Proof of 1 = 0. Mathematical fallacies

PeteLong: Ah ha! an inductive fallacy good. Probably 50% of the inductive proofs I got from students when I was TA as a grad student fell into this category. This one is good the trouble of course is in going from s(1) to s(2). :-)

This one also illustrates some thing that has interested me for a while and that is how the limitations of language effects our ability to comprehend a problem and formulate solutions. There is an old classic riddle concerning a missing dollar where by 3 guys buy a pizza between them at a hotel giving the porter a tip etc etc (most are familar with this one) and at the end of the transaction there seems to be a dollar missing. However hard I try I have never been able to explain to some what goes on here because of language, (and it not that I am inarticulate).

GfW

:0) thre guts pay 10 pound each

manager gives them 5 pounds back

waiter cant split 5 pounds three ways

waiter gives them all a pound and keeps two pounds

they all paid 10 pound and got a pound back 10 -1 =9 pounds

there were three of them 3X9=27 pounds

if they paid 27 pounds and the waiter kept 2 pounds 27+2=29

where did the other pound go??  30 -1 =29

That the one, :0)

PeteL

Yes thats the one, but however hard I try I can not explain it to my neighbour who although some what inumerate is not stupid. I think it a problem with the limits of language to describe often simple mathematical ideas precisely.

:0)

That the english problem, the spoken word is open to interpretation

eg

I have twenty sick sheep one of them dies how many do I have?

the answer is obviously 19

now ask someone the same question, they will say 25 :0)

Or

Fruit flies like bananas

this is obvious but spoken I could be talking about
the arodynamics of fruit OR eating habits of insects

The problem above (Im pants at mathematics by the way!) is more a bending of accepted mathematical rules, and executing mathematical statements in the wrong oder

for example 2+3X5 equals 25 right
wrong
dont believe me set your PC calculator to scientific
and do it again
The answer is 17

   My interest in this started when a number of yers ago I was hired by a lawyer to sort out a problem involving his client a landord who had not had his last months rent paid. The contract for the tenent stated  6 months at $200 a month. The documents clearly showed that the tenent had not paid the last months rent. This did not concern the tenent and he presented a complex arithmetic argument to 'demonstrate" that he was not in breach of his contract,  neither the lawyer or the landlord could find an error in it! I found the error quickly but could not explain it to them as neither could any of my colleagues.

  Legally the onus was on the Landlord to prove his case and the lawyer was worried that if it could not be explained to them what would happen in court. Eventually the tenent just paid up and it was found out he had been using this trick for some years, often getting out from paying his rent.

  What occured to me was if such  a simple problem could cause trouble what about maths in general?

Anyway back to the fallacies Let's have some more!!

:0)

dorf - Good one.

Let's be clear: If the host reveals another door that is a booby prize, and offers you a new choice, then you must change your choice.  THIS DOUBLES YOUR LIKELIHOOD OF WINNING!

??The bike moves forward?? the lower crank - on either side been driven backwards propells the bike forwards, unless of course Im a clown and have missed something obvious??

PL

deighton:  Interesting.  The force applied by the person causes an acceleration in the opposite direction!  The other horizontal force on the bike is the friction force applied by the ground to the bike.  And it is in the direction of the acceleration.  We need both linear and rotational principles to analyze this.  Both the force applied by the person and the force applied by the ground contribute to the total torque on the wheel.  Note: A sufficiently strong force applied by the person can overcome friction and cause a backward acceleration.

It would seem

>>Im a clown and have missed something obvious??

Im an Engineer, shesh backwards on the bottom crank has to equal forward on the top crank - accepting the bike has a chain and is connected the rear wheel (if the chain is going to the rear wheel) will drive forwards??

Now for one quite past a few....

Using the following partial deriviatives:

Int(f(x)*g(x))dx = f(x)*G(x) - Int(f'(x)*G(x))dx

where G(x): the primitive function of g(x)
     and f'(x): the derivative of f(x)

Now watch the following 'proof':

    Int(1/x^2 * 2x)dx  =  1/x^2 * x^2 - Int(-2/x^3 * x^2)dx
           {just take f(x) = 1/x^2 and g(x) = 2x}

    this yields:

    Int(2/x)dx  =  1 - Int(-2/x)dx  = 1 + Int(2/x)dx    

    substracting Int(2/x)dx on both sides yields:

    0 = 1

oR i SHOULD SAY PARTIAL INTEGRALS

I have not seen that one before, I like it. :-)

The integration fallacy is expressed more simply:

d(u+1) = du

Integrate both sides:

Int(d(u+1)) = Int(du)

u+1 = u

1 = 0

I've reposted meintsi's post below for reference.

