a=b a^2=ba a^2-b^2=ba-b^2 (a+b)(a-b)=b(a-b) a+b=b b+b=b 2b=b 2=1
Posted by Raia (Member # 4700) on :
Posted by Megachirops (Member # 4325) on :
Welcome to Hatrack!
Posted by saxon75 (Member # 4589) on :
I feel like an idiot, but I can't, for the life of me, figure out which step is wrong.
Posted by Raia (Member # 4700) on :
Woah, sorry, my brain was so defeated by that that I didn't even notice you were new...
Welcome!!! Posted by Megachirops (Member # 4325) on :
*pat pat*
Posted by Eruve Nandiriel (Member # 5677) on :
????? how the heck did you get from a=b to a^2=ba ?????
(and hi)
Posted by Raia (Member # 4700) on :
Eru, a^2 = a times a, right? So a times a = ba, because a=b.
Wow, I just explained something... mathematical??? What's WRONG with me???
Posted by Caleb Varns (Member # 946) on :
"a+b=b" does not fit.
Welcome to Hatrack.
[ October 21, 2003, 05:09 PM: Message edited by: Caleb Varns ]
Posted by Papa Moose (Member # 1992) on :
I won't spoil it, but if one pays close attention, it's clear which step is not legitimate. Welcome to Hatrack!
--Pop
Posted by saxon75 (Member # 4589) on :
That part's easy enough:
a = b
multiply both sides by a: a * a = b * a
which is equivalent to: a^2 = ba
The first four steps check correctly if you substitute a for b (so you just use a).
Edit: look what happens when I type slowly...
[ October 21, 2003, 05:07 PM: Message edited by: saxon75 ]
Posted by Jon Boy (Member # 4284) on :
If a = b, then a * a is the same as a * b.
Of course, the entire equation falls apart as soon as you give an actual value to the variables.
Edit: Dang. Beaten by about half a dozen different people.
[ October 21, 2003, 05:07 PM: Message edited by: Jon Boy ]
Posted by saxon75 (Member # 4589) on :
Oh duh... Posted by Raia (Member # 4700) on :
*types fast*
Posted by Eruve Nandiriel (Member # 5677) on :
*blushes* Oh. I didn't realise what the "^" was for. Thanks Raia.
Posted by Mormo (Member # 5799) on :
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.
Eruve, you just substitute a=b. a^2= aa =ba.
Welcome to the forum, Gosu.
[ October 21, 2003, 05:13 PM: Message edited by: Mormo ]
Posted by Dagonee (Member # 5818) on :
The problem is in the step going from
(a+b)(a-b)=b(a-b)
to
a+b=b
Since (a-b) = 0, division by (a-b) creates an unreal (not imaginary) result. So this step is not allowed.
In fact, the whole sequence is a decent proof that you can’t divide by 0.
Very nice,
Dagonee
Posted by Dagonee (Member # 5818) on :
Doh! Mormo beat me by 10 seconds.
[ October 21, 2003, 05:14 PM: Message edited by: Dagonee ]
Posted by saxon75 (Member # 4589) on :
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."
Posted by Mormo (Member # 5799) on :
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 Posted by Dagonee (Member # 5818) on :
Good point, saxon75. I haven't taken math in 11 years, so I mixed up the words.
Posted by Gosu (Member # 5783) on :
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
Posted by saxon75 (Member # 4589) on :
Just in case anyone cares...
The implicit step between steps 4 and 5 is:
[(a+b)(a-b)] / (a-b) = [b(a-b)] / (a-b)
Taking the limit of both sides gives:
lim ( [(a+b)(a-b)] / (a-b) ) = 2b a->b
and
lim ( [b(a-b)] / (a-b) ) = b a->b
Which is not particularly useful, but is illustrative of what happens when I'm bored at work.
Posted by Jon Boy (Member # 4284) on :
This thread almost makes me pine for calculus.
Posted by Robespierre (Member # 5779) on :
Is it possible that arithmatic and symbolic logic are not totally interchangable? I honestly don't know how to frame this whole thing.
Posted by saxon75 (Member # 4589) on :
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.
Posted by Bob the Lawyer (Member # 3278) on :
quote: This thread almost makes me pine for calculus.
Which is odd, because this is a lot more like algebra than it is like calculus.
Posted by saxon75 (Member # 4589) on :
Well, it was. Until I started indiscriminately taking limits.
Posted by Gosu (Member # 5783) on :
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.
