High School Proving a^0=1: Step-by-Step Guide

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The discussion centers on proving that a^0 equals 1, with various participants exploring definitions and mathematical properties. One argument suggests defining a^n as the product of a multiplied n times, leading to the conclusion that a^0 must equal 1 to maintain consistency in exponent rules. Others point out that this approach is more of a motivation for a definition rather than a formal proof, emphasizing that a^0 = 1 is a necessary definition for the laws of exponents to hold true for all integers. Additionally, the conversation touches on the importance of defining mathematical terms clearly before engaging in proofs. Ultimately, the consensus is that defining a^0 = 1 is logical and consistent within the framework of exponentiation.
  • #91
I always thought that a point is nil-dimensional.

Perhaps I didn't convey the meaning properly that a power is dimensionality... ( i hope you don't take all this so seriously, eh? )
 
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  • #92
PeroK said:
If we have a cube of side 2, it's volume is 8, Or, if we have a cube of bricks, with each side 2 bricks long, we have 8 bricks.

For a square of side 2, it's area is 4. Or, a square of bricks needs 4 bricks.

For a line of length 2, it's length is 2. Or, a line of bricks needs 2 bricks.

For a point, it's size is 0. Or, if we have a zero-dimensional array of bricks, then there are no bricks.

By this intuitive reasoning, we should have ##2^0 = 0##.

Very, very wrong.

2^4 is the Figurate Number represented by the vertices (points) of a hypercube ( Dimension 4 }

2^3 is the Figurate Number represented by the vertices (points) of a cube ( Dimension 3 )

2^2 ................ of a square.

2^1 ........... by the end points of a line. ( Dimension 1 )

And 2^0 is just the point or the atom itself of Nil-Dimension.
 
  • #93
rada you are quite right that we need the condition f(0) ≠ 0, and this is guaranteed by my assumption that f(x+y) = f(x).f(y) for all x,y, and f(n) = a^n for n = a positive integer and a > 0, since then f(1) = a ≠ 0, and thus f(x) ≠ 0 for all x. (if f(x) = 0, for some x, then for all y, f(y-x + x) = f(y-x).f(x) = f(y-x).0 = 0. i.e. thus either f(x) = 0 for all x or f(x) never equals zero.)
 

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