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Math's Puzzle!!! Solve that...
Old 02-17-2011, 10:49 AM Re: Math's Puzzle!!! Solve that...
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And in our next lesson we will move on to polynominals
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Old 02-17-2011, 09:46 PM Re: Math's Puzzle!!! Solve that...
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I would say square root isn't either.

Assuming they are, if ^(1/2) and ^(1/3) are distinct operators then why not any arbitrary ^(p)?. If this is the case then the problem becomes trivial. For any number N greater than 1, there is some p such that N^p = 6. In other words log_N(6) = p (log base n of six equals p). For example, 2^p = 6 when p ~ 2.585 (log_2(6) ~ 2.585).

Here is the answer for 8:
Code:
(8 + 8 - 8)^(log_8(6))
Actually, you make a good argument... cheers
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Last edited by smoseley; 02-17-2011 at 09:53 PM..
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Old 02-17-2011, 09:55 PM Re: Math's Puzzle!!! Solve that...
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Cubic root and square root are absolutely operators....

There are two basic symbols in Math - operators & values

If cubic root is not an operator, it's a value...

So tell me, what is the value of cubic root?
Roots are just the inverse of exponentiation, which is a binary operator (it takes two parameters, the base and the power). Asking what the value of cubic root is is analogous to asking the value of "2 +", you're missing the other operand.
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Old 02-17-2011, 10:17 PM Re: Math's Puzzle!!! Solve that...
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And in our next lesson we will move on to polynominals
please no..
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Old 02-18-2011, 12:54 PM Re: Math's Puzzle!!! Solve that...
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2+2+2=6
3*3-3=6
4*√4-√4 = 6
5/5+5=6
6*6/6=6
7-7/7=6
√√8+√√8+√√8 = 6
(9+9)/square root of 9 = 6
how about this
1 1 1=6
0 0 0 =6
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Old 02-18-2011, 03:11 PM Re: Math's Puzzle!!! Solve that...
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It's not possible with 0 and 1...
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Old 02-20-2011, 04:29 AM Re: Math's Puzzle!!! Solve that...
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A mathematical operator is simply a mapping -- it takes a "thing" and maps it to another "thing". There is not a requisite of having a specific number of operands, though I've never heard of an operand operating on a null set of operands.

For example, absolute value operates over the Real Numbers on a single operand (and returns a single value).

The root operand, does have 2 inputs (no matter how we choose to write the expression): To take the nth root of X means to find y s.t. y^n = X. Like the absolute operator being applied to real numbers, the inverse of taking a root has 2 answers in the real numbers. For example, the 2nd root of 4 is +/- 2 since (+/- 2)^2 = 4 -- it is only 2 if we are working in the positive integers.

When looking at such functions it is very important to consider the set of number you are working with: positive integers, real numbers, complex number... each set provides a sometimes varying set of answers.

I think of an operator as an algorithm: it simply defines an ordered procedure for one to follow.

Take the operator I had defined earlier: ⊕

Since I had defined x ⊕ x = 6, it made the issue trivial, but what kind of algorithm could represent such an operator? Well, if we look at it as a function, then we have ⊕(x,x) = 6 which represents a line in 3-space. Very similar to the 2-space line represented by f(x)=6 or y=6 which we learned in introductory algebra classes. I could have just as easily defined the 4-space function ⊕(x,x,x) = 6 which would still represent a line in 4-space.

NOTE: It's been awhile since my college abstract algebra classes, so I hope I didn't miss any glaring points!
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Old 02-21-2011, 04:21 AM Re: Math's Puzzle!!! Solve that...
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it good entertainment for people who want to fefresh.
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Old 02-21-2011, 04:36 AM Re: Math's Puzzle!!! Solve that...
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Wow. They're great. I tried 0 and 1 but i can't find any solution. and I think anything that use zero falls to the answer of likely 0, 1 or infinity.
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