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Topic: changing the fundamentals to Math and its cons. (Read 1586 times) 

BMAD
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changing the fundamentals to Math and its cons.
« on: May 24^{th}, 2014, 7:21am » 
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1. numerical value is comprised of set of one equal to that value 2. All numerical values should be written in a sequence using a comma to separate all numbers. 3. the addition of two numerical values is the listing of both sets of ones join by a comma. 4. The solution to an addition problem is the count of commas. E.g. 5 = {1,1,1,1,1} 6 ={1,1,1,1,1,1} 5+6 = {1,1,1,1,1 , 1,1,1,1,1,1} = 10 (because there are 10 commas) 5. All other mathematical operations are linked to addition in the manner as it was before. Solve: 1+0= 8  5 = 3 × 5 = 1 2 / 4 = Sqrt (36) 3+5×2= 3x6 = 21, x = Derivative of 3x Anti derivative of 3x

« Last Edit: May 24^{th}, 2014, 8:50am by BMAD » 
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BMAD
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Re: changing the fundamentals to and its consequ
« Reply #1 on: May 24^{th}, 2014, 7:28am » 
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When i say linked to addition I mean Multiplication can be considered repeated addition. Subtraction can be written as an equivalent addition statement Division is repeated subtraction And so on

« Last Edit: May 24^{th}, 2014, 8:50am by BMAD » 
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BMAD
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Re: changing the fundamentals to Math and its cons
« Reply #2 on: May 25^{th}, 2014, 8:07am » 
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1 + 0 = {1} = 0 (no commas)


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dudiobugtron
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Re: changing the fundamentals to Math and its cons
« Reply #4 on: May 25^{th}, 2014, 4:56pm » 
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So, basically, each time you add, you subtract 1 from the normal total? In that case, multiplication isn't commutative. 5 + 5 + 5 = 13 is different from 3 + 3 + 3 + 3 + 3 = 11. So we need to know which of those 3 x 5 means.


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BMAD
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Re: changing the fundamentals to Math and its cons
« Reply #5 on: May 25^{th}, 2014, 10:53pm » 
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on May 25^{th}, 2014, 4:56pm, dudiobugtron wrote:So, basically, each time you add, you subtract 1 from the normal total? In that case, multiplication isn't commutative. 5 + 5 + 5 = 13 is different from 3 + 3 + 3 + 3 + 3 = 11. So we need to know which of those 3 x 5 means. 
 Not quite: 5x3 would be either 5 ={1,1,1,1,1} 3= {1,1,1} So 5 x 3 = 5 + 5 +5 or = 3 + 3 + 3 + 3 + 3 5 case= {1,1,1,1,1 , 1,1,1,1,1 , 1,1,1,1,1} = 14 3 case={1,1,1 , 1,1,1 , 1,1,1 , 1,1,1 , 1,1,1} = 14 So remember it is the count of commas.


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towr
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Re: changing the fundamentals to Math and its cons
« Reply #6 on: May 25^{th}, 2014, 11:20pm » 
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The trouble is that means there's a difference between 5 + 5 +5 and (5 + 5) + 5. So how does that affect other sums with multiple operations? Do we have 3 + 5 × 2 = 3 + (5 × 2) = 3 + 9 = 11 or 3 + 5 × 2 = 3 + 5 + 5 = 12 ?


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BMAD
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Re: changing the fundamentals to Math and its cons
« Reply #7 on: May 25^{th}, 2014, 11:25pm » 
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To me in this set of rules. .. The order of operations do not get resolved step wise but in one swoop, the order just gives structure to which ones get combined first. Which shouldn't change the answers.


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dudiobugtron
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Re: changing the fundamentals to Math and its cons
« Reply #8 on: May 26^{th}, 2014, 12:04am » 
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12 / 4 : 9 + 4 = 12 so 12  4 = 9 5 + 4 + 4 = 12 so 12  4  4 = 5 similarly 12  4  4  4 = 1 So I guess that means 12 / 4 = 3 remainder 1. PS: re order of operations: What towr meant (I imagine) wasn't specifically about brackets and how to interpret them. It's that if you do 5 + 5 first, then take that answer and add 5 again, the answer you get is different from 5 + 5 + 5.

