Universe can't be infinite.

[TimeSeeker]

For example:

lim x->∞ (1/x) = lim x->∞ ( 2/x)

1/x <> 2/x

WELL DUH!!!

That is called a CONTRADICTION.

[devans99]

For example:

lim x->∞ (1/x) = lim x->∞ ( 2/x)

1/x <> 2/x

WELL DUH!!!

That is called a CONTRADICTION.

Exactly, so what you did in your 'proof' was not valid. Just because two things are equal when x=∞ does not mean they are equal for x=other values

[TimeSeeker]

Exactly, so what you did in your 'proof' was not valid. Just because two things are equal when x=∞ does not mean they are equal for x=other values

So WHY was it NOT VALID?

Point out the EXACT ERROR IN REASONING. Highlight it.

IF lim x->∞ (1/x) = lim x->∞ ( 2/x)
FROM THE LIMIT LAWS IT FOLLOWS
∴ 1/x = 2/x
∴ 1 = 2

Where is the error?

[TimeSeeker]

If A = B and A = C
∴ B = C (Transitive property)

IF f(x) = c and f(y) = c
∴ f(x) = f(y)
∴ x = y

Where is the error?

[devans99]

IF lim x->∞ (1/x) = lim x->∞ ( 2/x)
FROM THE LIMIT LAWS IT FOLLOWS - *** NO IN DOES NOT FOLLOW ***
∴ 1/x = 2/x
∴ 1 = 2

lim x->∞ (1/x) = lim x->∞ ( 2/x)
DOES NOT IMPLY
1/x = 2/x

All sorts of expressions for example evaluate to 0 in the limit:

1/x, 2/x, 1/x^2

But

1/x <> 2/x <> 1/x^2

[TimeSeeker]

IF lim x->∞ (1/x) = lim x->∞ ( 2/x)
FROM THE LIMIT LAWS IT FOLLOWS - *** NO IN DOES NOT FOLLOW ***
∴ 1/x = 2/x
∴ 1 = 2

lim x->∞ (1/x) = lim x->∞ ( 2/x)
DOES NOT IMPLY
1/x = 2/x

YES IT DOES!!!!

If A = C and B = C
∴ A = B (Transitive property)

IF f(x) = C and f(y) = C
∴ f(x) = f(y)
∴ x = y

Where is the error?

https://en.wikipedia.org/wiki/Isomorphism

[devans99]

IF f(x) = c and f(y) = c
∴ f(x) = f(y)
No not in general. You have evaluated f(x) and f(y) for specific values of y and x. That does not mean they are equal for all values of x and y.
∴ x = y

[TimeSeeker]

No not in general. You have evaluated f(x) and f(y) for specific values of y and x. That does not mean they are equal for all values of x and y.

OK, lets examine the particular then:

f(x) = 1/x
g(x) = 2/x

Can you think of any x (for x in the set of real numbers) where f(x) or g(x) cannot be evaluated?

The only problem is 0, right? And since we are dealing with lim x->∞ and ∞ != 0

Then I guess it's a valid operation?

f(x) = c
f(y) = c
From (1)
∴ f(x) = f(y)
∴ x = y

[devans99]

No not in general. You have evaluated f(x) and f(y) for specific values of y and x. That does not mean they are equal for all values of x and y.

OK, lets examine the particular then:

f(x) = 1/x
g(x) = 2/x

Can you think of any x (for X in the set of real numbers) where f(x) or g(x) cannot be evaluated?

The only problem is 0, right? So I guess it is true in general, except where x=0 ?

f(x) <> g(x) for all x apart from infinity. It undefined at 0.

So you can only write f(x)=g(x) when x=infinity.

[TimeSeeker]

So you can only write f(x)=g(x) when x=infinity.

I see.

OK. I won't write that.

What I am going to write then is this.

lim x->∞ (1/x) <> lim x->∞ (2/x)

Would you say this is true in general except at x = infinity, but since x never gets to infinity then it's OK ?

[devans99]

So you can only write f(x)=g(x) when x=infinity.

I see.

OK. I won't write that.

What I am going to write then is this.

lim x->∞ (1/x) <> lim x->∞ (2/x)

Would you say this is true in general except at x = infinity, but since x never gets to infinity then it's OK ?

1/x <> 2/x
is true for all x except infinity and 0. So we can write:

lim x->∞ (1/x) ~= lim x->∞ (2/x)

[TimeSeeker]

lim x->∞ (1/x) ~= lim x->∞ (2/x)

This is no good! It violates transitivity!!!

lim x->∞ (1/x) = ~0
lim x->∞ (2/x) = ~0
lim x->∞ (1/x) ~= lim x->∞ (2/x)

~0 ~= ~0

CONTRADICTION!!!!!!

[devans99]

~= means approximately equal. We usually write:

lim x->∞ 1/x = 0

But x tends to but never actually reaches infinity, so the right side never actually reaches zero, so IMO its more correct to write:

lim x->∞ 1/x ~= 0

So whenever a limit is evaluated, it’s correct to use the approximately equals sign (~=) rather than equals.

[TimeSeeker]

lim x->∞ 1/x ~= 0

OK fine.

lim x->∞ (1/x) ~= 0
lim x->∞ (2/x) ~= 0

is this correct ?

[devans99]

Yes.