The fallacy in this argument is a little tricky but here it is:

We were given that a = b.
If a = b then a - b = 0.
The second step places (a-b) in the denominator.
But (a-b) = 0 so the second step, and all further steps, are invalid; 0 in the denominator is not defined.




MEINTSI's POST (REPOST):

Here's your classic example:

Given a = b  then this implies

       a - b + b = b

Now divide both sides by (a-b) and we have:

       a - b + b  =        b
       _________    _____
         (a-b)             (a-b)

Reducing the left side:
             
       1 +      b     =       b
              _____       _____
              (a-b)          (a-b)

Then subtract b/(a-b) from both sides and you have

               1=0

:)

the problem, of course, with the integration "proof" is that you aren't accounting for the constant of integration:

Int(d(u + 1)) = Int(du)

doesn't give you:

u + 1 = u

it actually gives you:

u + 1 + c1 = u + c2

and c2 then must equal c1 + 1 and there is no problem whatsoever :-)

meintsi's "classic example" post... my math teacher for GTA (no, not grand theft auto! - geometry, trigonometry, and advanced algebra) actually taught that one, except he taught it as 1=2, not 1=0. So it looked something like this:

Given: a = b

a*b = b^2

and then something else happened... shoot! i forget how he did it lol but there was the same "dividing by 0" problem with it... yeah...

Here's one:

given a=b:

a^b = b^a

let b=0,

a^0 = 0^a

1 = 0

calculate integral of (x-1)^-2 from -3 to 2

note that the function is always positive, so the area lies above the x axis and is positive.

integrating the original (x-1)^-2 gives -(x-1)^-1 or -1/(x-1)

so plugging in the limits of integration

-1/(2-1) + 1/(-3-1) = -1.25

a negative area for a function that is always negative??????

that's crazy... it's caused, i believe by the "volcano" (vertical asymptote with upwards sloping sides) within that range... but it's interesting to note that that then shows that positive infinity = -0.25 :-)

Well, what's going on there is that the antiderivative F(x)=1/(1-x) not only goes to infinity at x=1, but goes to positive infinity from one side, and negative infinity from the other. Because there is no way to make those limits converge, the integral calculated across x=1 does not represent the area under the curve. One way to see what's going on is to break that integral up into parts that don't cross the singularity. For example, adding the integrals from x=0 to x=1-d and x=1+d to 2 for 0<d<1, we get:

(1/d - 1) + (-1 + 1/d) = 2/d - 2

which is strictly positive for 0<d<1. And as d->0, this goes to positive infinity, as we would expect.

Breaking up the integral and taking it in the limit that the size of the gap containing the singularity goes to zero is a more correct way of computing the area under the curve in that range.

-Seth

yes, that's the one! see201, that's the one that my GTA teacher taught me... nice to see it once again.

PeteLong, wytcom & deighton:

As far as the bicycle one is concerned, wytcom is correct assuming the gear ratio is typical. However, if the gear ratio is such that the pedals move faster than the bike, it will move forward instead. Just because you push backwards doesn't mean the bike will go that way, since the ground can still push the wheels forwards in response to the applied torque.

-Seth

This is essentially Zeno's Paradox. It stems from the (incorrect) assumption that there are a countably infinite (like integers) number of "frames" in any interval of time. Although this is usually resolved by assuming there are an uncountable number of frames (a continuum, like real numbers), my feeling is that a time interval cannot be divided into infinitely many sub-intervals in the first place. Not only does that eliminate the apparent paradox, but can also explain special relativity without a continuum and provides a link to the kinds of discreteness we find in quantum theory.

-Seth

The waiter and 3 guys problem is a simple one if u change ur point of view a bit.
Initially, when the guys paid 10 dollars each, in all they paid 30 dollars to the manager.
When the manager gave away 5 dollars to the waiter, the 3 guys had paid 25 dollars to the manager and 5 to the waiter. When the waiter returns 3 dollars, one can infer that the 3 guys paid 3x9 dollars (25 to the manager and 2 to the waiter).

half-empty is the same as half-full. therefore empty is the same as full.

