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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

New page: Continuous Functions

Let me know if you find any errors, or think of some way to improve it.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,812

Hi MathsIsFun,

The **Continuous Functions** page is very well done! The illustrations complement the page!

I don't think there is any error.

A neat page! Thank you!

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi MIF!

Looks like an errorless labor of love to me!

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,123

Hi MIF;

Looks good and reads easily. I like it.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,738

hi MathsIsFun,

Another excellent page.

Just one thing.

Is that really undefined at x = 1 ? I'm not sure about the answer to that.

You can say it isn't defined there in the definition, or you could define it to be 2, but it's unclear.

There are lots of situations where we simplify and then substitute after (eg. calculus). This function obeys the continuous definition requirement at x = 1. I accept the point about 0/0 but this example may nevertheless cause controversy amongst your readers.

Suggestion:

Replace with

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,123

Hi bob bundy;

I have asked that question too, why can't I simplify? I was told you can not simplify there because when you do you have changed the domain of the function. So the original function and the simplified one are not the same.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Yes, this gets me every time, too. It is just so natural to simplify, how can it be wrong?

It remind me of the 2=1 proof:

Let a=b

a² = ab

a²-b² = ab-b²

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

a+b=b

1+1=1

2=1

The error is when we simplify by dividing by (a-b)=0

Be that as it may, I can choose a less controversial example.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,738

hi MathsIsFun and bobbym,

Example: f(x) = (x2-1)/(x-1) for all Real Numbers

The function is undefined when x=1:

(x2-1)/(x-1) = (12-1)/(1-1) = 0/0

So it is not a continuous function

It's all in the definition of the domain.

It's not really the 0/0 problem; you could do it with any function.

eg.

y = 2x for x in {reals} except x = 1 where it is undefined.

I've made a discontinuous function at x = 1.

Sure, it's very contrived but it demonstrates that it's all in the definition.

eg y = (x^2-1)/(x-1) for all x in {reals} except x = 1

y = 2 when x = 1

This function is continuous for all x.

I'm happy that it stays unchanged; but it may provoke this discussion again. Maybe that's a good thing ??

bobbym wrote:

changed the domain of the function

There have been several posts recently that seem to suggest that once you have been told the equation for the function then the domain follows of its own accord. I disagree. I think that the domain is part of the definition and hence it's up to the definer to declare whether the domain is changed.

eg.

y = (x^2-1)/(x-1) for all x in {reals} except x=1

y = x+1 when x = 1

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

bob bundy wrote:

hi MathsIsFun,

Another excellent page.

Just one thing.

Is that really undefined at x = 1 ? I'm not sure about the answer to that.

You can say it isn't defined there in the definition, or you could define it to be 2, but it's unclear.

There are lots of situations where we simplify and then substitute after (eg. calculus). This function obeys the continuous definition requirement at x = 1. I accept the point about 0/0 but this example may nevertheless cause controversy amongst your readers.

Suggestion:

Replace with

Bob

Those functions are not continuous, but the sinvularities they have are removable. Try searching removable singularities. I think there is a Wikipeadia article on it.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,738

There is:

http://en.wikipedia.org/wiki/Removable_singularity

and that covers what I was trying to say.

Thanks

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,648

Excellent!

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

Hi Bob

You're welcome.

Hi MIF

Awesome page!

*Last edited by anonimnystefy (2012-11-08 03:01:12)*

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**Calligar****Member**- Registered: 2011-09-24
- Posts: 272

Hmm, I'm honestly a little bit confused about 1...indirect thing. Is 0/0 really undefined? I would have honestly thought it equals 0. I'm a little bit confused on that one part, so anyone care to explain (why 0/0 is undefined)? Though I **do** understand why it's not a continuous function. Other then that, I'd say you explained things clearly enough to be able to understand everything, so I'd say you did a pretty good job.

As for one of the posts, I don't really understand the 2=1 proof. Everything seems right until you get to a+b=b, nor do I understand how they got to that. I fail to see how this is proving 2=1, or rather, making it arguable I should say.