« Last Edit: May 26^{th}, 2014, 12:10am by dudiobugtron » 
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rmsgrey
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Re: changing the fundamentals to Math and its cons
« Reply #9 on: May 26^{th}, 2014, 7:30am » 
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So 2={1,1} 1={1} 1+1={1 , 1}=1 So {1} = {1 , 1} =/= {1,1}


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BMAD
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Re: changing the fundamentals to Math and its cons
« Reply #10 on: May 26^{th}, 2014, 7:34am » 
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In this world, 1 is the additive identity. So 8 + 1 = 8, 1 + 1 =1 and so on.


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dudiobugtron
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Re: changing the fundamentals to Math and its cons
« Reply #11 on: May 26^{th}, 2014, 3:12pm » 
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sqrt(36) isn't defined. There is no number x where you can add it to itself x times to get 36. 6+6+6+6+6+6 = 35 7+7+7+7+7+7+7 = 48 so sqrt(35) = 6, but sqrt(36) isn't defined. Unless you have a way of representing 'unnatural' numbers, you can't sqrt 36. on May 26^{th}, 2014, 7:30am, rmsgrey wrote:So 2={1,1} 1={1} 1+1={1 , 1}=1 So {1} = {1 , 1} =/= {1,1} 
 1 + 1 doesn't actually equal {1 , 1}; you just have to create the construction {1 , 1} in order to work it out.

« Last Edit: May 26^{th}, 2014, 3:15pm by dudiobugtron » 
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towr
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Re: changing the fundamentals to Math and its cons
« Reply #12 on: May 26^{th}, 2014, 10:08pm » 
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Maybe sqrt(36) = 5 {1,1,1,1,1,1, 1,1,1,1,1,1, 1,1,1,1,1,1, 1,1,1,1,1,1, 1,1,1,1,1,1, 1,1,1,1,1,1} [ sqrt ] > {1,1,1,1,1,1} [=]> 5

« Last Edit: May 26^{th}, 2014, 10:09pm by towr » 
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BMAD
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Re: changing the fundamentals to Math and its cons
« Reply #13 on: May 27^{th}, 2014, 5:09am » 
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Interestingly I had the same answer for sqrt (36) as tower originally considering it only a geometric interpretation. I failed to consider it a arithmetic interpretation like dudiobugtron. So I think they may be right in labeling this as undefinable given its two possible solutions (and even others may still exist).


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dudiobugtron
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Re: changing the fundamentals to Math and its cons
« Reply #14 on: May 27^{th}, 2014, 6:48pm » 
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In that case, I'm not entirely sure how to solve: 3x6 = 21 I was going to solve it by rearranging: 3x = 21 + 6 3x = 26 x = 26 / 3 x = 9 (because 9 + 9 + 9 = 26 ) Is that what you had in mind?  Also I think the derivative question is interesting. Of course, since there are no inbetween numbers, the graph of y = 3x is not anywhere continuous, and so can't really be differentiated. But assuming we're allowed to 'connect the dots' for the purposes of differentiating, then: The graph of y = 36 looks exactly like the normal graph of y = 3x  1. So, in that case, you'd think the gradient would be 3; since the slope is the same. However, it depends on how you define the gradient. If it is something like 'rise over run', then the formula might be: (y_{2}  y_{1}) / (x_{2}  x_{1}) when x = 2, y = 5 when x = 3, y = 8 So the gradient from 2 to 3 is: ( 8  5 ) / ( 3  2) = 4 / 2 = 2 remainder 1. (since 2 + 2 = 3, but 2 + 2 + 1 = 4) Let's pick a different point: when x = 1, y = 2 Then, the gradient from 1 to 3 is: (8  2) / (3  1) = 7 / 3 = 2 remainder 2. So it looks like defining the gradient that way doesn't always give the same answer. Anyone have any ideas on how to do it consistently?


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BMAD
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Re: changing the fundamentals to Math and its cons
« Reply #15 on: May 27^{th}, 2014, 7:00pm » 
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Yes.


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