1/2=1/2. therefore 0=1. Now it looks stupid :)

what a strange logic you have ;-)

1/2=1/2 thus 1=1 or 0=0 that's all you can get with adding or substracting :D

The story of Achille and the tortoise is the famous "paradox of Zénon"

All numbers (positive integers) are interesting.

1 is interesting as it is the only number by which you can multiply any other number without changing that other number's value.

If "x is interesting" for all x from 1 to (n-1) then n must also be interesting as otherwise it would have the interesting property of being the lowest boring number.

Interesting...  ;-)

Now try that for an uncountable set (like real numbers).

-Seth

I think I'll leave Induction to countable sets (and gynaecologists).....  :-)

I hope you all know those "proofs" are breaking Axiom laws.
None of those prove anything, they all have a MAJOR Axiom failure.

i'll use a few to demonstrate

see201:
((
a = b

 2
a  = ab

 2      2             2
a   -  b   = ab - b

(a+b)(a-b) = b(a-b)

a+b = b

since a=b: 2b = b

2 = 1
))
Axiom you're breaking is dividing bothsides by (a-b) if a = b, then you're dividing both sides by 0.


SethHoyt
((
given a=b:

a^b = b^a

let b=0,

a^0 = 0^a

1 = 0
))

a = b, so if b = 0 so does a.
complete the math it'll end up as 0^0 = 0^0


Paul
((
Int(d(u + 1)) = Int(du)

doesn't give you:

u + 1 = u

it actually gives you:

u + 1 + c1 = u + c2

))
I'm curious, doesn't the int(d(u+1)) actually give you  u + c1, since you have to take the derivative of u + 1 (d(u+1) = 1 -> int(i) = u + c) ?)  I haven't done this for a while...

anyway, no probs there anyway, since c could be any number =)



Makc
((
Half full and half empty therefore full = empty
))

Not true, put the wording in completely, Half-empty is a poorly stated "state" of complete.  It's halfway FROM being empty, and half-way From being full.
Their "direction" of speech implies one is going "up" and other going down.  1/2 + (c) = 1/2 - (c)
the c is the change.  They cancel out.





The induction proof on top, I can't explain why it's wrong, but the induction theory doesn't say what's for 1 applies for all, it states as far as I remember, that there's always a predecessor for all n'.

That's my 2cents :p
-Ixeus

Ixeus,

This question specifically asks for fallacies, which is why the "proofs" all have errors. Of course it cannot be proven that 0=1.

With regards to your comment on my "proof", you did not correctly point out the fallacy. You only showed a tautology, which proves nothing at all.

Can someone point out the error in my "proof"?

-Seth

You can take this further to prove -1 = 0
therefore 1 = -1

0^a = a^0
could say that 1 = 0
But if you subtract both sides a^0

0^a - (a^0) = 0

since 0^a always 0 no matter what a is therefore

-(a^0) = 0
-1 = 0
-1 = 1
since 0' = 1
-1 = 0
then -1' = 1

and by induction all numbers are the same :p

There's clearly a flaw in this logic too :)

Seth, I think your comments on Zeno's paradox are good.

Analytical philosophers, following Aristotle, distinguished between logical and actual infinity, divisibility, etc. As I recall, modern philosophers have tended to accept Zeno's paradox and other paradoxes related to infinity as valid LOGICAL statements but invalid empirical statements.

Just because it is mathematically possible for there to be an infinite population on the earth doesn't mean it is PHYSICALLY possible. Indeed, the fact that there is NOT an infinite population suggests that there are limits to the sustainable population on earth.

Similarly, it is certainly possible that the real world behaves as Zeno's turtle behaves; that is to say, a universe in which that description is true is certainly a possible universe. But it collides with the reality of THIS universe, where it appears that there is a [real, actual] lower limit to divisibility, and the limit is, as you pointed out, an essential feature of quantum mechanics, i.e. of reality.

Movement in THIS world is self-evident, which causes us to be shocked by Zeno's description of motion.

One might turn Zeno on his head, using his argument as the first premise to deduce the necessity of quantum phenomena:

If there were infinite divisibility of the world, there would be no motion.
There is motion.
Therefore the world is not infinitely divisible.

