2 @ 2 = 4

[gib65]

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.

[stopandthink]

I don't really see what's so fascinating, but maybe it's just me.

[imatfaal]

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]

[gib65]

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?

[Melissa Mika]

There's nothing into it.. Just 'playing cards. I often double check, therefor I am

[phillip1882]

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