posted
Since a=b, a-b=0. Therefore, (a+b)(a-b)=b(a-b)=0. or 0=0. Since a-b=0, you cannot factor it out of the above equation, and everything past this step is erroneous. Most proofs that 1=2 involve division by zero somewhere.
posted
Mmmm... While the division of 0 / 0 does not produce a real number answer, it's a little misleading to say it's "unreal." Better to say "undefined."
Posts: 4534 | Registered: Jan 2003
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posted
Hee hee! Just like those cheesy cable modem commercials, where the guy wins the auction because of his modem. We'll call it a tie, Dagonee. Morbo Posts: 327 | Registered: Oct 2003
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posted
To find that the weight of an elephant is equal to that of a mouse:
Let E = the weight of an elephant. Let m = the weight of a mouse. Let A = the average weight. Then E + m = 2A (E + m)(E - m) = 2A(E - m) E^2 - m^2 = 2AE - 2Am E^2 - 2AE = m^2 - 2Am E^2 - 2AE + A^2 = m^2 - 2Am + A^2 (E - A)^2 = (m - A)^2 E - A = m - A E = m
Posts: 102 | Registered: Oct 2003
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posted
Is it possible that arithmatic and symbolic logic are not totally interchangable? I honestly don't know how to frame this whole thing.
Posts: 859 | Registered: Oct 2003
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posted
That may be the case, but this thread is not an example thereof. The problems in this thread are more illustrative of the fact that mathematics, whether symbolic or arithmetic, has rules that need to be followed in both cases, not just one.
Posts: 4534 | Registered: Jan 2003
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posted
I havent even taken pre-calc yet. Though a teacher of mine did show me a calclus problem if you want to kill youself with it.
Posts: 102 | Registered: Oct 2003
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posted
Define function p as follows: p(n)= (10n-1*(3.14159265...M~n))/(10n-1) Where M~n is the n'th digit of pi. This function puts out a decimal approximation of pi in the form of an integer over an integer (rational number). It is easy to show this works for n=1: p(1)=(100*(3))/(100)=(1*3)/1=3/1 which is a rational number. Using the Principle of Mathematical Induction, one can prove this for all values of n: Assume p(n) is rational. p(n+1)=10(n+1)-13.14159265...M~(n+1)/10(n+1)-1 =10n3.14159265...M~(n+1)/10n =10n(3.14159265...M~n+10-nM~(n+1))/10n =10n(3.14159265...M~n+10n10-nM~(n+1))/10n =10n(3.14159265...M~n+10n10-nM~(n+1))/10n =10n(3.14159265...M~n+M~(n+1))/10n =10n3.14159265...M~n/10n+M~(n+1)/10n =10n-13.14159265...M~n/10n-1+M~(n+1)/10n =Rational Number+(M~(n+1))/10n =Rational Number+Integer/Integer =Rational Number+Rational Number =Rational Number Since p(n) is rational, the limit as n approaches infinity (which is equivalent to pi) is rational.
Posts: 102 | Registered: Oct 2003
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In the limit as n approaches infinity, the function just becomes p(n) = [(10n-1)*pi] / (10n-1)
which when you directly take the limit yields infinity / infinity, which is undefined. So you apply L'Hôpital's rule and take the derivative of the numerator and denominator and then take the limit. As follows:
p(n) = N(n) / D(n) where N(n) = (10n-1)*pi = 10pi*n - pi, and D(n) = 10n-1.
N'(n) = 10pi, D'(n) = 10
So:
lim N(n)/D(n) n->inf
is equivalent to
lim N'(n)/D'(n) n->inf
equals
lim (10*pi)/10 n->inf
equals
lim pi n->inf
equals pi, which is not rational.
Posts: 4534 | Registered: Jan 2003
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posted
There's nothing that says the limit of an infinite (convergent) sequence of rational numbers must be rational. If that were the case, there would be no such thing as an irrational number. When defined through set theory, they are equivalence-classes of infinite sequences of rational numbers which don't correspond functionally to rational numbers.
And when you took the square-root of each side, you didn't use absolute values.
posted
Your first equation I think is burdened by being defined as pi becasue the 10n-1 can simply be canceled out. In that case it would be hard to show that the limit is a rational number because for any n your answer would be pi, an irrational number.
Here is a problem I like. I use (integral of) to denote the integral sign in calculus. (integral of) (1/x)dx Using integration by parts we get: u=(1/x) du=-(1/x^2) v=x dv=1
so that: u'v=uv-v'u (integral of) (-1/x^2)*(x)dx=(1/x)*x-(int of)(1/x)dx
reducing we get: -(Int of)(1/x)dx=1-(int of)(1/x)dx by adding (int of)(1/x) to the other side we get: 1=0, this is of course repeatable for all numbers so that any number to infiniti can equal 0!
quote:I'm still confused as to where (a+b)(a-b)=b(a-b) came from.
The line before that was a^2-b^2=ab-b^2 If you look at the first part you can factor that into (a+b)(a-b) and on the other side, factor out a b so that you have b(a-b). I hope that helps.
Posts: 107 | Registered: Apr 2003
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posted
I wonder if this thread will get as long as the one arguing whether or not 1 and .999999... were the same thing.
Posts: 6213 | Registered: May 2001
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posted
My band director has a favorite saying: "To be early is to be one time, To be on time is to be late, To be late is to be dead." Last year one of our drum majors wrote on the board: Early = on time On time = late early = late Thus, even if you show up late you can still technically be considered early. Just thought it was funny.
Posts: 981 | Registered: Aug 2003
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quote:I'm still confused as to where (a+b)(a-b)=b(a-b)
If you factor a^2 - b^2, you get (a+b)(a-b). Try it in reverse, multiplying the two parenthetical expressions together and you'll see it's true. So that's where the left-hand expression comes from. The right-hand expression comes from a more obvious factoring of ba-b^2.
quote:In that case it would not be hard to show that the limist is a rational number because for any n your answer would be pi, a rational number.
Pi is not a rational number.
In regards to your integration, the integration-by-parts substitution is incorrect.
I'll use int() to indicate an integration:
int( [1/x]dx )
u = 1/x = x^-1, du = [-1/(x^2)]dx = -x^-2 dx v = x, dv = dx
int(u dv) = uv - int(v du)
therefore
int( x^-1 dx ) = (x^-1)(x) - int( -x^-2 * x dx)
therefore
int( x^-1 dx) = 1 - int( -x^-1 dx )
and
int( x^-1 dx ) = 1 + int( x^-1 dx )
The important thing here is that you can't just subtract the integrals from both sides, because they are indefinite integrals. So you have to remember to include the antiderivative plus a constant. What you actually get is:
ln(x) + C1 = 1 + ln(x) + C2
then you can subtract away
ln(x) + C1 - ln(x) = 1 + ln(x) + C2 - ln(x)
therefore
C1 = 1 + C2
which is true, since you can pick constants C1 and C2 such that the equation works.
Posts: 4534 | Registered: Jan 2003
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posted
Hey cool, someone saw it. The substitution on mine isnt wrong I just did it differently. As you showed you get the same thing so thats cool. But you are right, the only problem is that they are indefinite integrals. Of course...if c1=c2 then we are in trouble.
Edit: Wow I cant believe I said pi is rational. I meant irrational. Crap. I didnt even see that.
My band director had the same saying (only that was over 20 years ago.
Is Mr. Jones still teaching? Or is that just something that all Band Directors learn in Band Director School?
Posts: 258 | Registered: Jun 1999
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