I think this is the one for which Seth wants the fallacy (he also did a nice "missing constant of integration")

>> given a=b:

>> a^b = b^a

>> let b=0,

>> a^0 = 0^a

>> 1 = 0


a^0 is certainly 1 for all a > 0.  Is a^0 defined to be 1 or is it really a/a ?

Similarly 0^a = 0 for all a > 0.  But it is undefined (calculated as 1/0) if a is negative.  It is not clear that it is defined on the boundary, when a = 0.

I think this is really a very subtle "undefined number" or "divide by 0" falacy.  To say "a^0 = 0^a when a=0" seems self-evidently true but it is only valid if 0^0 has some meaning.  I think 0^0 is undefined and calculated as 0/0.  I suppose the invalid line is "let b = 0" (following "given a=b" and "a^b = b^a") as it implies that you can put a=b=0 into the formula, which requires 0^0 to be defined.

Ixeus - I don't think my induction falacy ("all numbers are interesting") is an explicit breach of the induction axiom or any other axiom.  I think it is more a case of using mathematics beyond the scope where it is valid.  The induction axiom has various forms but they are all basically equivalent to
"if      F(1) is true
  and  F(n) implies F(n+1) for all positive integers n
  then F(n) is true for all positive integers n"

Ixeus: Is this true, these are false 'proofs'! So these people have been posting rubbish, this is terrible and against the EE members agreement.  I will complaining to Admin, I have been misled!

GfW

heck no =o  I wasn't saying that :p  I was probably saying it wrong, I wasn't critisizing, I was letting those know what exactly the problem with those proofs were, I had my own "false" proofs before, since we had a rally at school to see who could come up with the most undetectable -false- proofs.  (one of the hardest was the $30 - $5 one, though there was no "false" proof, just incorrect mathematics, but so snidely worded that noone could catch why it was wrong, I remember the hardest one being something about logs y = 1/x + (1+ln(sin(ln(cot(x/5))))) (not sure if that's the real problem, I'll have to find it some day


Sorry if I mis-worded it

Company A makes a $20M loss in 2003
Company A makes a $100M profit in 2004

Whats the % profit change ?

I've seen methods which would record this as
(a) a 600% improvement in profit (100--20)/-20 * (-1)                  
(b) a 500% profit 100/-20 *(-1)

Cheers

Dave

richardjb
- I don't have the mathematical background to back it up or explain it but I remember my teacher saying that the "^0" is the "overiding" operation for 0^0, ie 0^0 is defined as 1.  The use up to the current limit of my proof skills (calculator) has so far supported this.
Thus...
a^0 = 0^a
   and a=0
0^0 = 0^0
1 = 1
ie - TRUE

a^0 is 1 for all positive a and for all negative a, so I guess it would be nice to call it 1 for a = 0.

Not that I have any problem with isolated singularities (I guess you can have a singularity which is not isolated ?).
The integral of x^n (with respect to x) is pretty well behaved for all n other than -1.

0^a is zero for all positive a and infinite for all negative a, so I guess you could regard it as passing through zero when a = 0.  The problem I have with this is I can also imagine it passing through any other number as well - so I can't help thinking of it as undefined.

Is the "rule" that "0^0 = 1" inferred from other axia (axxioms ?) or is it an axiom in itself.  In any case I shall always think of it as "Seth's rule" :-)

If it is an axiom (not possible to deduce what 0^0 must be from the rest of mathematics) then congratulations to Seth for finding a fallacy so subtle that it requires an axiom to fix it.

richardjb,

Yes, that is the one I was referring to, and you have the right idea as to why it is a fallacy.

It comes down to what 0^0 is defined to be, if it's defined at all. Typically, when 0^0 is defined, it's defined to be 1 because z^0 = 1 for all complex z other than zero (and for other reasons). So it is natural to include zero, so that z^0 = 1 for all z.

On the other hand, 0^z is not well defined except for positive real z, where 0^z = 0. Taking zero to a negative power is essentially dividing by zero. Thus, there's little use to define 0^0 to be 0, so that is not typically done.

So the answer is that one of the two reductions (a^0=1 and 0^a=0) is incorrect, depending on the definition used for 0^0, if one exists. Regardless, at most one definition for 0^0 can be used within the context of a given proof, forcing at least one reduction to be incorrect by the pigeonhole principle.

-Seth

y =a^x is a suprisingly complicated function, the complexity of which, from I can see, is completely ignored in highschool maths and not always covered adequately at the college level.

