Poll: Solve a simple math problem

[NightlyNews]

0=1-1
0=1-1+1-1
0=1-1+1-1+1-1
0=1-1+1-1+1-1+.....
0=1+(-1+1)+(-1+1)+....
0=1+0+0+0+...
0=1

The cardinality of the -1 group is one larger than the cardinality of the +1 group so they can't be canceled out equaling zero.

No matter how long the string goes on even infinitely the cardinality (number of numbers in a group) of the -1 group will always be one larger than the cardinality of the +1 group.

Pretty common sense, but thats the mathematical explanation as to why that's not true or any sort of flaw I mathematics.

 

[TheTejs]

Give it up, the people whom are determinated on telling you thats it's not 14 are either trolling or have misunderstood the basic princibles of math.

oh i remember its called SIMPIFICATION OF NUMBERS!

Gill is trolling, someone who does math at university level would now that simplification of numbers have nothing to do with this.

It's insane how much trolling that comes out of math equations...

 

[RHadley]

Seriously. Anything Multiplied by 0 equals zero. Think of it like the whole "This many lots of this" system. If you have no lots of 2 (2x0 - or 2*0), then you obviously have zero.

 

[gamefreakbsp]

Damn you PEMDAS......DAMN YOUUUUUUUUU!! The order of operations has foiled me again.

 

[Lukeje]

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

You're at the point now where you're not arguing about what I'm trying to explain, which makes this rather pointless. I'm trying to show the problem inherent with trying to treat a concept as a number. If you can't grasp that point, there's really nowhere to go from here.

You can't pretend to be treating infinity as a `number' if you start off by defining an operation which is fundamentally un-`numberlike' (in this case addition without the group properties). That is my point.

 

[Farotsu]

Not this again...

Anything multiplied by 0 is 0 but you only multiply the last 1 with 0 as per order of operations. So answer is not 0.

Order of operations also dictates that Parentheses > Exponents > Multiplying = Divison > Addition = Substraction. Which means that the order of operation when dealing with Multiplying and Division is handled from left to right. Same with Addition and Substraction. So the answer is not 4.

-1 cancels out one of the +1 and that 0 multiplier cancels out the last +1 which makes the answer 14.

Also: Anything divided by 0 is Undefined. You can't divide by 0 within our current math system and for a good reason. However lim x->0, 1/x->infinity is correct and is what makes people so confused about it.

 

[SeaCalMaster]

One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.

That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.

I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

Lesson time!

When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0. As for whether or not infinity + n = infinity, that depends on the exact definition of addition you're using. However, a definition that makes it true that infinity + n = infinity for all n breaks associativity.

Well i'm not sure but my maths text book says 1/0 is infinity so no ofense but i'll trust that over you.

Are you sure it doesn't just say that the limit of 1/x as x->0 is infinity? Infinity is a tricky concept...

Pretty sure as some questions in the book required you to use 'infinity = 1/0' in order to get to the correct answer.

Example please?

Riemann Sphere [http://en.wikipedia.org/wiki/Riemann_sphere]

AAH WHY DO THE ANALYSTS HAVE TO RUIN EVERYTHING
THE EXTENDED COMPLEX NUMBERS AREN'T EVEN A GROUP UNDER ADDITION

Also, it's not technically correct to say that the limit of 1/x as x->0 is infinity. The series diverges, so it's not even really true that the limit exists.

If you're using a set where -infinity = infinity (such as the Riemann Sphere), then 1/x in fact converges as x->0, as the limits from both sides are equal.

No, it doesn't, at least not under the standard metric. If we use the standard definition where |infinity - n| = infinity (if n != infinity), then the distance from 1/x to infinity for any (non-zero) value of x is infinite.

