1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
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1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Sorry, boring subject line, but that is actually the question. I was chatting with a friend of mine earlier discussing having seen this on one of the more mindless social media sites. We came up with "13" as the answer because of the order of precedence stripping off the (1*0) and leaving the rest to just be added. We were joined by another guy who made the following claim. Is he right and in what context?
in general mathematical theory  there are 13 axioms to the real number system and an order of precidence is not one of them
the two primary axioms state an associative property of addition and multiplication
There is no rule in mathematics however that says a) you have to read the equation from left to right, b) you have to interpret multiplication before addition, or c) you have to create intermediate values as part of finding a solution to the problem
so if you punch the formula in  as is  into a calculator you will always get 0 because * is the last operation a calculator will process and therefore the answer will always be 0
if you used a computer program to evaluate the answer then you will always get 13 as the answer because * is processed before + or 
therefore the 0 is taken out of the equasion immediately and all you are left with is a series of +/ that are interpreted from left to right
however if you use calculus to evalutate the expression the axioms of association for both multiplication and addition allow you to move the expression around into almost any order so long as the (1*0) remain together
That means you can actually come out with any answer from 1 to 13 and still be right
in general mathematical theory  there are 13 axioms to the real number system and an order of precidence is not one of them
the two primary axioms state an associative property of addition and multiplication
There is no rule in mathematics however that says a) you have to read the equation from left to right, b) you have to interpret multiplication before addition, or c) you have to create intermediate values as part of finding a solution to the problem
so if you punch the formula in  as is  into a calculator you will always get 0 because * is the last operation a calculator will process and therefore the answer will always be 0
if you used a computer program to evaluate the answer then you will always get 13 as the answer because * is processed before + or 
therefore the 0 is taken out of the equasion immediately and all you are left with is a series of +/ that are interpreted from left to right
however if you use calculus to evalutate the expression the axioms of association for both multiplication and addition allow you to move the expression around into almost any order so long as the (1*0) remain together
That means you can actually come out with any answer from 1 to 13 and still be right
 gmalivuk
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
It may not be an axiom defining the real numbers as a set, but it is part of the definitions of addition and multiplication of real numbers that multiplication takes precedence. So the *0 only annihilates the last 1, and the 1 only subtracts that individual 1, for a total of 13 according to the order of operations every elementary school student is taught.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
It can actually be derived from one of the 13 set axioms of the real numbers that multiplication takes precedence. (Although you only need 8 axioms really)
Ed. actually it will depend on how the axioms are written.
The question that actually causes problems is when you have an infinite series:
11+11+11+11+11+11+11...
This is actually undefined by mathematics as you can "prove" it is equal to 0 and that it is equal to 1/2. So if you want to impress your friends tell them about Grandi's Series.
http://en.wikipedia.org/wiki/Grandi%27s_series
Ed. actually it will depend on how the axioms are written.
The question that actually causes problems is when you have an infinite series:
11+11+11+11+11+11+11...
This is actually undefined by mathematics as you can "prove" it is equal to 0 and that it is equal to 1/2. So if you want to impress your friends tell them about Grandi's Series.
http://en.wikipedia.org/wiki/Grandi%27s_series
"Absolute precision buys the freedom to dream meaningfully."  Donal O' Shea: The Poincaré Conjecture.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
dirkm wrote:so if you punch the formula in  as is  into a calculator you will always get 0 because * is the last operation a calculator will process and therefore the answer will always be 0
Not on either of the calculators I own.
A calculator would do that if it's the kind that doesn't actually allow you to enter expressions, in which case it is calculating intermediate values for the addition steps before applying any of the multiplication. But in that case, you're no longer evaluating the original expression  the calculator can't evaluate multistep expressions at all. You're evaluating a whole bunch of individual steps and implicitly treating it as a larger calculation. If you were to rewrite the expression implicitly being evaluated in that case, it would have to have a bunch of parentheses, because you're evaluating things lefttoright without regard for order of operations.
Anyway, there's no rigorous requirement (AFAIK) that order of operations HAS to work the way it usually does. If you used an alternate notation where addition took precedence over multiplication, you could still do equivalent and valid math. You'd just have to parenthesize differently to perform the same calculations. Indeed, some notations work very differently but are still perfectly valid.
