Re: 4=3 A maths joke

In article
<ff67886f-043a-430b-9f23-349fcd344cac@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Tonico <Tonicopm@xxxxxxxxx> wrote:

On May 17, 7:38 am, Michael Press <rub...@xxxxxxxxxxx> wrote:
In article
<ee37a8e9-bd70-4a11-b16f-f05c76ac3...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,





 Tonico <Tonic...@xxxxxxxxx> wrote:
On May 16, 1:52 am, Michael Press <rub...@xxxxxxxxxxx> wrote:
In article
<dc9e218c-9429-48e2-a05b-d5ab1dfcb...@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,

 Tonico <Tonic...@xxxxxxxxx> wrote:
On May 15, 9:02 am, Albert <albert.xtheunkno...@xxxxxxxxx> wrote:
LOL This took me 5 minutes to work out what was wrong to my friend and
it was hilarious.
a+b=c where c > a and c > b
4a-3a+4b-3b = 4c-3c
4a+4b-4c=3a+3b-3c
4(a+b-c)=3(a+b-c)
4=3

Where's the flaw?

*****************************************************************

Want a slightly (VERY slightly) less boring and less known "hoax",
which at least uses some basic calculus? Here:

Let n be a natural number (non-zero, to be sure). As we all know, n^2
= n*n, and the right side can be put as n + n +....+n (n times), so:

**  n^2 = n + n +...+ n (n times). Now evaluate the derivative in both
sides, using the well known rule for the sum:

**  2n = 1 + 1 +...+ 1 (n times) = n , and thus

^^^^^^^^   ^^^^^^^^^^^^^
     OK    Bzzzzzzzzzzzzzzzzzzzzt!

           should be 1 + 1 +...+ 1 (n times) + n (1 times)

What's (x-a) * (x-b) * (x-c) * ... * (x-z) ?
Hide quoted text -

Hey! You plonked me.





Again the same thing as with Tommy: I don't have the faintest idea
what you wrote. Apparently you thing the hoax lays in some kind of
algebraic blunder. Well, it does not:

** (1) It is true that  n^2 = n*n = n +...+ (n times)

** (2) It is true that d(x^2)/dx = 2x

** (3) It is true that d(x+...+x)/dx = 1 +...+ 1 (as many times as in x
+...+x)

Now, the mistake is (don't read the following if you still want to
work your own solution!)
...
...
...
...
...
...
...
...
...
...
...

It is not true in general that x*x = x +...+ x (x times) for a REAL
variable!, and in fact the gist of the hoax lays in confusing between
discrete and continuous variables. One needs the latter to define
limits as done in the derivative's definition.

The above hoax was given in a calculus I (sceond semester) course of
which I was the instructor some years ago, and it created one serious
fuss back then within the class, since it is not as trivial as the
usual "proofs" that 1 = 2, or 0 = 1, etc.

I strongly disagree. d/dn is derivative with respect to n.
d/dn (a + a + ... + a (n times)) = a.
This all works. You have not demonstrated a problem with
the _calculation_ of derivatives. You have demonstrated
your hesitation to embrace all of what follows from the
differential calculus.

*****************************************************

Oh, If I just could know what you disagree with...Who said that what
you remark doesn't work? Who said I had demonstrated there's a
problem with calculation of derivatives?
About "my hesitation" I, once again, don't have the faintest idea what
you're talking about...

*****************************************************

d/dn (n + n + ... + n (n times)) is a pretty application
of Leibniz's theorem for differentiation of an integral.
Do you suppose Gottfried Wilhelm von Leibniz would hesitate
to differentiate the right side?

********************************************************

I really have not the slightest idea what good'ol Gottfried would do
in such a situation...perhaps he would begin talking about monads and
nonsenses like that? Can't say, really.

The problem, as was already noted, is that the equality
n^2 = n +...+ n (n times) used in the hoax can't be used with real
variables in general...wasn't this clear? Oh, well.

Poor Tommy, lost as usual, thought that he would apply the derivative
of a product and get a correct answer...but that was not how I
presented the hoax and has nothing to do with it. He could as well use
the chain rule or the second cohomology group.
How it was presented has an answer, and it already was given above.

I will find the derivative of the right side from the definition of derivative.
Suppose we increase n by a quantity h. Then we have

(n + h) + (n + h) + ... + (n + h) {n + h times}
= n + n + ... + n {n times}
+ n + n + ... + n {h times}
+ h + h + ... + h {n times}
+ h + h + ... + h {h times}

The quotient for the derivative is
n + n + ... + n {h/h times}
+ h/h + h/h + ... + h/h {n times}
+ h + h + ... + h {h/h times}

= n + n + ... + n {1 times}
+ 1 + 1 + ... + 1 {n times}
+ h + h + ... + h {1 times}
= n + n + h.

As h ->0, the quotient goes to 2.n.

The form n + n + ... + n {n times}
where the quantities are not integers is commonplace.
Suppose (as was the case at my secondary school)
a running track is a 1/5 mile oval. To run a 1/2 mile race
the contestants circled the track
1 + 1 + ... + 1 {2 1/2 times}.

A twelve furlong race on an eight furlong track:

8 furlong + 8 furlong + ... + 8 furlong {1 1/2 times} = 12 furlong.

--
Michael Press
.

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