08/16/2021, 10:56 PM

Hi, Tetration Community

is it active a user called Tommy, if it is the case then :

w = (-3 # +5 # *11 # @23)

reciprocalOf(w) = -(5068/228144) # +(5572/228144) # *(364/228144) # @16156/228144)

w·reciprocalOf(w) = -1 # +1 @ *1 # @2 = @1

path :

w = -a # +b # *c # @d

where # is addition symbol in p4

example with -3 # +5 # *11 # @23

1/w = 1/(-a # +b # *c # @d)

w~= -'a # +b # *'c # @d

where s' m <-> aditiveInverseOf( sm )

1/w = (1/w)(w~/w~) = w~/(ww~)

w~/(ww~) = (-'a # +b # *'c # @d ) / ( ( -a # +b # *c # @d) (-'a # +b # *'c # @d) )

w~/(ww~) = (-'3 # +5 # *'11 # @23) / ( ( -3 # +5 # *11 # @23)(-'3 # +5 # *'11 # @23) )

w~/(ww~) = (-'a # +b # *'c # @d ) / (+'a^2 # @b^2 # +'c^2 # @d^2 # @'2ac # +2bd )

w~/(ww~) = (-'3 # +5 # *'11 # @23) / ( +'9 # @25 # +'121 # @529 # @'66 # +230 )

Note that is enough with express the statement just with the signs @, + ,@' and +'

and is not necessary convert if remain symbol @' and +'...

w~/(ww~) = (-'3 # +5 # *'11 # @23) / ( @480 # +100 )

Now, we complete to make difference of squares in the denominator , but

one must be careful to complete it towards the greatest number, not the smallest

as (x # y)(x # y') if x>y, or as (x # y)(x' # y) if x<y

..in this case we keep @480 invariable in the difference of squares, since it s the greatest

(w~z)/(ww~z) = ( (-'3 # +5 # *'11 # @23)(@480 # +'100) ) / ( (@480 # +100)(@480 #

+'100) )

(w~z)/(ww~z) = ( (-'3 # +5 # *'11 # @23)(@480 # +'100) ) / ( @228144 )

(w~z)/(ww~z)=(-'1464 # +2440 #*'5368 # @11224 #*300 #@'500 # -1100

#+'2300)/@228144

(w~z)/(ww~z) = ( -5068 # +5572 # *364 # @16156 )/@228144

hammering the neutral-signed terms upwards

(w~z)/(ww~z) = -(5068/228144) # +(5572/228144) # *(364/228144) # @16156/228144)

(w~z)/(ww~z) = reciprocalOf( w )

https://groups.google.com/g/sci.math/c/m...IMz54hBwAJ

Best regards Grasshoppa

is it active a user called Tommy, if it is the case then :

w = (-3 # +5 # *11 # @23)

reciprocalOf(w) = -(5068/228144) # +(5572/228144) # *(364/228144) # @16156/228144)

w·reciprocalOf(w) = -1 # +1 @ *1 # @2 = @1

path :

w = -a # +b # *c # @d

where # is addition symbol in p4

example with -3 # +5 # *11 # @23

1/w = 1/(-a # +b # *c # @d)

w~= -'a # +b # *'c # @d

where s' m <-> aditiveInverseOf( sm )

1/w = (1/w)(w~/w~) = w~/(ww~)

w~/(ww~) = (-'a # +b # *'c # @d ) / ( ( -a # +b # *c # @d) (-'a # +b # *'c # @d) )

w~/(ww~) = (-'3 # +5 # *'11 # @23) / ( ( -3 # +5 # *11 # @23)(-'3 # +5 # *'11 # @23) )

w~/(ww~) = (-'a # +b # *'c # @d ) / (+'a^2 # @b^2 # +'c^2 # @d^2 # @'2ac # +2bd )

w~/(ww~) = (-'3 # +5 # *'11 # @23) / ( +'9 # @25 # +'121 # @529 # @'66 # +230 )

Note that is enough with express the statement just with the signs @, + ,@' and +'

and is not necessary convert if remain symbol @' and +'...

w~/(ww~) = (-'3 # +5 # *'11 # @23) / ( @480 # +100 )

Now, we complete to make difference of squares in the denominator , but

one must be careful to complete it towards the greatest number, not the smallest

as (x # y)(x # y') if x>y, or as (x # y)(x' # y) if x<y

..in this case we keep @480 invariable in the difference of squares, since it s the greatest

(w~z)/(ww~z) = ( (-'3 # +5 # *'11 # @23)(@480 # +'100) ) / ( (@480 # +100)(@480 #

+'100) )

(w~z)/(ww~z) = ( (-'3 # +5 # *'11 # @23)(@480 # +'100) ) / ( @228144 )

(w~z)/(ww~z)=(-'1464 # +2440 #*'5368 # @11224 #*300 #@'500 # -1100

#+'2300)/@228144

(w~z)/(ww~z) = ( -5068 # +5572 # *364 # @16156 )/@228144

hammering the neutral-signed terms upwards

(w~z)/(ww~z) = -(5068/228144) # +(5572/228144) # *(364/228144) # @16156/228144)

(w~z)/(ww~z) = reciprocalOf( w )

https://groups.google.com/g/sci.math/c/m...IMz54hBwAJ

Best regards Grasshoppa