But when you present a problem in what appears to be normal notation without mentioning otherwise, it is generally good to assume that you are actually using normal notation, which means the standard order of operations. The disagreements over problems like this in social media are mostly just a version of communicating badly.
No, even in theory, you cannot build a rocket more massive than the visible universe.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Well if you write the distributive law as follows: A*(B+C)=A*B+A*C I'm not sure you can get a consistent system unless multiplication takes precedence.
"Absolute precision buys the freedom to dream meaningfully."  Donal O' Shea: The Poincaré Conjecture.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
 dudiobugtron
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Whether you do addition or multiplication first, or just in order from left to right, is just a convention. It saves you having to write a lot of brackets.
But then, brackets are also a convention. As is conducting the operations from lefttoright. As is using those particular symbols to represent those particular mathematical concepts.
You could quite easily construct different conventions with exactly the same underlying maths. It's silly to argue about which convention is 'right' from a mathematical viewpoint. It's like trying to argue that English is the 'correct' language to do maths in, since the axioms are written in English.
Similarly, if you write it as B+C*A = (B*A)+(C*A) , it doesn't work unless you do multiplication and addition from left to right, as described in the OP.
So if there's confusion, you should write it as A*(B+C)=(A*B)+(A*C)
But then, brackets are also a convention. As is conducting the operations from lefttoright. As is using those particular symbols to represent those particular mathematical concepts.
You could quite easily construct different conventions with exactly the same underlying maths. It's silly to argue about which convention is 'right' from a mathematical viewpoint. It's like trying to argue that English is the 'correct' language to do maths in, since the axioms are written in English.
Frimble wrote:Well if you write the distributive law as follows: A*(B+C)=A*B+A*C I'm not sure you can get a consistent system unless multiplication takes precedence.
Similarly, if you write it as B+C*A = (B*A)+(C*A) , it doesn't work unless you do multiplication and addition from left to right, as described in the OP.
So if there's confusion, you should write it as A*(B+C)=(A*B)+(A*C)
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
His basic approach centers around
so he is saying that that, in (his perception of) fact, the *0 can be performed on any of the interim answers (creating a zero value out of those) and then you can proceed with the rest of the addition and subtraction. Because of this, the answer can be any integer from 1 to +13.
I found one other person at work who says that he can see how the first guy arrives at this answer (set). The way I found this guy is that I asked around if anyone knew a "dysfunctional theoreticalmath geek" (me being labeled one of the dysfunctional programming geeks). The guy conceded that this view is very impractical and he, himself, wouldn't use it for this type of equation.
So, the fundamental question here is:
There is no rule in mathematics however that says a) you have to read the equation from left to right, b) you have to interpret multiplication before addition, or c) you have to create intermediate values as part of finding a solution to the problem
so he is saying that that, in (his perception of) fact, the *0 can be performed on any of the interim answers (creating a zero value out of those) and then you can proceed with the rest of the addition and subtraction. Because of this, the answer can be any integer from 1 to +13.
I found one other person at work who says that he can see how the first guy arrives at this answer (set). The way I found this guy is that I asked around if anyone knew a "dysfunctional theoreticalmath geek" (me being labeled one of the dysfunctional programming geeks). The guy conceded that this view is very impractical and he, himself, wouldn't use it for this type of equation.
So, the fundamental question here is:
Is order of precedence part of math or is it a convenience that we have come to rely on?
 jestingrabbit
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
dirkm wrote:So, the fundamental question here is:Is order of precedence part of math or is it a convenience that we have come to rely on?
Its a mathematical convention, so its both.
Somewhat funny story: I recently had an expression of the form a/b*c in python, thought that it would be read as a/(b*c), but was actually read as (a/b)*c. Couldn't see it for ages, because I was reading it like a mathematician, instead of like a programmer.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
dirkm wrote:So, the fundamental question here is:Is order of precedence part of math or is it a convenience that we have come to rely on?
The underlying mathematical meaning of "1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0" changes depending on how we interpret it. But the underlying mathematical meanings themselves don't change, we're merely changing which meaning we chose to assign to that particular string.
