Is (1/♾️)+(1/♾️), Indeterminate?

Maybe.

For (1/♾️)+(1/♾️)

Will equal,

(1*1+1*♾️+♾️ 1)/(♾️)

Which is (♾️/♾️). 

Which is indeed indeterminate.

While (0/1)+(0/1) is not intermediate. 

Due to: lim x→infinity [ 1/x ] = 0

So lim x→infinity [ 1/x + 1/x ] = lim x→infinity [ 1/x ] + lim x→infinity [ 1/x ] = 0+0 = 0

That mean: 1/infinity + 1/infinity = 0

Is not intermediate.

Nope Sir,

1/infinity is positive 

-1/infinity is negative

0 is both.

Can you disprove that 1/infinity is positive?

1/infinity = 0, so it is 0 + 0 which is 0 right lol

UGH after my ge9metry lesson I had a few years back I’ve hated the word prove and explain

Nope,

(1/infinity)^-1 equals infinity/1 And not (-1/infinity)

While (0^-1) equals 1/0 both which is  + and - infinity. 

(1/infinity)^-1 will prove that 1/infinity is not 0, other than as limit, and is unique from -infinity^-1, and 0.

Infinity is actually not a real number but it’s just a symbol showing that sequence is convergence to infinity. Your viewed it as the sum of two fraction and proof like this. It mean that you multiplied by infinity both the numerator and the denominator. Therefore an error occurred.

No Sir, no error with the arithmetic occurred.

And if infinity is not negative or positive and  not a number whatsoever.

Then 1/"infinity" would equal just an unchanged 1.

Wait. If 1/infinity = 0.000...005 × infinity

So 1/infinity + 1/infinity = (0.000...005 + 0.000...005) × infinity = 0.000...01 × infinity [Unclear, can not be concluded]

However, value of 0.000...005 × infinity is interminate. So I still don't understand why you said that 1/infinity is equal to 0.000...005 × infinity

I think 1/infinity = 0.000...005 (not 0.000...005 × infinity)

1/infinity + 1/infinity = (1+1)/infinity = 2/infinity ?

I think Infinity is definitely positive. but not a real number.

I agree, Sir.

your prove?

1/infinity + 1/infinity

= (1×infinity / infinity×infinity) + (infinity×1 / infinity×infinity)

= ((1×infinity)+(infinity×1)) /infinity²

= infinity / infinity

Because of if you use the algebraic identity: a/b+c/d = ad/bd + bc/bd = (ad+bc) / bd That means you multiply by infinity both numerator and denominator.

It does prove that the sum, (1/♾️)+(1/♾️) can only be derived by dividing infinity into infinity.

Looks like that's what caused an error like 2/3 = 2×infinity / 3×infinity = infinity/infinity

in the same way as 1/infinity = 1×infinity / infinity×infinity

Compare it to a method which uses only the distribution property.

1/infinity + 1/infinity = (1+1)/infinity = 2/infinity