Irrational Square Roots
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From: Jesse F. Hughes on 13 Aug 2010 11:53 Ludovicus <luiroto(a)yahoo.com> writes: > On Aug 4, 5:17 pm, bill <b92...(a)yahoo.com> wrote: >> Conjecture: If N is an integer and Sqrt(N) is a >> decimal number; then Sqrt(N) is irrational. >> >> regards, Bill J > > It is not a conjecture but a theorem. Proof: > Given the equation N - x^2 = 0 ; N = integer > 0 > If testing the integers 1,2,3...x util x^2 > N > and not finding a solution, then sqrt(N) is irrational. That's gotta be the most original proof of this theorem thus far. Congrats! -- Jesse F. Hughes "[From now on] no simple dismissal of claims of counterexamples or counter proofs. I may not consider such claims, but I won't just dismiss them [...]" -- James S. Harris turns a new leaf
From: bill on 13 Aug 2010 15:30 On Aug 9, 3:45 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> wrote: > In article > <7d557612-615b-4a98-9f40-ca165762d...(a)s24g2000pri.googlegroups.com>, > > > > bill <b92...(a)yahoo.com> wrote: > > On Aug 4, 2:17 pm, 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. > > > > regards, Bill J > > > Some corrections: > > > Conjecture: If N is an integer and > > Sqrt(N) > floor[( Sqrt(N)], then Sqrt(N) is not rational. > > To you, it's a conjecture. To anyone whose mathematical education > has gotten up to where Euclid was, it's a simple theorem, indeed, > it's just a minor rewriting of the theorem we've been talking about > since the beginning of this thread. > > > Suppose that Sqrt(N) is rational. Then Sqrt(N) can > > be written as an integer + a fraction, say (I + F) > > Where, by "fraction", you mean, a rational number > strictly between zero and one. > > > Or (N) = I^2 + 2*I*F + F^2. Can (2*I*F + F^2) > > be an integer? Only if F = 0.9... . > > True, but I'd like to see you prove it! > > -- > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) To clarify what i was trying to say is; If F < 0.9... , then F^2 < 1. IMMHO I don't have to prove that If F = 0.9... , then F^2 = 1. Do I? BJ
From: Pubkeybreaker on 13 Aug 2010 16:06 On Aug 13, 3:30 pm, bill <b92...(a)yahoo.com> wrote: > On Aug 9, 3:45 pm, Gerry Myerson <ge...(a)maths.mq.edi.ai.i2u4email> > wrote: > > > > > > > In article > > <7d557612-615b-4a98-9f40-ca165762d...(a)s24g2000pri.googlegroups.com>, > > > bill <b92...(a)yahoo.com> wrote: > > > On Aug 4, 2:17 pm, 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. > > > > > regards, Bill J > > > > Some corrections: > > > > Conjecture: If N is an integer and > > > Sqrt(N) > floor[( Sqrt(N)], then Sqrt(N) is not rational. > > > To you, it's a conjecture. To anyone whose mathematical education > > has gotten up to where Euclid was, it's a simple theorem, indeed, > > it's just a minor rewriting of the theorem we've been talking about > > since the beginning of this thread. > > > > Suppose that Sqrt(N) is rational. Then Sqrt(N) can > > > be written as an integer + a fraction, say (I + F) > > > Where, by "fraction", you mean, a rational number > > strictly between zero and one. > > > > Or (N) = I^2 + 2*I*F + F^2. Can (2*I*F + F^2) > > > be an integer? Only if F = 0.9... . > > > True, but I'd like to see you prove it! > > > -- > > Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) > > To clarify what i was trying to say is; > > If F < 0.9... , then F^2 < 1. False. > > IMMHO I don't have to prove that > > If F = 0.9... , then F^2 = 1. > > Do I? I don't know. Do you?
From: Bart Goddard on 13 Aug 2010 16:15 bill <b92057(a)yahoo.com> wrote in news:ac70c4ba-0f12-49be-b8a0-bcd2152c0027 @x18g2000pro.googlegroups.com: >> > Or (N) = I^2 + 2*I*F + F^2. Can (2*I*F + F^2) >> > be an integer? �Only if F = 0.9... . >> >> True, but I'd like to see you prove it! >> >> -- >> Gerry Myerson (ge...(a)maths.mq.edi.ai) (i -> u for email) > > To clarify what i was trying to say is; > > If F < 0.9... , then F^2 < 1. > > IMMHO I don't have to prove that > > If F = 0.9... , then F^2 = 1. > > Do I? > Your first statement says "Only if F=.9...". Your last statement says "If F=.9..." These are not the same statements. In fact, they are, in a definite sense, opposites. -- Cheerfully resisting change since 1959.
From: Virgil on 13 Aug 2010 17:55
In article <4454057d-1d8d-4975-b2a8-f1112ced3f87(a)t20g2000yqa.googlegroups.com>, Ludovicus <luiroto(a)yahoo.com> wrote: > On Aug 4, 5:17�pm, bill <b92...(a)yahoo.com> wrote: > > Conjecture: �If N is an integer and �Sqrt(N) is a > > decimal number; then Sqrt(N) is irrational. > > > > regards, Bill J > > It is not a conjecture but a theorem. Proof: > Given the equation N - x^2 = 0 ; N = integer > 0 > If testing the integers 1,2,3...x util x^2 > N > and not finding a solution, then sqrt(N) is irrational. > Ludovicus A proper statement of that conjecture would be: If N is an integer and sqrt(N) is not an integer then sqrt(N) is irrational. The difficulty with the original statement is that "decimal numbers", by most interpretations, include integers. |