If I want to add two numbers (B and C) together and then multiply them by a third number (A), that is a different concept from multiplying two numbers together (A and B) and then adding on a third number (C). The two concepts remain different regardless of how we choose to communicate them.
Learning about Order of Operations is important for the study of maths, because you need to be able to communicate mathematical concepts, and you need to be able to do it well. Getting it wrong can have disastrous effects, and a good communication system can help to influence further mathematical developments.
So in that sense, it is an important part of mathematics. But only on an 'administrative' level. By changing the order of operations, you don't change anything about the actual meaning of the underlying mathematics.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Sort of repeating what's been said above: None of the axioms of the real numbers have anything to do with preference of operations. They define an object. The expressions we write down to stand for variables and constants in that object are completely separate from that object. [imath]\mathbb{R}[/imath] does not care how we choose to write down anything. If you wanted to be formal, you would end up using lots and lots of parenthesis. All the order of operations does is construct a shorthand.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
jestingrabbit wrote:Somewhat funny story: I recently had an expression of the form a/b*c in python, thought that it would be read as a/(b*c), but was actually read as (a/b)*c. Couldn't see it for ages, because I was reading it like a mathematician, instead of like a programmer.
Multiplication and division have equal precedence in the order of operations and are evaluated from lefttoright, so a mathematician should read that as (a/b)*c. Or perhaps you're expecting that / and ÷ are different operators that do the same binary calculation but with different precedence?
An interesting observation from one of my education textbooks (don't recall which  perhaps Molina's "The Problem with Math is English") is that the algebraic conventions are human decisions but not particularly arbitrary. For instance, if I were to say "I have two books and five pencils," there is an implied multiplication between "two" and "books" without a cue word like the "and" that signals addition, and also it would make no sense to us to try to interpret the phrase "books and five" before parsing those implied multiplications first. (As I write this, I am realizing that the book didn't go on to mention whether those grammatical conventions exist in other languages, specifically the languages associated with the cultures that formed and codified the algebraic conventions.)
 jestingrabbit
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Tirian wrote:jestingrabbit wrote:Somewhat funny story: I recently had an expression of the form a/b*c in python, thought that it would be read as a/(b*c), but was actually read as (a/b)*c. Couldn't see it for ages, because I was reading it like a mathematician, instead of like a programmer.
Multiplication and division have equal precedence in the order of operations and are evaluated from lefttoright, so a mathematician should read that as (a/b)*c. Or perhaps you're expecting that / and ÷ are different operators that do the same binary calculation but with different precedence?
I expect that a mathematician writing out a large product and divsion, where some things are in the numerator and some in the denominator, would typically choose to only write a single vinculum or solidus, and have everything in the numerator in one place, and everything in the denominator in the other.
I mean, when you write out the quadratic formula do you put brackets around the 2a?
ameretrifle wrote:Magic space feudalism is therefore a viable idea.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Frimble wrote:It can actually be derived from one of the 13 set axioms of the real numbers that multiplication takes precedence. (Although you only need 8 axioms really)
Ed. actually it will depend on how the axioms are written.
The question that actually causes problems is when you have an infinite series:
11+11+11+11+11+11+11...
This is actually undefined by mathematics as you can "prove" it is equal to 0 and that it is equal to 1/2. So if you want to impress your friends tell them about Grandi's Series.
http://en.wikipedia.org/wiki/Grandi%27s_series
It's actually "equal" to any rational number.
For example, if we let S = 1  1 + ... = sum_{n=0}^{inf}(1)^{n} then we can take two numbers:
S = S_{0} + S_{1} where
S_{0} = sum_{n=0}^{inf}(1)^{4n} + sum_{n=0}^{inf}(1)^{4n+1} and
S_{1} = sum_{n=0}^{inf}(1)^{4n+2} + sum_{n=0}^{inf}(1)^{4n+3}
But clearly S_{0} = S_{1} = S.
We can keep dividing it, getting us S = kS. But S on the left is equal to 1 (via 1 + (1 + 1) + ...) , and so S on the right is equal to 1/k. We can then use S = mS for another integer m and get S = m/k, thus it covers all rational numbers.
