Comments on: ha! http://easyquestion.net/thinkagain/2010/02/01/ha/ mathematical dialogues aimed to confuse Thu, 02 Feb 2012 18:09:47 +0000 hourly 1 http://wordpress.org/?v=3.3.1 By: mathmom http://easyquestion.net/thinkagain/2010/02/01/ha/#comment-43285 mathmom Wed, 03 Feb 2010 02:58:59 +0000 http://easyquestion.net/thinkagain/?p=2678#comment-43285 I'm not following this part of your argument: R2 implies that 0 ha a = xa for some constant x __ [1] I’m not following this part of your argument:

R2 implies that 0 ha a = xa for some constant x __ [1]

]]> By: Richard Sabey http://easyquestion.net/thinkagain/2010/02/01/ha/#comment-43276 Richard Sabey Tue, 02 Feb 2010 20:37:39 +0000 http://easyquestion.net/thinkagain/?p=2678#comment-43276 Building on Mathmom's work: R2 implies that 0 ha a = xa for some constant x __ [1] (a ha b) - b = 0 ha (a-b) [by R1 with c=-b] thus a ha b = x(a-b) + b __ [2] Thus x(a-b) + b = a ha b = b ha a (by R3) = x(b-a) + a (by [2]) thus x(2a-2b) = a-b thus x = 1/2 thus a ha b = (a-b)/2 + b (by [2]) = (a+b)/2. Building on Mathmom’s work:

R2 implies that 0 ha a = xa for some constant x __ [1]

(a ha b) – b = 0 ha (a-b) [by R1 with c=-b]
thus a ha b = x(a-b) + b __ [2]

Thus
x(a-b) + b = a ha b
= b ha a (by R3)
= x(b-a) + a (by [2])
thus x(2a-2b) = a-b
thus x = 1/2
thus a ha b = (a-b)/2 + b (by [2])
= (a+b)/2.

]]> By: mathmom http://easyquestion.net/thinkagain/2010/02/01/ha/#comment-43215 mathmom Mon, 01 Feb 2010 04:52:17 +0000 http://easyquestion.net/thinkagain/?p=2678#comment-43215 Let's name the rules R1: (a ha b) + c = (b + c) ha (a + c) R2: 0 ha (a + b) = (0 ha a) + (0 ha b) looking at R1 with c=0 we see that a ha b = b ha a let's call that R3 23 ha 77 = (22+1) ha (76+1) (rewrite) = 22 ha 76 + 1 (R1 backwards) if we apply R1 backwards a total of 23 times we have 23 ha 77 = 0 ha 54 + 23 For any x: 2x ha x = x ha 0 + x (x applications of R1 backwards) = 0 ha (-x) + 2x (2x applications of R1 backwards) equating the last 2 lines and re-arranging, we have x ha 0 = 0 ha (-x) + x use R3 on LHS: 0 ha x = 0 ha (-x) + x rewrite x as (2x + (-x)): 0 ha (2x + (-x)) = 0 ha (-x) + x using R2: 0 ha (2x + (-x)) = 0 ha 2x + 0 ha (-x) equating those two expressions for 0 ha (2x + (-x)): 0 ha (-x) + x = 0 ha 2x + 0 ha (-x) therefore, for all x: 0 ha 2x = x So, 23 ha 77 = 0 ha 54 + 23 = 27 + 23 = 50 double checking: use 77 applications of R1 backwards to get 23 ha 77 = -54 ha 0 + 77 = 0 ha (-54) + 77 (R3) = -27 + 77 = 50 In my explorations, I also proved that x ha x = x for all x (use x applications of R1 backwards, and recall that 0 ha 0 = 0) But I couldn't come up with a formula for 0 ha y with y odd. If ha can operate on non-integers ("number" above is vague, so I guess it can) then obviously the same formula applies. Let’s name the rules
R1: (a ha b) + c = (b + c) ha (a + c)
R2: 0 ha (a + b) = (0 ha a) + (0 ha b)

looking at R1 with c=0 we see that a ha b = b ha a
let’s call that R3

23 ha 77
= (22+1) ha (76+1) (rewrite)
= 22 ha 76 + 1 (R1 backwards)

if we apply R1 backwards a total of 23 times we have
23 ha 77 = 0 ha 54 + 23

For any x:
2x ha x
= x ha 0 + x (x applications of R1 backwards)
= 0 ha (-x) + 2x (2x applications of R1 backwards)

equating the last 2 lines and re-arranging, we have
x ha 0 = 0 ha (-x) + x
use R3 on LHS:
0 ha x = 0 ha (-x) + x
rewrite x as (2x + (-x)):
0 ha (2x + (-x)) = 0 ha (-x) + x

using R2:
0 ha (2x + (-x)) = 0 ha 2x + 0 ha (-x)

equating those two expressions for 0 ha (2x + (-x)):
0 ha (-x) + x = 0 ha 2x + 0 ha (-x)
therefore, for all x: 0 ha 2x = x

So,
23 ha 77 = 0 ha 54 + 23
= 27 + 23
= 50

double checking: use 77 applications of R1 backwards to get
23 ha 77 = -54 ha 0 + 77
= 0 ha (-54) + 77 (R3)
= -27 + 77
= 50

In my explorations, I also proved that x ha x = x for all x (use x applications of R1 backwards, and recall that 0 ha 0 = 0)

But I couldn’t come up with a formula for 0 ha y with y odd. If ha can operate on non-integers (“number” above is vague, so I guess it can) then obviously the same formula applies.

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