Any maths teachers on this forum?

Posted by: mista h on 29 April 2014

Posted on: 11 June 2014 by TomK
A couple of years after graduating I had an interview with IBM where I had to take an IQ test which included a question like this. Afterwards during the actual interview I was asked about the test and mentioned that, as a mathematician it was impossible for me to predict the behaviour of a potentially infinite sequence from the first six or so terms. That didn't go down well at all. I didn't get the job and I was told later that at that time at least IBM didn't like rogues and this was seen as rogue behaviour. Or maybe I just fekked it up .
 
Originally Posted by Don Atkinson:

I recal at university, one lecturer asking us for the next number 1, 2, 4, 8, 16, 32.................

 

Most of us opted for 64.

 

"Nope !" said the lecturer. "Each number in the sequence is simply "numerically bigger" than the preceding one, so any number 33 or larger will do !"

 

Not everything is logical or "bleeding obvious" - says a lot about this forum.......................

 

 

 

Posted on: 12 June 2014 by Big Bill
Originally Posted by Kevin Richardson:

If 0 does not exist then

x-x = y -->

x+(x-x) = x + y -->

x = x + y and x - y = x -->

x(x-y) = (x+y)(x-y) -->

x^2 - xy = x^2 - y^2 -->

y^2 = xy ->

y = x  ->

x = x +x ->

x = 2x ->

1= 2 an impossibility as Russel proved

Nice but is x - x = y a valid equation?  Well yes it is but only ever when y = 0, so x = x + y and x - y = x become x = x & err x = x.  Not a very useful equation and not only that x = 2x is NOT an impossibility when x = 0.

 

Note also x = 2x --> 1 = 2 becomes false too when x=0, it implies that 0=0!

 

So the equation is only valid when y=0 and is not impossible when x=0.  So as long as x=0 we can say that zero definitely does not exist.  When x <> 0 then we can say zero definitely does exist.  Or is it the other way around?

 

This is the Big Bill paradox!

 

Is this another teapot?

Posted on: 12 June 2014 by Big Bill
Originally Posted by TomK:
A couple of years after graduating I had an interview with IBM where I had to take an IQ test which included a question like this. Afterwards during the actual interview I was asked about the test and mentioned that, as a mathematician it was impossible for me to predict the behaviour of a potentially infinite sequence from the first six or so terms. That didn't go down well at all. I didn't get the job and I was told later that at that time at least IBM didn't like rogues and this was seen as rogue behaviour. Or maybe I just fekked it up .
 
Originally Posted by Don Atkinson:

I recal at university, one lecturer asking us for the next number 1, 2, 4, 8, 16, 32.................

 

Most of us opted for 64.

 

"Nope !" said the lecturer. "Each number in the sequence is simply "numerically bigger" than the preceding one, so any number 33 or larger will do !"

 

Not everything is logical or "bleeding obvious" - says a lot about this forum.......................

 

 

 

Quite correct IBM was (is, or is it 100% public now?) run by Mormon's and they would definitely have not liked your response I think.  They were a stuffy lot and they all used to wear the same suits.  I can remember going to their execs canteen in one of the big buildings they used to have (Basingstoke I think) and having a really good meal but having the point made quite forcibly to me that I couldn't have any wine!!!!!   I don't drink wine at lunchtime working or not but it was the way the point was made - very weird.

Talk to youngsters today and they don't understand how dominant IBM were back in the day.  They thought they could tell the market what to do and the market would follow.  The number of times IT managers would say a computer can be anything but has to be blue - or something similar.

I believe their relative downfall was when they tried to force everyone to use the MCA architecture which of course was licensed to them.  Ahh the famous Bus wars, which suddenly came to an end when Intel suddenly (it seemed at the time) released motherboards and useful cards using the PCI architecture.

In the future our children will ask what did we do in the Bus Wars?

 

Not quite sure how big they are now, I know they branched out successfully into areas like Point-of-Sale.  I guess I should look at their current Market Cap.

Posted on: 12 June 2014 by mista h

Something to keep you amused when its half time at the football tonite !!

Posted on: 12 June 2014 by Kevin Richardson
Originally Posted by Big Bill:
Originally Posted by Kevin Richardson:

If 0 does not exist then

x-x = y -->

x+(x-x) = x + y -->

x = x + y and x - y = x -->

x(x-y) = (x+y)(x-y) -->

x^2 - xy = x^2 - y^2 -->

y^2 = xy ->

y = x  ->

x = x +x ->

x = 2x ->

1= 2 an impossibility as Russel proved

Nice but is x - x = y a valid equation?  Well yes it is but only ever when y = 0, so x = x + y and x - y = x become x = x & err x = x.  Not a very useful equation and not only that x = 2x is NOT an impossibility when x = 0.

