Is the Is definition of a ‘point’ in geometry contradictory? - Page 3

Micrm@ss · September 25th, 2018, 11:30 AM

Quote:

Originally Posted by Devans99 $View Post$

Before working out new axioms, we need a clear definition of a point:

‘A point is a one dimensional object with an infinitesimal but non-zero length.’

Where an infinitesimal is a quantity approaching but never reaching 0.

I hope with a definition like this most of the existing axioms are still ok?

You seriously are going to attack the use of "infinity" in mathematics, and then go on define a point as something of INFINITESIMAL length?

Are you serious right now? Really?

Micrm@ss · September 25th, 2018, 11:33 AM

Quote:

Originally Posted by Micrm@ss $View Post$

You seriously are going to attack the use of "infinity" in mathematics, and then go on define a point as something of INFINITESIMAL length?

Are you serious right now? Really?

Because obviously anybody will see the problem with this, except the OP, let me ask him this:

Take a point according to your definition. It has an infinitesimal radius d. So what is 1/d? What does that equal according to you?

Devans99 · September 25th, 2018, 11:42 AM

1/d where d is an infinitesimal is an expression that tends to but never actually reaches infinity. So at best we could write 1/d ~ oo (because actual infinity does not exist).

Micrm@ss · September 25th, 2018, 12:12 PM

Quote:

Originally Posted by Devans99 $View Post$

1/d where d is an infinitesimal is an expression that tends to but never actually reaches infinity. So at best we could write 1/d ~ oo (because actual infinity does not exist).

OK.

So in classical geometry that everybody is doing, we have circles of a definite radius, say 2 or 3 or 10.

In your geometry, you say that al of that is very confusing, and you instead replace it with circles whose radius aren't like numbers anymore, but expressions that tend to infinity but not reach it?

How are you going to draw such a circle with a 'radius that tends to infinity and not reach it'? How are you going to show a picture of it in textbooks? Say what you want about our flawed math system, but when we talk about a circle with radius 1, we can actually DRAW it.

Micrm@ss · September 25th, 2018, 12:14 PM

Also, let y = 1/d be your number that tends to infinty but not reaches it.

What is y+1? Is y+1>y? Is y+1=y?

cjem · September 25th, 2018, 12:14 PM

Quote:

Originally Posted by Devans99 $View Post$

actual infinity

Uh oh, looks like you forgot to use the proper capitalization here. Time to say a hundred hail Marys.

Micrm@ss · September 25th, 2018, 12:22 PM

Here is a fun one:

If x is a real number such that its integer part is even, then we set f(x) = 0.
If x is a real number such that its integer part is odd, we set f(x) = 1.

Clearly, we have like f(3.14)=1, since the integer part is 3, which is odd.
Or we have like f(6.4) = 0 since the integer part is 6, which is even.

So, clearly the integer part of d is 0, so f(d) = 0.
What is f(1/d)?

Or what is sin(1/d)? Is it positive or negative? 0?

Devans99 · September 25th, 2018, 12:34 PM

Quote:

Originally Posted by Micrm@ss $View Post$

Here is a fun one:

If x is a real number such that its integer part is even, then we set f(x) = 0.
If x is a real number such that its integer part is odd, we set f(x) = 1.

Clearly, we have like f(3.14)=1, since the integer part is 3, which is odd.
Or we have like f(6.4) = 0 since the integer part is 6, which is even.

So, clearly the integer part of d is 0, so f(d) = 0.
What is f(1/d)?

Or what is sin(1/d)? Is it positive or negative? 0?

f(1/d) is dependant on the actual value of d which is used for the calculation.

Sin(1/d) is also likewise dependant.

Micrm@ss · September 25th, 2018, 12:40 PM

Quote:

Originally Posted by Devans99 $View Post$

f(1/d) = 0 always since zero is even.

Huh? What kind of reasoning is this?

f(1/2) = 0 since 2 is even?
f(1/3) = 1 since 3 is odd?
I don't think you understood this well....

Micrm@ss · September 25th, 2018, 12:41 PM

Anyway is 1/d bigger than every number?

Is (1/d) + 1 > (1/d)?

September 25th, 2018, 11:42 AM	#23
Devans99 Senior Member Joined: Jun 2018 From: UK Posts: 103 Thanks: 1	1/d where d is an infinitesimal is an expression that tends to but never actually reaches infinity. So at best we could write 1/d ~ oo (because actual infinity does not exist).
$Devans99 is offline$

September 25th, 2018, 12:14 PM	#25
Micrm@ss Senior Member Joined: Oct 2009 Posts: 544 Thanks: 174	Also, let y = 1/d be your number that tends to infinty but not reaches it. What is y+1? Is y+1>y? Is y+1=y?
$Micrm@ss is offline$

September 25th, 2018, 12:22 PM	#27
Micrm@ss Senior Member Joined: Oct 2009 Posts: 544 Thanks: 174	Here is a fun one: If x is a real number such that its integer part is even, then we set f(x) = 0. If x is a real number such that its integer part is odd, we set f(x) = 1. Clearly, we have like f(3.14)=1, since the integer part is 3, which is odd. Or we have like f(6.4) = 0 since the integer part is 6, which is even. So, clearly the integer part of d is 0, so f(d) = 0. What is f(1/d)? Or what is sin(1/d)? Is it positive or negative? 0?
$Micrm@ss is offline$

September 25th, 2018, 12:41 PM	#30
Micrm@ss Senior Member Joined: Oct 2009 Posts: 544 Thanks: 174	Anyway is 1/d bigger than every number? Is (1/d) + 1 > (1/d)?
$Micrm@ss is offline$

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