There are always other variables. -[unknown]

But Nature flies from the infinite, for the infinite is unending or imperfect, and Nature ever seeks an end. -Aristotle

Everything makes sense, one only needs to figure out how. -[unknown]

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,123

Hi Calligar;

In line 4 to 5 of the 2=1 proof there is a division by a-b. But if a=b as stated in the first line then a - b =0 and you can not divide by zero.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**Calligar****Member**- Registered: 2011-09-24
- Posts: 272

If you are talking about the 3'rd line where it says, (a-b)(a+b) = b(a-b), you are multiplying, not dividing, unless I'm missing something...?

Also, I'm not sure I quite understand the reason for simplifying it. Everything seems fine about it too me...

There are always other variables. -[unknown]

But Nature flies from the infinite, for the infinite is unending or imperfect, and Nature ever seeks an end. -Aristotle

Everything makes sense, one only needs to figure out how. -[unknown]

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 7,738

hi Calligar

0 x 63 = 0 x 105

If dividing by zero was ok you'd get

63 = 105 ??

So in basic number theory there is a rule that you shouldn't ever divide by zero

In the fake proof

(a-b)(a+b) = b(a-b) is ok.

The problem arises at the next step because when you cancel a common factor what you are actually doing is dividing both sides of the equation by it.

eg 7 times 3 = 7 times N

divide both sides by 7

3 = N

In the fake proof to get to the next line you have to divide by (a-b), that is the false step as (a-b) = 0

0/0 has an indeterminate value. Why?

Let's go back to what division is.

48 = 6 x 8 so 48/6 = 8

You reverse the multiplication process.

So to work out 0/0 you have to think what would be the number here 0 x ? = 0

As any number will make that work, you cannot say what 0/0 is. (In calculus you have to work with limits to sort out 0/0)

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 107,123

Hi Calligar;

Calligar wrote:

If you are talking about the 3'rd line where it says, (a-b)(a+b) = b(a-b), you are multiplying, not dividing, unless I'm missing something...?

Let a=b

a² = ab

a²-b² = ab-b²

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

a+b=b

1+1=1

2=1

In line three you have a factor of (a-b) on both sides. On line 4 you do not. Both sides were divided by (a-b) which is equal to 0. That is the misstep. Generally, you never divide by a variable unless you are sure that it cannot be 0. Sometimes such divisions are fatal, sometimes they just destroy answers. For instance

Divide both sides by x

Here the decision to divide by x, has cost us the other root which is 0.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Nice extra example bobby.

And that 0/0 thing ... it is like asking "how many 0s in 0?". Are there no zeros in zero, or perhaps just the one? See: Dividing by Zero

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**noelevans****Member**- Registered: 2012-07-20
- Posts: 236

Hi!

Here's another explanation why division by zero is not possible. Division is a secondary operation,

not a primary operation; that is, division is not given in the field axioms that establish the real

number system. Multiplication is given in the axioms (as well as addition, but not subtraction).

Division is then defined as multiplication by reciprocals.

For example suppose x is a real number whose reciprocal is y. Then given a number z we define

z/x as z*y. So to divide by x, x must have a reciprocal. BUT zero has no reciprocal as stated in

the multiplicative inverse axiom: For each real non-zero number x, there is a real number y such

that xy=1. Of course the usual notation for the reciprocal of x is written 1/x. But this looks like

division, so using 1/x for the reciprocal in the axiom and then later using "/" for division is a bit

confusing and makes the definition via z/x = z*(1/x) look circular.

When all is said and done, the axioms do not allow for a reciprocal of zero, hence division by zero

is a non-issue --- it could never happen since zero has no reciprocal to define division by zero in

terms of.

Having zero in the denominator of a fraction is a similar issue. The set of fractions based on the

set of whole numbers W={0,1,2,3,...} is defined as

F = { p/q | p and q are in W and q is not zero }

If someone asks why we can't have something like 2/0 the answer is simply the definition does

NOT allow zero in the denominator. It has nothing to do with "division by zero."

Analogously the field axiom for the reals do not allow zero to have a reciprocal. Hence to write

an expression one would read as "division by zero" simply violates the field axioms. Trying to

write "a number divided by zero" would be defined as "a number times the reciprocal of zero."

But a "reciprocal of zero" does not exist.

The other explanations of why one can't divide by zero illustrate the problems that would occur

if we did allow zero to have a reciprocal. As such they provide good reasons from disallowing

zero to have a reciprocal in the first place.