 

Note also x = 2x --> 1 = 2 becomes false too when x=0, it implies that 0=0!

 

So the equation is only valid when y=0 and is not impossible when x=0.  So as long as x=0 we can say that zero definitely does not exist.  When x <> 0 then we can say zero definitely does exist.  Or is it the other way around?

 

This is the Big Bill paradox!

 

Is this another teapot?

But that was the point of the exercise.  Proving 0 exists by showing the contradiction produced in its absence.  We could say w=x , x - w = y... if 0 does not exist, then y != 0 [and of course x !=0]

Posted on: 12 June 2014 by Big Bill
Originally Posted by Kevin Richardson:
Originally Posted by Big Bill:
Originally Posted by Kevin Richardson:

If 0 does not exist then

x-x = y -->

x+(x-x) = x + y -->

x = x + y and x - y = x -->

x(x-y) = (x+y)(x-y) -->

x^2 - xy = x^2 - y^2 -->

y^2 = xy ->

y = x  ->

x = x +x ->

x = 2x ->

1= 2 an impossibility as Russel proved

Nice but is x - x = y a valid equation?  Well yes it is but only ever when y = 0, so x = x + y and x - y = x become x = x & err x = x.  Not a very useful equation and not only that x = 2x is NOT an impossibility when x = 0.

 

Note also x = 2x --> 1 = 2 becomes false too when x=0, it implies that 0=0!

 

So the equation is only valid when y=0 and is not impossible when x=0.  So as long as x=0 we can say that zero definitely does not exist.  When x <> 0 then we can say zero definitely does exist.  Or is it the other way around?

 

This is the Big Bill paradox!

 

Is this another teapot?

But that was the point of the exercise.  Proving 0 exists by showing the contradiction produced in its absence.  We could say w=x , x - w = y... if 0 does not exist, then y != 0 [and of course x !=0]

But you might as well say y = 0 and then expanding 0 to x - x we get y = x - x.  You are starting off with what you want to prove, that is my point.

Posted on: 12 June 2014 by Don Atkinson

Bill, Kevin,

 

Most of us on this forum argue about something.

 

You two(*) are argueing about nothing.

 

(*) and yes, I am well aware that in this context, "two" doesn't exist either................

Posted on: 12 June 2014 by Jasonf

As the great British painter, Francis Bacon once said in an interview with Melvin Bragg,

 

"I am optimistic...about nothing"

 

One of my very favourite quotes.

 

Jason.

Posted on: 12 June 2014 by winkyincanada
Originally Posted by Big Bill:
Originally Posted by Kevin Richardson:
Originally Posted by Big Bill:
Originally Posted by Kevin Richardson:

If 0 does not exist then

x-x = y -->

x+(x-x) = x + y -->

x = x + y and x - y = x -->

x(x-y) = (x+y)(x-y) -->

x^2 - xy = x^2 - y^2 -->

y^2 = xy ->

y = x  ->

x = x +x ->

x = 2x ->

1= 2 an impossibility as Russel proved

Nice but is x - x = y a valid equation?  Well yes it is but only ever when y = 0, so x = x + y and x - y = x become x = x & err x = x.  Not a very useful equation and not only that x = 2x is NOT an impossibility when x = 0.

 

Note also x = 2x --> 1 = 2 becomes false too when x=0, it implies that 0=0!

 

So the equation is only valid when y=0 and is not impossible when x=0.  So as long as x=0 we can say that zero definitely does not exist.  When x <> 0 then we can say zero definitely does exist.  Or is it the other way around?

 

This is the Big Bill paradox!

 

Is this another teapot?

But that was the point of the exercise.  Proving 0 exists by showing the contradiction produced in its absence.  We could say w=x , x - w = y... if 0 does not exist, then y != 0 [and of course x !=0]

But you might as well say y = 0 and then expanding 0 to x - x we get y = x - x.  You are starting off with what you want to prove, that is my point.

You two are in furious agreement. But one of you doesn't know it.

Posted on: 12 June 2014 by Don Atkinson
Originally Posted by Big Bill:
 

 

 

 

Quite correct IBM was (is, or is it 100% public now?) run by Mormon's and they would definitely have not liked your response I think.  They were a stuffy lot and they all used to wear the same suits. 

oooppps...................I didn't see the second "m" first time around............still seemed to make sense somehow !!