It's a bit silly.
But can you get irrational numbers?
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
I don't think that works. You can't change the order of the numbers in an infinite sum.
"Absolute precision buys the freedom to dream meaningfully."  Donal O' Shea: The Poincaré Conjecture.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Frimble wrote:I don't think that works. You can't change the order of the numbers in an infinite sum.
Not unless it's absolutely convergent. Which means that it converges to a finite value if all the terms are made positive.
11+11+1... is not absolutely convergent, because making all the terms positive gives 1+1+1+1+1... which is divergent.
1  0.5 + 0.25  0.125 + 0.0625... is absolutely convergent, because making all the terms positive gives 1 + 0.5 + 0.25 + 0.125 + 0.0625... which converges to 2. You can reorder the terms and do a bunch of other things you might want to do with an infinite series, and it will still make sense.
The preceding comment is an automated response.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Thanks, I forgot about the obvious exception.
"Absolute precision buys the freedom to dream meaningfully."  Donal O' Shea: The Poincaré Conjecture.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
"We need a reality check here. Roll a D20."  Algernon the Radish
"Should I marry W? Not unless she tells me what the other letters in her name are" Woody Allen.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
jestingrabbit wrote:Tirian wrote:jestingrabbit wrote:Somewhat funny story: I recently had an expression of the form a/b*c in python, thought that it would be read as a/(b*c), but was actually read as (a/b)*c. Couldn't see it for ages, because I was reading it like a mathematician, instead of like a programmer.
Multiplication and division have equal precedence in the order of operations and are evaluated from lefttoright, so a mathematician should read that as (a/b)*c. Or perhaps you're expecting that / and ÷ are different operators that do the same binary calculation but with different precedence?
I expect that a mathematician writing out a large product and divsion, where some things are in the numerator and some in the denominator, would typically choose to only write a single vinculum or solidus, and have everything in the numerator in one place, and everything in the denominator in the other.
I mean, when you write out the quadratic formula do you put brackets around the 2a?
The vinculum is fine because the scope is clear from the context. I avoid the solidus whenever I can unless I can use additional typography to indicate the scope, and to be honest I really am gratuitous in the use of parentheses when there is the potential for more than one interpretation. ^{1+2}⁄_{4+8} is vaguely unsettling but I suppose it is unambiguous.
As for the quadratic formula? Yes. A compiler or a graphing calculator would expect to see brackets around the 2a, and would cheerfully give the wrong answer if they weren't there (and a wasn't 1). Therefore, my students had better be used to seeing them there whenever they use a solidus or an obelus to denote the division. (And Wikipedia tells me that ISO has deprecated the poor obelus. Someone should tell the calculator manufacturers.)
I'm sympathetic to the issue. To carry on Molina's argument that I mentioned in my previous post (and my memory is becoming clearer that it is Molina's argument), this is the one rule of operator precedence does not match our linguistic expectations, if you were to imagine the sentence "eight apples were divided between two friends". Or, indeed, it would seem pathological to interpret 8x ÷ 2x as 4x², so it would seem to be an unstated rule that we do treat the implicit multiplication between a variable and its coefficient as having a higher precedence than an explicitly stated division.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Frimble wrote:It can actually be derived from one of the 13 set axioms of the real numbers that multiplication takes precedence. (Although you only need 8 axioms really)
I'm not sure I believe that. Well ... actually, I'm sure I don't believe that!
After all, if I write 2^3^4 that means (by convention) 2^81. But there's no fundamental reason why the convention couldn't have been to interpret that as 8^4. The orders of precedence are conventions, not required by the axioms.
There's no axiom that tells you how to interpret 2 + 3 * 4. If you believe otherwise, please supply the demonstration or reference.