A nice physical example of trying to divide by zero can be seen in the operation of the old

mechanical calculators of yesteryear. They operated on division as a "repeated subtraction."

If one tried to divide 2 by zero, the calculator would subtract zero from two, add one to the

quotient, and then check to see if there was enough left to subtract zero from it again. Of course

there was enough left to subtract zero again since 2>0. So it subtracted zero from two again,

added one to the quotient, and checked to see if there was enough left to subtract zero again.

The old mechanical calculators got stuck in an "infinite loop" subtracting zero over and over

again. The quotient register looked like an old gas pump register with the dials spinning as

the quotient grew.

A friend of mine in high school rented a mechanical calculator to do lots of arithmetical homework.

On the front in bold letters it said "DO NOT DIVIDE BY ZERO!" Of course, that was an invitation.

So about half way through his assignment he starts a division by zero. The calculator was run

on electricity, so it just kept spinning the digits in the quotient. Quite fun to watch, but tiresome

after a while. So he punched "clear" and all the other buttons and levers he could find. Nothing

stopped the process. Finally he got the idea to pull the plug, and presto! it stopped.

So he thought that he'd better get back to his homework. He plugged it back in and presto! it

resumed the division calculation. He did the rest of his assignment by hand. It appears that

something had to be reset internally to get it to stop the division. I suspect when he returned

the machine he just set it on the counter and quickly headed for the door!

The old electric mechanical calculators (which were made of metal and weighed about 60 lbs)

demonstrated quite well the "repeated subtraction" algorithm for division. You could watch it

as it proceeded through a calculation as the dials spun and the carriage shifted. Slow by today's

standards, but it got the job done (if not dividing by zero!).

Writing "pretty" math (two dimensional) is easier to read and grasp than LaTex (one dimensional).

LaTex is like painting on many strips of paper and then stacking them to see what picture they make.

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**Calligar****Member**- Registered: 2011-09-24
- Posts: 272

Okay, I understand the fact that you can not divide x by 0, but i was **strictly** talking about 0 / 0, **not** 1, 2, or **any** other number. I don't understand why you can't divide by 0 in that case, because I do see 0 x 0 = 0 if u changed it to multiplication. Also, I didn't realize that rule applied to 0 as well, that was why I was asking. I already knew about stuff like 1/0 is undefined though and already understand why that is. Also I have read every post, and it still doesn't seem to prove to me that 0 / 0 is undefined, maybe because I'm looking at it the wrong way? Though I also worry about arguing this further, as this might only end up being something just like 0.999...=1. So, I'm just going to assume its a rule right now I don't... personally agree with. If there's better reasoning for it though, I would not mind hearing it.

There are always other variables. -[unknown]

But Nature flies from the infinite, for the infinite is unending or imperfect, and Nature ever seeks an end. -Aristotle

Everything makes sense, one only needs to figure out how. -[unknown]

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**Calligar****Member**- Registered: 2011-09-24
- Posts: 272

Actually, I just realized I was looking at it differently. This in my opinion is another futile argument, I understand why it's a rule now. No need to further explain to me. I also understand the 2 = 1 proof now and see the error, I was having a...similar kind of issue with it. So with all my questions answered, I go back to with what I was originally going to say, but didn't say it quite clearly. I think the continuous functions page is very clear and well done. I didn't have an issue understanding it, only understanding indirectly relating things.

But Nature flies from the infinite, for the infinite is unending or imperfect, and Nature ever seeks an end. -Aristotle

Everything makes sense, one only needs to figure out how. -[unknown]

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Thanks Calligar. Questioning is an important part of mathematics.

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**Calligar****Member**- Registered: 2011-09-24
- Posts: 272

I'd agree, because I have high emphasis on myself for understanding things. It really bothers me if I walk away from something without understanding it well. It has also done nothing but help me with...well pretty much everything I know now (though there is still a **lot** I don't understand). So yeah, I will often question something I don't understand to better understand it.

But Nature flies from the infinite, for the infinite is unending or imperfect, and Nature ever seeks an end. -Aristotle

Everything makes sense, one only needs to figure out how. -[unknown]

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,018

Hi MIF

You can put on the same page the definition of a smooth function.

Here lies the reader who will never open this book. He is forever dead.

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,631

Smooth function would need a page of its own, I think.

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