gib65 Posted October 18, 2012 Share Posted October 18, 2012 Does anybody else find this fascinating--that if both operands are 2 for any "growth" operator, then the result will always be 4. By a "growth" operator, I mean +, x, power, etc.--and operation whose result is greater than its operands (as opposed to "shrinking" operators like -, /, square root, etc.). 2 + 2 = 4 2 x 2 = 4 2 ^ 2 = 4 2 # 2 = 4 (where we could suppose x # y means x raised to itself y times). Wouldn't this trend go on indefinitely? Couldn't we say that 2 @ 2 = 4 where @ is any "growth" operator whatever? And wouldn't a similar rule apply where 4 % 2 = 2 where % is any "shrinking" operator whatever? NOTE: I'm aware that I've defined "growth" operations poorly--for example, 10 x 0.5 = 5 (5 is hardly greater than 10)--but I think it's good enough to get the idea across that +, x, ^ seem to be one group of operators that have a common feature in their results, and similarly for -, /, square root. 1 Link to comment Share on other sites More sharing options...
stopandthink Posted October 19, 2012 Share Posted October 19, 2012 I don't really see what's so fascinating, but maybe it's just me. Link to comment Share on other sites More sharing options...
imatfaal Posted October 19, 2012 Share Posted October 19, 2012 Does anybody else find this fascinating--that if both operands are 2 for any "growth" operator, then the result will always be 4. By a "growth" operator, I mean +, x, power, etc.--and operation whose result is greater than its operands (as opposed to "shrinking" operators like -, /, square root, etc.). 2 + 2 = 4 2 x 2 = 4 2 ^ 2 = 4 2 # 2 = 4 (where we could suppose x # y means x raised to itself y times). Wouldn't this trend go on indefinitely? Couldn't we say that 2 @ 2 = 4 where @ is any "growth" operator whatever? And wouldn't a similar rule apply where 4 % 2 = 2 where % is any "shrinking" operator whatever? NOTE: I'm aware that I've defined "growth" operations poorly--for example, 10 x 0.5 = 5 (5 is hardly greater than 10)--but I think it's good enough to get the idea across that +, x, ^ seem to be one group of operators that have a common feature in their results, and similarly for -, /, square root. Not really sure it is very interesting I am afraid. Additionally one could think difficulties - we could even consider an operator that is in actual use.If you use calculators or spreadsheets you will know the operator e. 5e3, 1.5e25 etc - computers use the operator e to stand in for multiply the first operand by ten and repeat as many times as the second operand 5e3 = 5x10x10x10 = 5000. 2e2 = 2x10x10 = 200. I am also not sure about your use of # operator - if x # y means x raised to itself y times then surely it means x^(x^y) 2^2^2 = 16 [math] x hash y = x^{x^y}[/math] [math] 2 hash 2 = 2^{2^2} = 16 [/math] Link to comment Share on other sites More sharing options...
gib65 Posted October 19, 2012 Author Share Posted October 19, 2012 I am also not sure about your use of # operator - if x # y means x raised to itself y times then surely it means x^(x^y) 2^2^2 = 16 Oh, yeah, well I kind of meant it in the same sense as when we say that a x b means "a added to itself b times" when we really mean "a added to itself b-1 times". Is a + a the same as a being added to itself twice or is it being added just once? Link to comment Share on other sites More sharing options...
Melissa Mika Posted November 3, 2012 Share Posted November 3, 2012 There's nothing into it.. Just 'playing cards. I often double check, therefor I am Link to comment Share on other sites More sharing options...
phillip1882 Posted November 10, 2012 Share Posted November 10, 2012 (edited) while somewhat interesting in the sense that 2 is the only number with this property, (2+2 = 4; 2*2 = 4; 2^2 = 4) it's not much challenge to define a growth operator where this is not true. for example, let's say # is 2*a +b -1. any number can be reached with smaller values, assuming 1 is given. 1 # 1 = 2 1 # 2 = 3 2 # 1 = 4 2 # 2 = 5 and so on. if you desire the commutative property this is only slightly more challenging. let # = (a << 1) x b |(b << 1) x a -1 where << is left shift, x is xor, and | is or. 1 # 1 = 2 1 # 2 = 4 1 # 3 = 6 2 # 2 = 5 Edited November 10, 2012 by phillip1882 Link to comment Share on other sites More sharing options...
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