Irrational Square Roots
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From: bill on 4 Aug 2010 17:17 Conjectrure: If N is an integer and Sqrt(N) is a decimal number; then Sqrt(N) 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. regards, Bill J
From: Virgil on 4 Aug 2010 17:56 In article <6f46e88a-7728-4bed-acd3-59671f5b37fa(a)g21g2000prn.googlegroups.com>, bill <b92057(a)yahoo.com> wrote: > Conjectrure: If N is an integer and Sqrt(N) is a > decimal number; then Sqrt(N) is irrational. > > Let R be any real number with a square root. I.e., any non-negative real. > If Sqrt(R) is a finite decimal, then [Sqrt(R)]^2 > cannot be an integer. Therefore, Sqrt(N) is an > infinite decimal. > > regards, Bill J If N is an integer and sqrt(N) is not, then then Sqrt(N) is irrational.
From: sttscitrans on 4 Aug 2010 19:56 On 4 Aug, 22:17, bill <b92...(a)yahoo.com> wrote: > Conjectrure: If N is an integer and Sqrt(N) is a > decimal number; then Sqrt(N) 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. 4.0 is a real number with sqrt = 2.0 2.0 is a finite decimal 2.0x2.0 = 4.0, an integer
From: Gerry Myerson on 4 Aug 2010 20:17 In article <6f46e88a-7728-4bed-acd3-59671f5b37fa(a)g21g2000prn.googlegroups.com>, bill <b92057(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. > 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." You might also want to note that there are infinite decimals that are not irrational, e.g., .33333333333333.... -- Gerry Myerson (gerry(a)maths.mq.edi.ai) (i -> u for email)
From: bill on 4 Aug 2010 21:38
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. > > -- > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) |