There are some good proofs here, the one I was hoping to see because I have forgotton it and am too lazy to work out again is the one (I think there are 2 ) based on the power series expansion of ln(1+x).

Unless I get another flurry of posts I will shut this one down soon.

Thanks to all

GfW

This is a follow up to PeteLong's puzzle about the 3 customers and the waiter.

I think that I can explain what goes wrong with the logic.

First of all, note that everything up to the last couple of lines is perfectly correct. No tricks so far. :~)

The trick lies in the very last line:

> where did the other pound go??  30 - 1 =29

This line implies that there's a pound missing. This is just a bare faced lie that has nothing to do with the rest of the story. It may as well have said "so why is the moon made of cheese".

We get confused because we've just been made to think of the number 29. This is because the word "and" often implies addition. When we read the sentence "they paid 27 pounds and the waiter kept 2 pounds" we're tempted to think "27 + 2". This is reinforced by the equation that follows.

We should really be thinking of 25 at this point because the sentence reminds us that the waiter got 2 pounds out of the total payment of 27 pounds. In other words, 27 - 2 = 25.

The more I look at the story, the more sophisticated I believe it is. It's like a good magic trick; everything that the magician shows you is designed to make you think about the wrong thing.

regards,
Gremlin

I think I got this one from EE

The Paradox of Achilles and the Tortoise

Achilles runs a race with the tortoise. He runs 10 times as fast as the tortoise. The tortoise has a 100 yards' start.
Now, Achilles runs 100 yards and reaches the place where the tortoise started. Meanwhile the tortoise has gone a tenth as far as Achilles, and therefore is 10 yards ahead of Achilles.
Achilles runs this 10 yards. Meanwhile the tortoise has run a tenth as far as Achilles, and therefore is 1 yard infront of Achilles.
Achilles runs this 1 yard, meanwhile the tortoise has run a tenth as far as Achilles, and is therefore a 10th of a yard infront of Achilles.
Achilles runs this 10th of a yard, meanwhile the tortoise has run a tenth as far as Achilles, 100th of a yard.
Achilles runs this 100th of a yard. Meanwhile the tortoise has run 1000th of a yard.
Continuing like this Achilles will never catch up with the tortoise but will get closer and closer to him.
Now, we know that Achilles will overtake the tortoise, but this statment proves otherwise.

LOL

Now you know a thread is too long when posts get repeated verbatim from earlier in the same thread yet claim to be new to it...

CaptainCyril, see the above comment in this thread:

Comment from thebigroll
Date: 09/12/2003 08:23PM EDT

and my discussion immediately following.

-Seth

LOL

Yes, Seth. I agree.

Blame it on the new look :). Man! My eyes hurt.

I had it saved and then pasted it here not knowing it was posted. :)

Quick answer please:

  if 1½ chicken can lay 1½ eggs in 1½ days, how many eggs do 1 chicken lay in 1 day ?

regards JakobA

Dear JakobA,

English mistake!

You should ask "how many eggs does 1 chicken lay" and not "do 1 chicken lay".

regards,

Cyril

Zero or 1. It changes from day to day. :~)

Of course, if they're mathematical chickens and not real ones than they lay fractional eggs.

if 1.5 can lay 1.5 in 1.5, the rate is    1.5 eggs / 1.5 days    which equals    1 egg / 1 day

so 1.5 can lay 1 in 1

and 1 can lay X in 1

therefore:

1.5 / (1/1) = 1 / (X/1)
1.5 = 1/X
3/2 = 1/X
2/3 = X/1

1 can lay 2/3 in 1

or something like that... yay

I know this is an old thread, but I have one.


There is a infinite line in space.

On the SAME Plane, there is point, where another infinite line intersects the point AND intersects the other line perpendicular.


______________________|_______________________
                                       |
                                       |
                                       !
                                       |
                                       |

NOW, the line that intersects the point begins to rotate, making the intersection of the two lines further from the point.
Since the line in rotating at the point, at some 'time' the two line must become parallel.  

How far from the point will the intersection of the two lines be just before they are parallel.

                                     

Fun -> find dx/dy of
               

  [ln(ln(sin^2(x+1)))]                (sin sqr)
 x

Where [ ] represents the greatest integer function

This is joke actually :)

Salary theorem states that " Engineers , Doctors , Teachers can never earn as much as business executives and sales people."

This theorem can now be supported by a mathematical equation based on the following two postulates.