Here's the WIkipedia article. It indicates that the rules of precedence are conventions. It says nothing about axioms.
http://en.wikipedia.org/wiki/Order_of_operations
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
The distributivity axioms, as usually stated, need the standard precedence.
ie. a(b+c)=ab+ac is an axiom. If addition would take precedence over multiplication, than the parentheses are superfluous, but worse:
assume a,b,c and b+c nonzero.
a(b+c)=a(b+a)c, [multiplicative inverses exist]
b+c=(b+a)c, [multiplication distributes over addition]
b+c=b(c+a)c [addition is commutative]
b+c=b(a+c)c [multiplication commutes]
b+c=(a+c)bc [multiplication distributes]
b+c=a(b+c)bc [multiplication commutes]
b+c=(b+c)abc [multiplicative inverse of b+c exists]
1=abc
therefore a=b=c=2 provides a contridiction.
perhaps more convincing: a=1 shows that any nonzero elements of our "field" must be eachothers inverses this can only happen if 1 is the only element of group of units. (unitgroup? I had group theory in dutch... the group of elements which have a multiplicative inverse.)
Therefore the only field which could satisfy this is F_2.
But that is in fact not the case, 1(1+0)=1!=0=1(1+1)0.
So there is no field such that a(b+c)=a(b+a)c. This shows that doing addidition before multiplication is in general not admissible in a field. (Proving from the axioms that a field requires doing multiplication first, is left as an exercise for the reader)
ie. a(b+c)=ab+ac is an axiom. If addition would take precedence over multiplication, than the parentheses are superfluous, but worse:
assume a,b,c and b+c nonzero.
a(b+c)=a(b+a)c, [multiplicative inverses exist]
b+c=(b+a)c, [multiplication distributes over addition]
b+c=b(c+a)c [addition is commutative]
b+c=b(a+c)c [multiplication commutes]
b+c=(a+c)bc [multiplication distributes]
b+c=a(b+c)bc [multiplication commutes]
b+c=(b+c)abc [multiplicative inverse of b+c exists]
1=abc
therefore a=b=c=2 provides a contridiction.
perhaps more convincing: a=1 shows that any nonzero elements of our "field" must be eachothers inverses this can only happen if 1 is the only element of group of units. (unitgroup? I had group theory in dutch... the group of elements which have a multiplicative inverse.)
Therefore the only field which could satisfy this is F_2.
But that is in fact not the case, 1(1+0)=1!=0=1(1+1)0.
So there is no field such that a(b+c)=a(b+a)c. This shows that doing addidition before multiplication is in general not admissible in a field. (Proving from the axioms that a field requires doing multiplication first, is left as an exercise for the reader)
 gmalivuk
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Sure, because as usually stated they assume everyone's in agreement about the order of operations. But you could just as easily sayflownt wrote:The distributivity axioms, as usually stated, need the standard precedence.
Product[a,Sum[b,c]] = Sum[Product[a,b],[Product[a,c]]
Or even simply that a(b+c) = (a*b)+(a*c)
The axiom doesn't actually say anything different than it did before, it simply no longer assumes agreement about order of operations. Which it shouldn't, because as previously mentioned those are merely a communication convention, rather than an intrinsic part of the axiom system.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
Frimble wrote:I don't think that works. You can't change the order of the numbers in an infinite sum.
You're right, and that's why it doesn't equal 0, 1 or even 1/2.
I just mean that if it made sense to be able to do those operations, you can show it equals any rational number (which tells you there's a mistake there somewhere). I think you can also get algebraic rational numbers by multiplying things together, but transcendental numbers are probably impossible.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
So, so far, I don't think I've seen anyone that agrees that the aforementioned equation is answered with anything more than 13, much less
[1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13].
The original postulation is that this set of numbers are correct possible answers to the equation.
Does anybody want to take this part on directly?
[1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13].
The original postulation is that this set of numbers are correct possible answers to the equation.
Does anybody want to take this part on directly?

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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
mrmitch wrote:Frimble wrote:I don't think that works. You can't change the order of the numbers in an infinite sum.
You're right, and that's why it doesn't equal 0, 1 or even 1/2.
I just mean that if it made sense to be able to do those operations, you can show it equals any rational number (which tells you there's a mistake there somewhere). I think you can also get algebraic rational numbers by multiplying things together, but transcendental numbers are probably impossible.
