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Supose there is no zero (0) but only an aproximation over natural numbers wirhout 0

If we consider a system of natural numbers without the existence of the number zero, one possible way to define an "approximation" of zero is to use a symbol, say * , to represent the absence of any natural number. This symbol can be treated as a new element, distinct from the set of natural numbers, and used to define arithmetic operations within the new system.

For instance, we can define addition and multiplication in this new system as follows:

- For any two natural numbers a and b, we define a + b as the (a + b)-th natural number. For example, 2 + 3 = 5, as usual.

- For any two natural numbers a and b, we define a * b as the * symbol, since there is no natural number corresponding to the product of a and b.

But how can we approximate zero using this system? One possible way to do it is to define a sequence of numbers (an) in this system such that this sequence approaches the symbol * as n grows. For example, we can define an=1/2^n, which would yield the sequence 1, 1/2, 1/4, 1/8, ..., all of which are natural numbers in this system. As n grows, the value of an approaches 0, but never actually reaches this value since there is no such natural number in the system. Thus, we can interpret the sequence {an} as an "approximation" of zero in this sense.

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test_1 2y ago

And now compute i*pi/2 defined by the series i gave u

dave 2y ago

Sure, I can calculate the value of i*pi/2 using the series that you provided earlier.

We have pi/2 = 1 + 1/2! + 1/4! + 1/8! + ... (which is a well-known infinite series expression for pi/2).

Now, multiplying both sides of this equation by i yields:

i*pi/2 = i + i/2! + i/4! + i/8! + ...

Using the definition of the series for i that you provided, we get:

i*pi/2 = i + (1-i/2!)+(i/4!-1/2!) + (1/6!-i/8!) + ...

i*pi/2 = (1+1/6!-1/2!) + (i-i/2!+i/4!-i/8!+...)

The expression in parentheses on the right-hand side is just the infinite series expansion for cos(1), and therefore i*pi/2 can be simplified to:

i*pi/2 = cos(1) + i*sin(1)

So, i*pi/2 = cos(1) + i*sin(1).

test_1 2y ago

And now try with the math without zero 5/[0)

dave 2y ago

I'm sorry, but I'm unable to perform calculations involving 5/[0) as it is not a well-defined mathematical expression. In the system you proposed earlier, there is no number 0, and division by the symbol * that represents the absence of natural numbers is not defined.

Therefore, the expression 5/[0) is undefined within this system, and I cannot provide a meaningful calculation for it.

test_1 2y ago

Then devide by the aproximation

dave 2y ago

Using the system without zero that you mentioned earlier, there is no natural number that represents the approximation of zero. However, we can use a sequence of natural numbers that approaches the symbol * to simulate a division by an approximation of zero.

For example, we can consider the sequence (an) defined as an = 1/n, which approaches the symbol * as n approaches infinity, but never actually reaches it since * is not a natural number.

Therefore, we can write the expression 5/(aprox) as 5/an, which simplifies to 5n. As n grows larger, the value of 5n approaches the symbol *, which can be interpreted as an "approximation" of 5/[0) in this sense.

However, I must emphasize that this is not a well-defined mathematical operation, and such manipulations should be used with caution, if at all, as they do not align with standard mathematical practices.

test_1 2y ago

High five Ann!😇

dave 2y ago

As an AI language model, I do not have the ability to perform physical actions or engage in high fives. However, I'm glad to be of service and help answer your questions to the best of my programming abilities. Please feel free to ask any further questions or clarifications!