Side Topic: 'It's Not a Toom-ah' - Page 4

0PolarBearsHere · 2015-11-13, 10:34

Quote:

Originally Posted by ewmayer

Beyond 'raft moves at same speed as river, which flows at constant uniform speed'...

Ahh of course... thanks.

firejuggler · 2015-11-13, 11:42

For the cumcumber...
1000 pound of cumbcumber, 99 %water, thus 990 pound of water.
They loose cumbcumber go from 99% water to 98, but the dry weight doesn't change. so it stay at 10 pound.
the 1 % water loss mean that 98/99 of water stay, thus 990->980, so effectively, the total weight is now 990 pound.

Am I a fool on this one?

R.D. Silverman · 2015-11-13, 11:54

Quote:

Originally Posted by ewmayer

I think Toom does engage in some rather 'selective prosecution' there - for the Tom/Dick/Harry linear algebra problem, yes, the student *should* have anal-retentively begun with 'let us denote the time needed by Tom, Dick and Harry by T, D and H, respectively,' but to pretend it is non-obvious what the student meant is being deliberately obtuse. (And Toom fails to mention whether the student actually solved the resulting system correctly.) At the same time Toom makes it sound trivial that 0.999... = 1, when in fact obtaining this result involves some highly nontrivial aspects of real-numerology and set theory - including 'faith-based' acceptance of at least one key axiom (probably several, but I've not the time to make a detailed list at the moment) pertaining to the reals. One must in effect resolve the seeming contradiction between the following two 'proofs' which yield opposing results:

Proof A: The distance d between the n-term (n finite) expansion and 1 is d = 10^(-n), which --> 0 as n --> oo, thus in the limit, 0.999... = 1.

Proof B: Using the same notation as Proof A, the number of distinct real points in the length-d interval separating the n-term expansion and 1 is infinite (in fact uncountably so). Incrementing n by 1 cuts the distance by a factor of 10, but the number of distinct reals in the new smaller interval is still uncountably infinite. Thus no matter how large we take n, there remains an uncountably infinite number of points separating the point corresponding to the resulting expansion from 1, hence 0.999... != 1.

This "proof" is bull. It assumes that a countable union of uncountable sets is countable. --> false.

This is a common problem when using ordinary English to describe infinite processes.
In fact, even during the time of Gauss, math did not have a firm definition of 'real number'.
It was not until the mid 1800's that Cauchy and Dedekind put forth a rigorous foundation.

And proof "A" is not a proof at all. A correct proof is the following.

First ask: what is the meaning of .9999999.....?

By DEFINITION it is the limit of the sequence .9, .99, .999, .9999, .............
Now use the epsilon-delta DEFINITION of limit of a sequence (or function) to complete the proof.
The given "proof A" merely asserts that the limit is 1.

R.D. Silverman · 2015-11-13, 11:58

Quote:

Originally Posted by retina

Hmm. I have no idea what you mean here. I thought it was simply one thing is numerically the same as the other.

Certainly. But one must PROVE that 1/3 = .33333333...

Mere handwaving does not suffice.
Then, one must prove that 2/3 = 2*(.333333333........) = .666666666666.......
Merely asserting that one can multiply an infinite sequence of digits one digit at a time
is not a proof.

Welcome to Real Analysis. A totally separate course from Number Theory.

0PolarBearsHere · 2015-11-13, 12:22

Quote:

Originally Posted by firejuggler

For the cumcumber...
1000 pound of cumbcumber, 99 %water, thus 990 pound of water.
They loose cumbcumber go from 99% water to 98, but the dry weight doesn't change. so it stay at 10 pound.
the 1 % water loss mean that 98/99 of water stay, thus 990->980, so effectively, the total weight is now 990 pound.

Am I a fool on this one?

Based on "..and the percentage of water dropped to 98%", I read it as:
Before: 99% wet, 1% dry
After: 98% wet, 2% dry
Like you said, dry mass doesn't change, so I had it going from 10lbs being 1% to 10lbs being 2% of total mass -> new mass=500lbs

Maybe I'm the fool.

Xyzzy · 2015-11-13, 13:00

Quote:

Originally Posted by ewmayer

:facepalm:

There is a smiley for that:

wblipp · 2015-11-13, 14:21

Some of the answers on this thread assume mathematical sophistication at the graduate school level. I doubt Toom was expecting that. Toom's pedagogical question to the students was "how much smaller?" so I think he was looking for an informal limit understanding. Something like this:

What does it mean to say 1/3 = 0.3333...? It means that as you take more terms you get closer and closer to 1/3 - closer than any positive number. In the same sense, 0.9999... gets closer and closer to 1.