Posted by saxon75 (Member # 4589) on :
Shoot.
Posted by Gosu (Member # 5783) on :
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.
Posted by Gosu (Member # 5783) on :
I think neh?
Posted by saxon75 (Member # 4589) on :
Not exactly.
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.
Posted by Papa Moose (Member # 1992) on :
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.
--Pop
Posted by Julie (Member # 5580) on :
I'm still confused as to where (a+b)(a-b)=b(a-b) came from.
Posted by Jerryst316 (Member # 5054) on :
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!
[ October 21, 2003, 07:17 PM: Message edited by: Jerryst316 ]
Posted by Papa Moose (Member # 1992) on :
+C
Posted by Jerryst316 (Member # 5054) on :
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.
Posted by Papa Moose (Member # 1992) on :
I wonder if this thread will get as long as the one arguing whether or not 1 and .999999... were the same thing.
Posted by Jerryst316 (Member # 5054) on :
LOL sorry papa but we all know they are the same thing.
Posted by Julie (Member # 5580) on :
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.
Posted by Megachirops (Member # 4325) on :
Really Papa? People argued this fact?
Sad.
[ October 21, 2003, 06:59 PM: Message edited by: Megachirops ]
Posted by saxon75 (Member # 4589) on :
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.
Posted by saxon75 (Member # 4589) on :
Stupid slow fingers...
Posted by Jerryst316 (Member # 5054) on :
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.
[ October 21, 2003, 07:10 PM: Message edited by: Jerryst316 ]
Posted by Evie3217 (Member # 5426) on :
*pokes head in*
*head explodes*
Posted by Mr.Funny (Member # 4467) on :
.999999 repeating DOES equal 1!
Posted by pwiscombe (Member # 181) on :
Julie,
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?
Posted by Ryan Hart (Member # 5513) on :
::head follows Evie::
Posted by fugu13 (Member # 2859) on :
e^(i*pi) = -1 e^(2*i*pi) = (-1)^2 ln e^(2*i*pi) = ln 1 2*i*pi = 0 2*pi = 0 (division by i, a non-zero constant) pi = 0 (division by 2) 1 = 0 (division by pi)
Posted by Papa Moose (Member # 1992) on :
x is not a distinct solution of ln e^x in the complex field, but I like yours best so far.
Posted by rivka (Member # 4859) on :
Dear God, is there nowhere to get away from studying for my actuary test today?
Um, thanks? Posted by Ralphie (Member # 1565) on :
The internet is hard today.
Posted by rivka (Member # 4859) on :
*sits back to watch the race to the OoC thread* Posted by fugu13 (Member # 2859) on :
Ye-p, it's due to the complex areas where it breaks down. Also, log 1 != 0 when you consider complex values.
Posted by kerinin (Member # 4860) on :
its been awhile since i saw this, so forgive me for being sort of vague, but i saw a really neat investigation of what happens when you try to differentiate the function
f(x)= x^x
which if i remember correctly got made complex (f(x)=x^z). Anyway, the function requires some really crazy differentiation which i understood a few years ago but since i've been avoiding math for some time now i don't remember well enough to explain.
anyway, the point is that the derivative of the function at 0 ends up being either 0 OR 1, depending on how you look at it.
maybe someone who has taken some math classes more recently than i will recognize this and be able to explain it better...
[edit] i remember now, it's because 0 raised to any number is 0, but any number raised to the 0 power is 1, which creates a contradiction at 0 for the original function
[ October 22, 2003, 12:02 AM: Message edited by: kerinin ]
Posted by Ralphie (Member # 1565) on :
You know, this is why I believe that no matter what I say it's immediately taken OoC.
Did I say the interpole was hard today? Did I say the world wide bulge? Did I make any reference to the .package?
Ralphie, totally innocuously, said the internet was hard today. And very humorously, I might add.
Sheesh. Deal.
Posted by Megachirops (Member # 4325) on :
quote:Did I say the interpole was hard today? Did I say the world wide bulge? Did I make any reference to the .package?
Well, this wouldn't be out of context. It would just be your usual leering innuendo.
Posted by fugu13 (Member # 2859) on :
I think the function just isn't differentiable at that point.
Posted by rivka (Member # 4859) on :
I just thought it was funny. I think I would have regardless of who said it.
And why, oh innocuous one, are you not on AIM?
Posted by Ralphie (Member # 1565) on :
I've already spent too much time on the interpole.