 

[Shreddie]

The problem with the answer being 4 since 10-6+1*0 = 4 is that (1+1+1+1+1+1+1+1+1+1)-(1+1+1+1+1+1)+(1*0) does not equal the original problem. The -(1+1+1+1+1+1) part is the same as (-1)*(1+1+1+1+1+1) which equals(-1-1-1-1-1-1). Therefore (1+1+1+1+1+1+1+1+1+1)-(1+1+1+1+1+1)+(1*0) = 1+1+1+1+1+1+1+1+1+1-1-1-1-1-1-1+1*0, which is completely different than the original.

 

[Lukeje]

One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.

That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.

I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

Lesson time!

When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0.

...unless infinity = -infinity of course.

 

[SeaCalMaster]

One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.

That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.

I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

Lesson time!

When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0.

...unless infinity = -infinity of course.

No, actually. If infinity = -infinity, then infinity - infinity is still 0, even though it would, in fact, mean that infinity + infinity = 0.

 

[Lukeje]

One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.

That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.

I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

Lesson time!

When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0.

...unless infinity = -infinity of course.

No, actually. If infinity = -infinity, then infinity - infinity is still 0, even though it would, in fact, mean that infinity + infinity = 0.

The problem, I think, is that the identity becomes non-unique when infinity is involved. The property infinity + n = infinity may be recognised as being of the same form as the definition of an identity element: k + 0 = k. Thus from the `point of view' of infinity every other element of the set is the identity element. Thus subtraction cannot be uniquely defined, and both the definitions of infinity-infinity are valid (and there are in fact infinitely many other valid definitions). Sorry if this is slightly hand-wavy (or nonsensical); some of it's kind of `stream of consciousness' stuff.

 

[SeaCalMaster]

One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.

That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.

I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

Lesson time!

When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0.

...unless infinity = -infinity of course.

No, actually. If infinity = -infinity, then infinity - infinity is still 0, even though it would, in fact, mean that infinity + infinity = 0.

The problem, I think, is that the identity becomes non-unique when infinity is involved. The property infinity + n = infinity may be recognised as being of the same form as the definition of an identity element: k + 0 = k. Thus from the `point of view' of infinity every other element of the set is the identity element. Thus subtraction cannot be uniquely defined, and both the definitions of infinity-infinity are valid (and there are in fact infinitely many other valid definitions). Sorry if this is slightly hand-wavy (or nonsensical); some of it's kind of `stream of consciousness' stuff.

If infinity + n = infinity for all n (including infinity itself), there is no value for which infinity + n = 0, and so infinity doesn't have an additive inverse. Then, the expression x - infinity is meaningless in much the same way that the expression x/0 is meaningless.

 

[DanDanikov]

With real numbers, dividing by zero has no meaning. You get a complex infinity if you're having fun with a Riemann sphere, but that's going in deep with complex numbers.

Again, this highlights the important of context, axioms, and assumptions. The 'problem' states nothing of the sort, hence the multiple valid answers. There is a visual 'trick' (the minus sign snuck in there) that might catch people out, but that hardly makes them stupid, just inattentive.

Think 0 is a wrong answer? Take out a pocket calculator and input the question as stated. You'll get 0. Are you going to now start saying calculators are wrong? Plug it into Wolfram Alpha. You get 14. They can't both be right...

Unless they have different notions of operator precedence. 'Calculator' maths isn't wrong, it's just different. Stop being operator precedencists!

 

[Harlief]

reading left to right, you get zero; but looking at all the discussion about what the answer actually is should show you how badly phrased this equation is.

No.

It shows how badly 50% of those who have answered the poll are at reading mathematical equations, and/or their misunderstanding of the Order of Operations.

My point was that this is the mathematical equivalent of Engrish. Nothing about this equation is simple, it is intentionally confusing. You wanna talk about BEDMAS? Where are the brackets? This equation could definitely use some for purposes of clarity.

 

[madwarper]

My point was that this is the mathematical equivalent of Engrish. Nothing about this equation is simple, it is intentionally confusing.

I completely understood the point you were trying to make. I also know that its completely incorrect.

