Discussion:
zero exponent
Don H
2008-04-17 19:40:24 UTC
1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
Is this axiomatic, or is there a proof?
==============================
Virgil
2008-04-17 21:29:09 UTC
Post by Don H
1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
Is this axiomatic, or is there a proof?
==============================
It follows from the rule a^x/a^y = a^(x-y), for a^y =/= 0.
P***@aol.com
2008-04-18 14:06:25 UTC
Post by Don H
1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
Is this axiomatic, or is there a proof?
==============================
It isn't a math fact --- it's a definition --- made so as to be
consistant and for convenience when writing infinite series etc.
Don H
2008-04-20 19:21:20 UTC
Post by P***@aol.com
Post by Don H
1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
Is this axiomatic, or is there a proof?
==============================
It isn't a math fact --- it's a definition --- made so as to be
consistant and for convenience when writing infinite series etc.
# And so is an arbitrary definition?
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
Post by Don H
Post by P***@aol.com
Post by Don H
1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
Is this axiomatic, or is there a proof?
==============================
It isn't a math fact --- it's a definition --- made so as to be
consistant and for convenience when writing infinite series etc.
# And so is an arbitrary definition?
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 ?
The standard inductive (or, if you prefer, recursive) definition of
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
Post by Virgil
Post by Don H
Post by P***@aol.com
Post by Don H
1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
Is this axiomatic, or is there a proof?
==============================
It isn't a math fact --- it's a definition --- made so as to be
consistant and for convenience when writing infinite series etc.
# And so is an arbitrary definition?
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 ?
The standard inductive (or, if you prefer, recursive) definition of
For any real/complex number x, x^1 = x and x^(n+1) = x^n * x.
# Thanks.
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
Post by Don H
Post by Virgil
Post by Don H
Post by P***@aol.com
Post by Don H
1^0 = 1 ; 2^0 = 1; 3^0 = 1 ; etc.
Is this axiomatic, or is there a proof?
==============================
It isn't a math fact --- it's a definition --- made so as to be
consistant and for convenience when writing infinite series etc.
# And so is an arbitrary definition?
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 ?
The standard inductive (or, if you prefer, recursive) definition of
For any real/complex number x, x^1 = x and x^(n+1) = x^n * x.
# Thanks.
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?
Of course! Otherwise 2^2 would be the same as 3^2.