Postulate

1: Knowledge is Power

2: Time is money

As every student knows that

Power = Work\Time

Since knowledge = Power

Time = Money

Solving for money we get

Knowledge = Work/money

Money = Work/Knowledge

Thus as knowledge approaches zero, Money approaches infinity, regardless of the amount of my work done.

CONCLUSION: The less you know the more you make.


Programmer0 :)

Here are the ones I know that have not been mentioned:

           x*x   =          x + x + ... + x + x       (addition x times)

Then take the derivative with respect to x:

 d/dx( x*x )  = d/dx( x + x + ... + x + x )    (addition x times)
 d/dx( x^2 ) = d/dx(x) + d/dx(x) + ... + d/dx(x) + d/dx(x)
          2x      = 1 + 1 + ... + 1 + 1               (addition x times)

At this point 1 added to itself x times is 1*x:

          2x      = x

Divide both sides by x:

          2        = x/x
          2        = 1



And then there is this one:

   -1 = (-1)^3 = (-1)^(6*0.5) = ((-1)^6)^(0.5) = (1)^(0.5) = 1



And then this one:

  e^(ix) = cos(x) + i sin(x)

  where x=2Pi

  e^(i2Pi) = cos(2Pi) + i sin(2Pi)

  e^(i2Pi) = 1 + 0

  e^(i2Pi) = 1

Now take the natural log of both sides:

  ln(e^(i2Pi)) = ln(1)

  i2Pi = 0

Then divide both sides by i2Pi:

  (i2Pi)/(i2Pi) = 0/(i2Pi)

  1 = 0



Or this one:

  e^(ix) = cos(x) + i sin(x)

  where x=2Pi

  e^(i2Pi) = cos(2Pi) + i sin(2Pi)

  e^(i2Pi) = 1 + 0

  e^(i2Pi) = 1

Now raise both sides to the power of i:

  (e^(i2Pi))^i = 1^i

  e^(i2Pi*i) = 1

  e^(-1*2Pi) = 1

Now we can take the log without it being a complex number:

  ln(e^(-1*2Pi)) = ln(1)

  -2Pi = 0

Divide both sides by -2Pi

  (-2Pi)/(-2Pi) = 0/(-2Pi)

  1 = 0



Check these sites for more:

http://www.math.toronto.edu/mathnet/falseProofs/
http://mathforum.org/dr.math/faq/faq.false.proof.html
http://mathworld.wolfram.com/Fallacy.html

And a couple from the Math Forum archives some that have
commentary that may have already been talked about here:

http://mathforum.org/library/drmath/view/53907.html
http://mathforum.org/library/drmath/view/57924.html
http://mathforum.org/library/drmath/view/53764.html
http://mathforum.org/library/drmath/view/57110.html
http://mathforum.org/library/drmath/view/57116.html

You can't divide by 0 in Math!
                       
                          0/12 = 0
but you can't do 12/0

Sure you can.  The answer is an indeterminate value but you can do the division.  Otherwise, explain L'Hopital's Rule.

Just because you can't say what the result will be doesn't mean the operation itself is invalid.  Assuming that the result makes sense is where these proofs fall down.

For example, start with this:

        (a-b)         b              b
        ____  +  ____   =  _____
        (a-b)      (a-b)         (a-b)

Now, the next step should be the substitution of (a-b)/(a-b) = 1 which does make sense if (a-b) is not zero.  The invalid step is assuming that you know the result of (a-b)/(a-b) is always 1 which isn't necessarily true when (a-b) equals zero.  It might be or might not - that's what indeterminate means...

Jerry

"Now, the next step should be the substitution of (a-b)/(a-b) = 1 which does make sense if (a-b) is not zero"
  (6-5)                 5                   5
 ______     +    _____     =      _____
  (6-5)               (6-5)              (6-5)

1   5    5
_ +_ = _
1   1    1

6/1 = 5/1
    False , it doesn't make sense. did you meant times instead of plus?

Yeah, I did.  Thanks for noticing.  I have a typing impediment - or is it a proof-reading deficiency?  Infallibility is not one of my better features...

Jerry

GfW, do you think you can help explain infinity to this person?  http:/Q_21266168.html

ozo, I have just noticed this and I have added a comment.

good fallacy discussion
http://nrich.maths.org/discus/messages/8577/7236.html?1071320334