You wanna talk about BEDMAS? Where are the brackets?

Well, I prefer PEMDAS, but if YOU want to use BEDMAS, then by all means...

First, there aren't any brackets. So, you skip the .
Also, there aren't any exponents listed, which you didn't mention, and the [E] is also skipped.
Then, you get to the [D/M], there is Multiplication and it's preformed left to right.
Finally, there's [A/S] and it's preformed left to right.

This equation could definitely use some for purposes of clarity.

Could it have been written more clearly? Yes.
Did it need to have been written more clearly? NO!

 

[Lukeje]

One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.

That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.

I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

Lesson time!

When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0.

...unless infinity = -infinity of course.

No, actually. If infinity = -infinity, then infinity - infinity is still 0, even though it would, in fact, mean that infinity + infinity = 0.

The problem, I think, is that the identity becomes non-unique when infinity is involved. The property infinity + n = infinity may be recognised as being of the same form as the definition of an identity element: k + 0 = k. Thus from the `point of view' of infinity every other element of the set is the identity element. Thus subtraction cannot be uniquely defined, and both the definitions of infinity-infinity are valid (and there are in fact infinitely many other valid definitions). Sorry if this is slightly hand-wavy (or nonsensical); some of it's kind of `stream of consciousness' stuff.

If infinity + n = infinity for all n (including infinity itself), there is no value for which infinity + n = 0, and so infinity doesn't have an additive inverse. Then, the expression x - infinity is meaningless in much the same way that the expression x/0 is meaningless.

You're still hanging on to the idea of a unique identity. The equation you are looking for is x - infinity = identity. 0 is not the only number with the property of an identity with respect to subtraction of infinity, thus one must specifically define subtraction in this case; one cannot derive it from the properties of the `other numbers'. I chose to define it in a way that stayed consistent, you defined it in a way that lead to contradictions. Either way, we have to define subtraction, you can't just assume it to be a general `property of numbers' (my original point, which seems to have gotten lost along the way). It is generally left undefined (see e.g. http://en.wikipedia.org/wiki/Riemann_sphere ). Think about it; in order to define subtraction of infinity from infinity you either have to give up the uniqueness of the identity element (a property you seem rather fond of) or get massive inconsistencies (infinity = 0, etc.).

 

[Maze1125]

One can of course make the assumption that infinity - infinity = 0. This leads to contradictory results and thus seems like a bad candidate, as you've discovered.

That is an actual way to construct a mathimatical proof, which was the entire point in the first place. If you treat infinity like a number and attempt to perform normal aritmetic operations on it (such as infinity - infinity = 0), you get invalid results. Ergo, infinity can not be treated that way.

I'm not saying that infnity - infinity = 0 is valid. I'm saying that it would be valid if infinity were a number, which, as demonstrated, doesn't work.

No. You seem to have this strange idea of `number'. If it were being treated as a `number', then infinity + n =/= infinity. This is a property of general numbers. If you have to specify a special rule for addition of infinity, then why should it behave `normally' under subtraction from itself? You have only proved that a system with subtraction of two infinities defined to be zero is inconsistent. This is not the same as saying that all sets of numbers including infinity are inconsistent.

Lesson time!

When we say "x-y", what we mean by that is "Add x to the additive inverse of y." The additive inverse of y, by definition, is the number which, when added to y, gives the additive identity (normally called 0) as a result. Therefore, if the expression "infinity - infinity" has any meaning at all, it cannot be equal to anything other than 0. As for whether or not infinity + n = infinity, that depends on the exact definition of addition you're using. However, a definition that makes it true that infinity + n = infinity for all n breaks associativity.

Lesson time!

In number systems where infinity is a member, infinity is almost always considered to not have an additive inverse.
Just like 0 is an exception and doesn't have a multiplicative inverse in the Real numbers.

Hence, just as 0/0 zero is undefined, infinity - infinity is undefined.
The first doesn't stop 0 from being a useful number, nor does the latter stop infinity from it.