There are rigourous ways to give those summations values. The easiest way is to say s=11+11+..., so 2s=1+(1+1)+(11)+(1+1)+...=1 (where the second value in the parenthesis is the second s), so s=1/2. Another way to do it is to take the average of all the values in the sequence so far, and take that to the limit, and set the sum to the value it converges to. One way to tell if a way to assign a value to a sum is consistent is to try it on a series that you know converges and you know the value, and see if it gives the right answer. For example, the "taking the average so far" way will always work with a convergent series, because if the series converges, then obviously the average converges. However, rearranging the numbers does not work, because it does not even work when the sum converges conditionally (as snowyowl pointed out).
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
 jestingrabbit
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
tomtom2357 wrote:The easiest way is to say s=11+11+..., so 2s=1+(1+1)+(11)+(1+1)+...=1 (where the second value in the parenthesis is the second s), so s=1/2.
That second equality uses an infinite rearrangement of the sum, which is liable to change the result when manipulating conditionally convergent sequences, so its definitely suspect. On the other hand, if we assert that if S = s0 + s1 + s2 + ... then
 for all constants C, C*S = C*s0 + C*s1 + C*s2 + ... and
 if t(i) = s(i+1) and T = t0 + t1 + t2 + ..., then S = s0 + T
Then we can conclude that if S = 11+11... then S = 1 + (1+11+1...) = 1  S, which has the finite solution S = 1/2. Both of my assumptions are valid for any sequence, not just unconditionally convergent ones, so it seems to me that there is a stronger claim on this being the "right" answer for what it converges to, rather than any other number, rational or not.
ameretrifle wrote:Magic space feudalism is therefore a viable idea.

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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
It is less an infinite rearrangement than termwise addition, which you can do with an infinite series. The first term of the series is just 1, the second term is the sum of the first and second terms of the original series, the third is the sum of the second and third terms of the original series, and so on. The new sum is the termwise addition of 2 of the old sums. You can do that with convergent series, so I don't see why it doesn't work with divergent series.
I have discovered a truly marvelous proof of this, which this margin is too narrow to contain.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
So, can anyone suggest where I could get an answer? This isn't an infinite series. It is a fixed length set of numbers with specific operators between them. There are also only a finite set of permutations, assuming that permutations are permissible in a question stated like this.
The tangents brought up here are interesting, but not directly related to the question. <sigh>
The tangents brought up here are interesting, but not directly related to the question. <sigh>
 gmalivuk
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
You got the answer. Like, several times already.
The reason people are now going on tangents is because there's nothing more to be said about the original question.
The reason people are now going on tangents is because there's nothing more to be said about the original question.
Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
"Answer"? Of course the answer is 13, or else it'd be impossible or very hard to do any calculation on this system. I believe some have said that order of operations are conventions, not axioms, and they are made for convenience.
 dudiobugtron
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Re: 1+1+1+1+1+1+1+1+1+11+1+1+1+1+1*0
dirkm wrote:So, so far, I don't think I've seen anyone that agrees that the aforementioned equation is answered with anything more than 13, much less
[1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13].
The original postulation is that this set of numbers are correct possible answers to the equation.
Does anybody want to take this part on directly?
dirkm wrote:So, can anyone suggest where I could get an answer?
OK I'll bite.
No one is taking that part on directly, because it is not a mathematical argument. It's just a handwaving argument using some mathematical jargon:
dirkm wrote:however if you use calculus to evalutate the expression the axioms of association for both multiplication and addition allow you to move the expression around into almost any order so long as the (1*0) remain together
That means you can actually come out with any answer from 1 to 13 and still be right
But I'll try to give an answer anyway:
In reality, the 'axiom of association' doesn't let you move the brackets around however you like when adding and subtracting. Even if you use 'calculus'. (It does let you move them around when adding, though.)
Also (as people have already explained at length above), even though the order of operations isn't set in stone by the axioms, you still have to agree on a particular order of operations before you write the axioms. If you use a different order of operations, you need to go back and rewrite the axioms so they are still true using the new OoO.
If you get that guy to write out his argument as a mathematical proof (starting from the axioms), you yourself will be able to see where it falls down. And if not, post it here and we'll be happy to show you.
So:
The short answer is no. For any particular order of operations you choose, at most one of those answers will be correct.
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