a)u38g2000prh.googlegroups.com>, bill <b92057(a)yahoo.com> wrote: > On Aug 4, 5:17�pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > wrote: > > In article > > <6f46e88a-7728-4bed-acd3-59671f5b3...(a)g21g2000prn.googlegroups.com>, > > > > �bill <b92...(a)yahoo.com> wrote: > > > Conjectrure: �If N is an integer and �Sqrt(N) is a > > > decimal number; then Sqrt(N) is irrational. > > > > If you manage to give a precise definition of decimal number > > and do it in such a way that integers don't count as decimal > > numbers, you will find that your "conjectrure [sic]" was proved > > 2000 years ago. > > This proof? is for the dilletante. I know that it was proved for N = > 2. I did not that it had been proved > for all integers. Then you have a lot of catching up to do. It has been proved that if n is an integer and the cube root of n is not an integer then the cube root of n is irrational. Similarly for the 4th root, 5th root, etc. Indeed, it has been proved that if you have a polynomial with integer coefficients and leading coefficient one, and you have a root of the polynomial and the root is not an integer, then the root is irrational. > > > Let R be any real number with a square root. > > > If Sqrt(R) is a finite decimal, then [Sqrt(R)]^2 > > > cannot be an integer. Therefore, Sqrt(N) is an > > > infinite decimal. > > > > Again, you had better give a precise definition of "finite decimal." > > I'll rely on any of the several published definitions > of "finite decimal" or "terminating decimal" or > "regular number". "Regular number" I never heard of. "Terminating decimal;" by the definitions with which I am familiar, 2.00000... is a terminating decimal. So, if R is 4, then the square root of R is a terminating decimal, but the square of the square root of R is an integer. So, if "finite decimal" means the same as "terminating decimal," then your statement is flat out wrong. > > You might also want to note that there are infinite decimals > > that are not irrational, e.g., .33333333333333.... > > True, but the squares of such "rational" numbers are > never exact integers. No kidding. I'm just pointing out that when you say "square root of N is an infinite decimal," you are not saying "square root of N is irrational." -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: Butch Malahide on 5 Aug 2010 00:18 On Aug 4, 7:17 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> wrote: > In article > <6f46e88a-7728-4bed-acd3-59671f5b3...(a)g21g2000prn.googlegroups.com>, > > bill <b92...(a)yahoo.com> wrote: > > Conjectrure: If N is an integer and Sqrt(N) is a > > decimal number; then Sqrt(N) is irrational. > > If you manage to give a precise definition of decimal number > and do it in such a way that integers don't count as decimal > numbers, you will find that your "conjectrure [sic]" was proved > 2000 years ago. Are you sure? 2000 years ago seems early for such a general result. And yet, I recall reading that Theodorus is supposed to have proved the irrationality of the square roots of nonsquare integers up to and including 17, sometime--let me look it up--in the 5th century BC. Yeah, I guess another 4 or 5 centuries would be enough time for somebody to work out the generalization. Do you know who it was?
From: sttscitrans on 5 Aug 2010 07:08 On 5 Aug, 02:38, bill <b92...(a)yahoo.com> wrote: > On Aug 4, 5:17 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > wrote: > > > In article > > <6f46e88a-7728-4bed-acd3-59671f5b3...(a)g21g2000prn.googlegroups.com>, > > > bill <b92...(a)yahoo.com> wrote: > > > Conjectrure: If N is an integer and Sqrt(N) is a > > > decimal number; then Sqrt(N) is irrational. > > > If you manage to give a precise definition of decimal number > > and do it in such a way that integers don't count as decimal > > numbers, you will find that your "conjectrure [sic]" was proved > > 2000 years ago. > > This proof? is for the dilletante. I know that it was proved for N = > 2. I did not that it had been proved > for all integers. > > > > > > Let R be any real number with a square root. > > > If Sqrt(R) is a finite decimal, then [Sqrt(R)]^2 > > > cannot be an integer. Therefore, Sqrt(N) is an > > > infinite decimal. > > > Again, you had better give a precise definition of "finite decimal." > > I'll rely on any of the several published definitions > of "finite decimal" or "terminating decimal" or > "regular number". > > > You might also want to note that there are infinite decimals > > that are not irrational, e.g., .33333333333333.... > > True, but the squares of such "rational" numbers are > never exact integers. > The square of 2.000000... is 4.00000.. Is 4.000.. and exact integer or not ?
From: bill on 5 Aug 2010 17:55
On Aug 4, 7:47 pm, Bart Goddard <goddar...(a)netscape.net> wrote: > bill <b92...(a)yahoo.com> wrote in news:d698bbe6-93d4-4b13-bf88- > d81debce6...(a)u38g2000prh.googlegroups.com: > > >> Again, you had better give a precise definition of "finite decimal." > > > I'll rely on any of the several published definitions > > of "finite decimal" or "terminating decimal" or > > "regular number". > > So pick one, and then use it to tell us whether > 3.99999..... is a finite decimal. > > -- > Cheerfully resisting change since 1959. From Mathworld: "A regular number, also called a finite decimal (Havil 2003, p. 25), is a positive number that has a finite decimal expansion." I don't know what "....." means. If it means "expand infinitely", then 3.99999..... is not a finite decimal. |