Well i'm not sure but my maths text book says 1/0 is infinity so no ofense but i'll trust that over you.

Are you sure it doesn't just say that the limit of 1/x as x->0 is infinity? Infinity is a tricky concept...

Pretty sure as some questions in the book required you to use 'infinity = 1/0' in order to get to the correct answer.

Example please?

Riemann Sphere [http://en.wikipedia.org/wiki/Riemann_sphere]

AAH WHY DO THE ANALYSTS HAVE TO RUIN EVERYTHING
THE EXTENDED COMPLEX NUMBERS AREN'T EVEN A GROUP UNDER ADDITION

Also, it's not technically correct to say that the limit of 1/x as x->0 is infinity. The series diverges, so it's not even really true that the limit exists.

If you're using a set where -infinity = infinity (such as the Riemann Sphere), then 1/x in fact converges as x->0, as the limits from both sides are equal.

No, it doesn't, at least not under the standard metric. If we use the standard definition where |infinity - n| = infinity (if n != infinity), then the distance from 1/x to infinity for any (non-zero) value of x is infinite.

So you're claiming that f(x) = x -/> infinity as x -> infinity?

Pro-tip, the definition of tending to infinity is different to the definition of tending to a finite number.

1/x -> infinity as x -> 0 from above, and 1/x -> -infinity as x -> 0 from below.
So, if infinity = -infinity, the limits from both sides are the same, and so, by the definition of convergence, the function converges at 0.

 

[akfg666]

The answer is zero because the * mean times by and anything times by zero is...zero!

It would if everything but times zero was in parenthesis. Sadly, it isn't, and thus, amazingly enough, you ONLY MULTIPLY THE NUMBER THAT IS BEING MULTIPLIED. As in, the last 1. Order of operations, people.

I believe the phrase mindfuck applies here! o_0

 

[Maze1125]

Think 0 is a wrong answer? Take out a pocket calculator and input the question as stated. You'll get 0. Are you going to now start saying calculators are wrong? Plug it into Wolfram Alpha. You get 14. They can't both be right...

When I put it into my calculator, I get 14.
My suggestion, get a better calculator.

'Calculator' maths isn't wrong,

There's no such thing. Calculators are just tools, they do what you tell them. If you tell your calculator to do operations in a particular order, that's what it'll do, but there's nothing stopping you from telling it a different order. You're the one in control, not the calculator. It's up to you to pick the right order.

 

[OldRat]

The answer is zero because the * mean times by and anything times by zero is...zero!

It would if everything but times zero was in parenthesis. Sadly, it isn't, and thus, amazingly enough, you ONLY MULTIPLY THE NUMBER THAT IS BEING MULTIPLIED. As in, the last 1. Order of operations, people.

I believe the phrase mindfuck applies here! o_0

See, I'll show you:

This is how you'd get a zero: (1+1+1+1+1+1+1+1+1+1-1+1+1+1+1+1+1)*0

Note the parenthesis, meaning everything inside overrides the usual order of operations (or is the first thing in order of operations, however you prefer). This way, you would get a zero.

HOWEVER, this is not the case. The case is: 1+1+1+1+1+1+1+1+1+1-1+1+1+1+1+1+1*0
No parenthesis, no special priority for the filthy masses of addition (and one miserly case of substraction)

Order of operations goes (as you really should already know and I shouldn't need to explain, since this is taught in elementary schools): parenthesis, exponents and roots, multiplication and division, and finally addition and subtraction. Going from left to right in case of equals. You solve the things in that order.

So, the first thing you solve is the multiplication. One times zero. That's zero. So the whole thing looks like this: 1+1+1+1+1+1+1+1+1+1-1+1+1+1+1+1+0
And the rest is just counting 1's.

Got it now?

And on a sidenote, I'm still honestly, actually amazed people manage to screw this up. THIS, of all things. Such a simple, elementary school level thing.