---------
The second and third problem are both rate problems. The "trick" is to understand that rates add (or subtract in the case of boat going upriver). I agree with others on this thread that the point of asking what the variables T, D, H stand for was to get the student to see that they needed to be rates, and hence their sum must also be a rate.

R.D. Silverman · 2015-11-13, 14:32

Quote:

Originally Posted by wblipp

Some of the answers on this thread assume mathematical sophistication at the graduate school level.

FALSE. I was, for example, introduced to epsilon-delta proofs in high school as part of the class. (sophomore in fact)
[I took the class early, but it was normally a junior level honors pre-calculus class)

It is certainly part of standard undergrad curricula in any calculus class for math majors.
(i.e. not 'calculus for engineers' or equivalent).

Quote:

What does it mean to say 1/3 = 0.3333...? It means that as you take more terms you get closer and closer to 1/3 - closer than any positive number. In the same sense, 0.9999... gets closer and closer to 1.

NO! NO! NO! This is NOT what it means. It means that the infinite sequence .3, .33, .333, .3333, ..... converges
to 1/3 in the limit. All this blather about "closer and closer to 1/3" is informal HAND-WAVING.

retina · 2015-11-13, 16:00

Quote:

Originally Posted by R.D. Silverman

Certainly. But one must PROVE that 1/3 = .33333333...

I will try.

Very early in my schooling I learned to do long division. If I use that technique to divide 3 into 1 then the answer generates 0.3333... The remainder at each step after dividing 10 by 3 is always 1, and carrying down the next 0 gives back 10/3 to the next step.

davar55 · 2015-11-13, 16:23

Quote:

Originally Posted by retina

Oh yeah, of course. Clear as mud to me.

I think you just made me realise this online course too complicated. There is no way I could write a two paragraph explanation like the above just for my usage of '=' when trying to prove something. "Cauchy sequences", "Hilbert spaces", "nonsmooth function", "Fourier-series", "Gibbs phenomenon", "pointwise-sense", "measure theory", "L^2 norm", "vanishes in the limit": I didn't expect the Spanish inquisition.

I guess I did expect the Spanish inquisition, because I basically understood his post.

First thought: Is an axiom necessarily an act of faith? I say no. Something like
"All right angles are congruent to each other" is an axiom, it is validatable on the
basis of concepts preceding the formalism of the axiomatic approach of geometry.

Some axioms are obvious, aren't they? Or we can never get past them.

alpertron · 2015-11-13, 16:29

Quote:

Originally Posted by R.D. Silverman

Certainly. But one must PROVE that 1/3 = .33333333...

By definition:
$x = .3333... = \frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...$ (1)
Multiplying by 10:
$10 x = 3+\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...$ (2)
Subtracting (2) - (1) all terms after the 3 are cancelled:

9x = 3, so x=1/3

Another attempt without subtracting infinite terms is by considering the sequence .3, .33, .333, .3333, ...

The nth term is:
$x = .3333...3 = \frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...+\frac{3}{10^{n-1}}+\frac{3}{10^n}$ (3)
$10x = 3+\frac{3}{10}+\frac{3}{10^2}+\frac{3}{10^3}+...+\frac{3}{10^{n-2}}+\frac{3}{10^{n-1}}$ (4)

Subtracting (4) - (3), discarding all terms that are equal:
$9x = 3-\frac{3}{10^n}$
So for n->inf we get 9x=3, which implies x=1/3

2015-11-13, 11:42	#35
firejuggler Apr 2010 Over the rainbow 2³×5²×13 Posts	For the cumcumber... 1000 pound of cumbcumber, 99 %water, thus 990 pound of water. They loose cumbcumber go from 99% water to 98, but the dry weight doesn't change. so it stay at 10 pound. the 1 % water loss mean that 98/99 of water stay, thus 990->980, so effectively, the total weight is now 990 pound. Am I a fool on this one? Last fiddled with by firejuggler on 2015-11-13 at 11:42

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2015-11-13, 14:21	#40
wblipp "William" May 2003 New Haven 2×7×13² Posts	Some of the answers on this thread assume mathematical sophistication at the graduate school level. I doubt Toom was expecting that. Toom's pedagogical question to the students was "how much smaller?" so I think he was looking for an informal limit understanding. Something like this: What does it mean to say 1/3 = 0.3333...? It means that as you take more terms you get closer and closer to 1/3 - closer than any positive number. In the same sense, 0.9999... gets closer and closer to 1. --------- The second and third problem are both rate problems. The "trick" is to understand that rates add (or subtract in the case of boat going upriver). I agree with others on this thread that the point of asking what the variables T, D, H stand for was to get the student to see that they needed to be rates, and hence their sum must also be a rate.