Support NeoGAF

[Raistlin]

There is no equals sign so I do not have to perserve any properties on both sides of the equation.

0 is done like this:
2+2+2+2 //Could be anything else like 2/2-2*2
(2+2)+(2+2)
(2/2)+(2/2)
(2/2)-(2/2)


22 is not allowed because it requires the concatenation operator.

Yes, it does use concatentation ... but we are letting that go in this case.

We are looking at this as though each character in the formula is atomic; ie. each separate character can be either a 2, +, -, *, -, (, or ), also with the caveat that there MUST be four instances of the 2 character. Any mathematical conventions (concatenation, etc.) that occur based on the grouping of these characters is inconsequential.

It is the same logic that I use for superscripting. Any implied operations are fine, as it doesn’t violate the atomic characters used.

Think of it as string in programming. When creating the string, we are simply grouping a series of characters. As long as we only use characters from the available set (and make sure to use four instances of the 2 character), we are okay.

Once the string has been completed, then and only then, do we actually treat it as a mathematical formula. At this point, any conventional mathematical operators implied by the above grouping is acceptable, as the creation of the ‘string’ did not break a rule.

 

[Raistlin]

() by itself has no meaning so 2^() has no meaning.

I'm not sure if I agree - while it is a stretch, I think there is some validity to it.

Numbers are conceptual. Any digit is equivalent to itself +/- an infinite string of 0's.

1 = 1+0 = ... = 1+0+0+0+0+0+0+0+0, etc.

This equivalency is obviously only set into motion when a value is present. You do not conceptually view nothingness as being equivalent to ‘nothing’ +0+0..., but if a value is present this relationship is understood.

In the case of parentheses, a side effect of their being used is the understanding that a value IS present (and any operators contained within - implied or not - must be calculated left to right before the value can be used with any values/operators outside the parentheses). With the implication that a value is present, we are left with nothingness - an abstract conceptual non-value (something with no intrinsic value unto itself), and an infinite string of +0's. The parentheses force the determination of a value, which in this case simply uses the operable values that are implied. Since the only operable values are an infinite string of 0+0, that is all we can use.

Therefore when solving, we find ourselves with 2^(0+0...) = 2^0 = 1

 

[NotMSRP]

() is an operation.
(args)
Since () has no arguments to it, it is invalid.
0 is the numeral representation of nothingness.
{}, the empty/null set also represents nothingness but in the context of nothing is in the container.
The only allowed implicit operation is the ()s: 5*3+2 => (5*3)+2 => ((5*3)+2)

 

[Stuggernaut]

This thread has been fun....but after grilling my daughter, it just got de-railed.

She said that squares don't count as 2's

And with that

(2² + 2²) - 2/2 = 7

I think I am going to ground her for the weekend for leaving that bit of info out.

On a side note...you guys are pretty damn good at math...lol

 

[Matlock]

I think I am going to ground her for the weekend for leaving that bit of info out.

Make her read Konex posts!

 

[Raistlin]

() is an operation.
(args)
Since () has no arguments to it, it is invalid.
0 is the numeral representation of nothingness.
{}, the empty/null set also represents nothingness but in the context of nothing is in the container.
The only allowed implicit operation is the ()s: 5*3+2 => (5*3)+2 => ((5*3)+2)

This is a good argument against it - though from a conceptual mathetmatics view, I think I gave some valid reasons why there could be implied arguments in empty parentheses.

What's weird, is that if you use a calculator or math SW program ... () will give you 0. Not sure why, but it accepts it as a valid representation?



Regardless ... this has been an interesting thread.

 

[Raistlin]

This thread has been fun....but after grilling my daughter, it just got de-railed.

She said that squares don't count as 2's

And with that

(2² + 2²) - 2/2 = 7

I think I am going to ground her for the weekend for leaving that bit of info out.

On a side note...you guys are pretty damn good at math...lol

So I guess squares were part of the operation list?

:lol

We've been powned!!1

 

[hobart]

Of course squares were allowed... it's part of the riddle to be able to see beyond the linear formation of normal equations.

As far as "^" goes... you guys are silly... I cannot type superscripts and, thusly, used the carot (^) to denote a power/exponent.

I'll ask some math majors around about ()... if it equals zero... we didn't even need the painfully easy ((2^2) + (2^2)) - (2/2) = 7.

2s don't count in powers! BAH!!!

 

[BTMash]

I'll ask some math majors around about ()... if it equals zero... we didn't even need the painfully easy ((2^2) + (2^2)) - (2/2) = 7.

One easier way around that one is (2^(2+2) - 2)/2 = 7.

Edit: Should have read the clarification by Mr Pockets when squares don't count. >_<

 

[hobart]

One easier way around that one is (2^(2+2) - 2)/2 = 7.

If 2's do not count in the superscript... then you are only using 3. But yah... that's an attractive answer for sure.

 

[mashoutposse]

One easier way around that one is (2^(2+2) - 2)/2 = 7.

Edit: Should have read the clarification by Mr Pockets when squares don't count. >_<

2^4 is actually 2*2*2*2, so....

 

[Lord Error]

This thread has been fun....but after grilling my daughter, it just got de-railed.

She said that squares don't count as 2's

Well that was lame, that the actual answer was "you can use this another operator that we didn't mention, and oh, the number 2 used in that operator doesn't count".

I liked the "Writing 2 on the opposite side of the paper" answer a lot better than this. Even the "((2++) + (2++)) + (2/2)=7" answer holds more water, because you at least don't use any more 2s than what's initially allowed, even though you still introduce a new operator (increment).

 

[BTMash]

2^4 is actually 2*2*2*2, so....

I'm confused...2*2*2*2 = 16
16 - 2 = 14
14/2 = 7 @_@

I do think the (2++) + (2++) - (2/2) solution is amazing