Posted on: 12 June 2014 by Big Bill
Originally Posted by Don Atkinson:
Originally Posted by Big Bill:
 

 

 

 

Quite correct IBM was (is, or is it 100% public now?) run by Mormon's and they would definitely have not liked your response I think.  They were a stuffy lot and they all used to wear the same suits. 

oooppps...................I didn't see the second "m" first time around............still seemed to make sense somehow !!

We re not arguing!

 

I was playing Devil's advocate, not for one moment do I believe that 0 or 1 does not exist.  But if you are trying to prove something then your opening premise cannot be the very thing you are trying to prove!  And I am unanimous on that!

 

What I was hoping was that someone would bring up our decimal number system (or even hexadecimal, binary or base 111) in which zero is absolutely essential and without it we would still be trying to do maths with the Roma system.

 

Because, unlike t'Romans our decimal system is positional.  We can write 1024 but that is not the same as 124 is it.  You see no zero no decimal system.

 

Now can anyone come up with something else that relies on the number zero?

Posted on: 12 June 2014 by Big Bill
Originally Posted by Char Wallah:

Bill, Kevin,

 

if your wife left you alone for one hour, and you had the choice between two porn videos, each with the running time of one hour, you would have watched neither by the time your wife returned.

That sounds like Schrodinger's Cat or should I say Schrodinger's Pussy?  In fact can I say Pussy on this forum?

Posted on: 12 June 2014 by Big Bill

"Now can anyone come up with something else that relies on the number zero?"

 

Apart from Millwall's average score last season.

Posted on: 12 June 2014 by sjbabbey
Originally Posted by Char Wallah:

If you undid the knot in your bellybutton; would your bottom fall off?

Maybe not but if wasn't for your bottom your belly would burst.

Posted on: 13 June 2014 by Big Bill
Originally Posted by Wat:

I think i have proved my conjecture

 

Conjecture: All people are stupid

 

Let us assume any set of n people is stupid

 

Now consider the set of n+1 people.
This set is the union of the sets 1 ... n and 2 ... n+1
Now as we assume any set of n people is stupid and 

both the sets 1 ... n and 2 ... n+1 have n members

then if our assumption is true we have the union of two sets of stupid people.
Hence the union is another set of stupid people.

 

Recapping if any set of n people is stupid then we have proved
the set of n+1 is stupid.

 

Now consider n=1, well I am stupid to post this nonsense
So n=1 is undisputedly true.

 

But if it is true for the set of n=1, we have shown it is true for the set of n+1=2
And if it is true for n=2, we have shown it is true for n+1=3
And so on by induction it must be true for all set sizes of n up to infinity

 

Thus the infinite set of all people is stupid


Or in other word all people are stupid

 

QED.

Wat I think in writing that post you have certainly proved if the set contains 1 certain member your assumption is true!

 

Not my member you understand!

Posted on: 13 June 2014 by Big Bill
Originally Posted by Char Wallah:

Were you watching the world cup last night? Mmm?

Nah, with all the crapping on that's been on the TV and Radio recently I am fed up with it already, but I guess I will start watching later on.  My old school never played foottie just Rugby and Cricket and they are my greatest loves sportwise still.

Posted on: 13 June 2014 by Big Bill
Originally Posted by Char Wallah:

Never mind, the football season starts again in two months.

I'm a Millwall supporter and, therefore, by definition know nothing about football anyway.  We narrowly escaped last season and let's hope we do a bit better next season.

Posted on: 13 June 2014 by Big Bill
Originally Posted by Char Wallah:

Pardon. Is this another quantum physics paradox? 

It might very well turn out to be one.

 

C'mon you Lions!

Posted on: 13 June 2014 by Big Bill
Originally Posted by Wat:
Originally Posted by Big Bill:
Originally Posted by Char Wallah:

Never mind, the football season starts again in two months.

I'm a Millwall supporter and, therefore, by definition know nothing about football anyway. 


What not even the late Harry Cripps?

You had a good team with Keith Weller, Derrick Posse, Barry Bridges & Gordon Bolland .... 

Our 'arry, what a great man he was, he did so much for youth football before he sadly dies at such a young age.  The man is a legend!

I can remember when Posse and Weller came from Tottenham to Millwall, they were a brilliant buy.  I can remember being over The Old Den for a match at season's end (I think it was Hull City) and Keith Weller hit a ball so hard that hit the post and said goalposts were vibrating for a few seconds.  The goalie turned around to the fans shaking his head and laughing.