Don H

2008-04-17 19:40:24 UTC

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1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.Raw Message

Is this axiomatic, or is there a proof?

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Discussion:

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Don H

2008-04-17 19:40:24 UTC

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1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.Raw Message

Is this axiomatic, or is there a proof?

==============================

Virgil

2008-04-17 21:29:09 UTC

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Raw Message

1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.

Is this axiomatic, or is there a proof?

==============================

P***@aol.com

2008-04-18 14:06:25 UTC

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Raw Message

1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.

Is this axiomatic, or is there a proof?

==============================

consistant and for convenience when writing infinite series etc.

Don H

2008-04-20 19:21:20 UTC

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Raw Message

1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.

Is this axiomatic, or is there a proof?

==============================

consistant and for convenience when writing infinite series etc.

What is a number n to the power x?

The x indicates how many n must be multiplied together.

Thus,

n^x = n by n (x times)

2^2 = (2, twice) = 4

2^1 = (2, once) = 2

2^0 = (2, no times) = 1 ?

- but if you multiply 2 by itself, no times, you've still 2 ?

Whereas, if you multiple 0 by 0, two times, you've still 0 ?

Virgil

2008-04-20 19:35:53 UTC

Permalink

Raw Message

1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.

Is this axiomatic, or is there a proof?

==============================

consistant and for convenience when writing infinite series etc.

What is a number n to the power x?

The x indicates how many n must be multiplied together.

Thus,

n^x = n by n (x times)

2^2 = (2, twice) = 4

2^1 = (2, once) = 2

2^0 = (2, no times) = 1 ?

- but if you multiply 2 by itself, no times, you've still 2 ?

Whereas, if you multiple 0 by 0, two times, you've still 0 ?

powers with positive integer exponents is:

For any real/complex number x, x^1 = x and x^(n+1) = x^n * x.

Don H

2008-04-21 17:56:43 UTC

Permalink

Raw Message

1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.

Is this axiomatic, or is there a proof?

==============================

consistant and for convenience when writing infinite series etc.

What is a number n to the power x?

The x indicates how many n must be multiplied together.

Thus,

n^x = n by n (x times)

2^2 = (2, twice) = 4

2^1 = (2, once) = 2

2^0 = (2, no times) = 1 ?

- but if you multiply 2 by itself, no times, you've still 2 ?

Whereas, if you multiple 0 by 0, two times, you've still 0 ?

For any real/complex number x, x^1 = x and x^(n+1) = x^n * x.

However, consider two rabbits, say, Rob and Ruby, living in a Platonic

relationship (2^0), who then decide to copulate, initially without result

(2^1), but finally generating two kids (Richard, Rachel) (2^2), who all then

continue to propagate, exponentially = 2 x 2 x 2 x 2...

Likewise, two non-entities, Zero and Zilch, who, no matter how many times

they try, produce nothing - or nothing evident to our eyes.

Does it depend on what you start out with, as to what the result will be?

Virgil

2008-04-21 19:03:55 UTC

Permalink

Raw Message

1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.

Is this axiomatic, or is there a proof?

==============================

consistant and for convenience when writing infinite series etc.

What is a number n to the power x?

The x indicates how many n must be multiplied together.

Thus,

n^x = n by n (x times)

2^2 = (2, twice) = 4

2^1 = (2, once) = 2

2^0 = (2, no times) = 1 ?

- but if you multiply 2 by itself, no times, you've still 2 ?

Whereas, if you multiple 0 by 0, two times, you've still 0 ?

For any real/complex number x, x^1 = x and x^(n+1) = x^n * x.

However, consider two rabbits, say, Rob and Ruby, living in a Platonic

relationship (2^0), who then decide to copulate, initially without result

(2^1), but finally generating two kids (Richard, Rachel) (2^2), who all then

continue to propagate, exponentially = 2 x 2 x 2 x 2...

Likewise, two non-entities, Zero and Zilch, who, no matter how many times

they try, produce nothing - or nothing evident to our eyes.

Does it depend on what you start out with, as to what the result will be?

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