A Calculus of Number Based on Spatial Forms - University of ...

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Inverse promotion carries the inverse over the modier form, as =(a[]):
This theorem translates to standard numbers as the conversion between the multiplicativeinverse
<strong>of</strong> a value and the additiveinverse <strong>of</strong> its exponent. Inverse promotion
for multiplication appears in standard and boundary numbers as:
1=x y = x ,y versus ()=(([[x]][]))
5.4.2 Rationals
Because the boundary calculus now includes the multiplicative inverse, integers can
be extended to rationals. The multiplicative inverse <strong>of</strong> a is simply (): For
n; d 2 I with d 6= , the element ([n]) is in Q, the set <strong>of</strong> rationals.
The rationals, a; b 2 Q add by collection ab and are closed under addition.
Pro<strong>of</strong>.
Rational numbers a; b 2 Q take the form a=([i]) and b=([k]);
where i; k; j; l 2 I with j; l 6= . The sum <strong>of</strong> these rationals is ab=([i])([k]);
which reduces to rational c=([m]); where m=([i][l])([k][j]) and n=([j][l]);
by the following derivation.
( [i] )( [k] ) Given
( [i][l] )( [k][j] < [j]>) Inversion
( [i][l] )( [k][j] < [j] [l] >) Inverse Collection
([([i][l])])([([k][j])]) Involution
([([i][l]) ([k][j])]) Distribution
([ m ]) Rewrite
Since the product <strong>of</strong> two integers in I is an integer, n is an integer. Because j and
l are non-void, their product is non-void. Products ([i][l]) and ([k][j]) are also
integers. Since the sum <strong>of</strong> two integers is an integer, m=([i][l])([k][j]) is also an
integer. Therefore, c is rational, c 2 Q, and addition is closed under the rationals.
Every elementinQhas an additiveinverse, found simply by wrapping the inverse
boundary around it.
Rationals, a; b 2 Q multiply by the same form used for natural numbers, ([a][b]):
The rationals are closed under multiplication.