Er, net.
edit: btw, Rivka - no real offense taken. Every once in a while I have to doth protest my skanky image.
[ October 22, 2003, 02:27 AM: Message edited by: Ralphie ]
Posted by A Rat Named Dog (Member # 699) on :
Wait a second. If a set of factors equal zero when multiplied together, that doesn't mean that any one of them could be zero if you just divide both sides by the others. It only means that at least ONE of the factors is zero.
Still doesn't fully answer your paradox, but still ... Math makes no sense. When I think about how many ridiculous things astronomers believe about black holes and the like just because they can make it happen with abstract mathematics ...
Posted by fugu13 (Member # 2859) on :
And yet a lot of those ridiculous things have turned out to be right: Hawking radiation, for instance.
Posted by prolixshore (Member # 4496) on :
Oh wait, i know this one, three guys go to a hotel and pay 30 dollars for the room, ten bucks each.....
--ApostleRadio
Posted by Tristan (Member # 1670) on :
... But then the manager discovers that he's made a mistake, and the room only costs 25$, and gives five dollar to the porter to return to the guests. The porter, not being very good at math nor a very honest person, considers it too much trouble to divide five $ on three persons, and gives back one $ to each of the guests and keeps two for his own pockets. The guests have now paid nine $ each. Three times nine is 27. The porter has two. That makes 29.
Hey, where did the last dollar go?!
[ October 22, 2003, 05:31 AM: Message edited by: Tristan ]
Posted by fugu13 (Member # 2859) on :
Behind the back of the trickster breaking out that old chestnut.
Posted by Tristan (Member # 1670) on :
Well, we non mathematicians must have something to play with, too.
Posted by Lalo (Member # 3772) on :
quote:... But then the manager discovers that he's made a mistake, and the room only costs 25$, and gives five dollar to the porter to return to the guests. The porter, not being very good at math nor a very honest person, considers it too much trouble to divide five $ on three persons, and gives back one $ to each of the guests and keeps two for his own pockets. The guests have now paid nine $ each. Three times nine is 27. The porter has two. That makes 29.
Hey, where did the last dollar go?!
Whoa. That's kinda trippy. I mean, despite having an American ejukashun, I know $25 + $3 = $28 -- not $27 -- but I'm still following the logic of the paragraph above.
A few more of these, mixed with weed and airplane model glue, and I'm well on my way to unravelling the secrets of the universe...
Posted by fugu13 (Member # 2859) on :
$30 paid. $25 stays with the hotel. $5 goes with the clerk. clerk gives $3 to men, keeps $2. hotel has $25, men have $3, clerk has $2. $25 + $3 + $2 = $30. All the money is accounted for.
Posted by Tristan (Member # 1670) on :
Well, that's all well and good, fugu13, but it does not really answer the question. The fun thing is to figure out exactly where and how the one buck disappears in the alternative way of counting that the problem provides. It isn't exactly high mathematics but that makes it perfect for those who, like me, aren't mathematicians but enjoy logical problems nevertheless.
[ October 22, 2003, 06:28 AM: Message edited by: Tristan ]
Posted by prolixshore (Member # 4496) on :
I'm just amazed that people responded to my post. But then again, that stupid problem always gets someone to talk.
--ApostleRadio
I love hatrack. You can't get this kind of conversation many other places.
Posted by fugu13 (Member # 2859) on :
The problem occurs with adding the $2. The $2 should be subtracted, just as the $3 was (though in a roundabout way, by phrasing it as an addition of what the three men had each paid), resulting in the price remaining with the hotel of $25, just as is expected.
Posted by Fyfe (Member # 937) on :
How fond I am of that proof. Actually, the one I saw was slightly different; it was in Zero: The Biography of a Dangerous Idea, and it started out by letting a and b equal one. Ultimately it proved that 1=0. Then, as I recall, the book went on to show that Winston Churchill was a carrot.
Jen
Posted by docmagik (Member # 1131) on :
Another way to look at it is this:
By keeping the two dollars, the desk clerk made the price of the room 27 dollars. This fits with the part of the riddle where you multiply three times nine to get the twenty seven dollars.
But the problem is in the next step. You can't add the two dollars to the 27 again, because it's already included in what the clients paid--that two dollars came out of the 27 dollars.
Rather, you have to add the three dollars that was given back to the customers--which don't even figure in to the original riddle, but were obviously part of the original transaction.
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