Paging Dr. Chomsky
We need a linguist up in here
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bonifaceviii
Canada2890 Posts
April 08 2011 18:36 GMT
#1601
Paging Dr. Chomsky We need a linguist up in here | ||
mpupu
Argentina183 Posts
April 08 2011 18:38 GMT
#1602
On April 09 2011 03:31 rackdude wrote: The operation of division is defined through the group which contains the set of the Integers and the operation division. In fact, the division on integers group is an Abelian group. Therefore the operator is a function from a pair integers to the set of integers that obeys the group axioms (which is really simple to prove and if you've ever taken math before you've already proved this so I'm letting this go as true). Not only is this true on the division operator, for each of the standard operators that you study in elementary school "math", the operator is part of an Abelian group. However, what this implies is that the input must be a pair of integers. 48÷2(9+3) does not give two integers, it gives 48 and some 2(9+3). However, if we denote that it is the value of 2(9+3), then we'd say 48÷(2(9+3)) and that is something division is defined on. However, we can also say it's (48÷2)(9+3) which is another way that division would be defined. While this doesn't have any relation to the original argument, I have to take you up on what you wrote because it is at least confusing and at worst, wrong. The integers with division as an operation are not an group, much less an Abelian group. The only operation for which this is true is addition. | ||
quiggy
Canada58 Posts
April 08 2011 18:38 GMT
#1603
48/2(9+3) Let 48/2= a a(9+3) a9+a3 12(48/2) 288 | ||
SharkSpider
Canada606 Posts
April 08 2011 18:41 GMT
#1604
On April 09 2011 03:35 Pufftrees wrote: Show nested quote + On April 09 2011 03:27 SharkSpider wrote: Some people are taught to approach these problems in a way that results in 2 Where do they teach Math incorrectly? I would like to know, seriously. Universities, mostly. Spend years writing division as big lines and never, ever bothering with multiplying fractions by numbers (ie, one big line, all the time) and that / becomes a symbol for isolating two sides of an expression and dividing them. I tossed this to a few people taking math at Waterloo and UT (arguably the two top math schools in Canada) and most people said it looked like a two. If we talk math problems on facebook or MSN (and we do), posting something like in the OP would mean 2 or 1/(2x) every time. That's just how it's developed, and our professors would never waste time clarifying the official arithmetic. I'll still maintain that most people discussing this topic don't know how to define "2" in any mathematical way. | ||
mpupu
Argentina183 Posts
April 08 2011 18:44 GMT
#1605
On April 09 2011 03:38 quiggy wrote: Basic level math and algebra doesn't work here. Most of you are trying to apply you lack of knowledge to this. Anyone with anything beyond highschool its clearly: 48/2(9+3) Let 48/2= a a(9+3) a9+a3 12(48/2) 288 When you're substituting 48/2 with a, you're applying Leibniz rule. You have to be careful when doing such substitutions, as evidenced by the following example: 2/2 = 1 therefore 12/24 = 114 (i.e. replace 2/2 by 1 in 12/24) This is clearly wrong. The same thing happens with the original expression if you interpret the implicit product as having higher precedence than the division. This interpretation is not right or wrong, because it's a writing convention instead of a fundamental axiom. | ||
rackdude
United States882 Posts
April 08 2011 18:46 GMT
#1606
On April 09 2011 03:27 Danjoh wrote: Show nested quote + On April 09 2011 03:17 rackdude wrote: On April 09 2011 03:05 dbosworld wrote: Sooooo What is this correct answer? Why do people always think there is one correct answer when it comes to math... the moment you start doing math you realize how wrong that is... + Show Spoiler + There is no correct answer here. There is a correct answer for each interpretation of the input, but since the input is ambiguous in the most proper sense (it would require one more set of brackets), then depending on how you put the missing set of brackets there is a correct answer. A lot of people have an interpretation of the input that reads from left to right and applies the order of operations and thus brackets like (48÷2)(9+3), others (including a lot of calculators) interpret the input when involving division signs as top divided by bottom until there's a space (like in mathematica), and thus they would bracket it as (48)/(2(9+3)). For either of those you will get a correct answer, but for the original question the correct answer is that you have a syntax error and thus are just using shorthand and if people misinterpret what you meant by your shorthand then you should probably change the way you wrote the problem. Soo.... anyone dare to use that argument to get their scores up on a math test? The different methods I've tried all resulted in 288, if you for some reason got a syntax error, you probably just need to add the * between the 2 and the (. And I see some arguments that ÷ is different from /. There isn't. My keyboard has the symbolr ÷, but types out /, my calculator has the symbol ÷, but types out /, my friends calculator has the symbol ÷ and also types out ÷. And his calculator also calculated 48÷2(9+3) as 288. Would you also argue that 2*2 is different from 2x2? (assuming that in this case x is the operator and not the variable) Uhh... axiom of choice ring a bell? What about not excepting the excluded middle for a given proof? How have you not taken math and gotten into an argument with a teacher over that... So there are two ways to show you're wrong. First of all, lets go back to logic 101: look back at Godel's Incompleteness Theorems and you will see you're flat out wrong. As a consequence of these theorems there will be problems defined by our set theory that we cannot prove with our set theory unless our axioms are inconsistent. If that doesn't do it for you, here's an example. I remember at least one problem on an analysis test that was to get as far as you can for proving the limit of a function at irrational and rational numbers that is defined as: 1/q for all rational number in the form of p/q where q is in lowest form. 0 for all irrational numbers. And then you go along and then you go aha, if and only if it holds true that the delta as you approach the rational number converges "quicker" than the values of 1/q, then you have a non-zero limit for rational numbers. And then you we were allowed to look stuff up and low and behold there was this cool theorem I found that would give a limit for at least a subset of the rational numbers (I think it was the Loiuville numbers) that states that if you accept the axiom of choice then you will have a convergence. If I recall correctly I think it also proved that there can exist no proof without the axiom of choice. However, not all mathematicians accept the axiom of choice. That's like the first thing you learn in an actual math class. So is it true or false that it converges at Louiville numbers? So, is there always an answer to every math question as it's stated without any extra conditions? If you still say yes... well then, I don't know what to say to you. | ||
jinorazi
Korea (South)4948 Posts
April 08 2011 18:46 GMT
#1607
anyone who has math experience will look at this and see "2(9+3)" as a whole and will result in 2 or take ÷ / as fraction. then take a second look and will realize its literally written 48 / 2 * 12 which will give 288. whichever it is, the expression isn't clear cut for it to result in 288 or 2, correct equation should be used, (48/2)(9+3) and 48/(2(9+3)) respectively to avoid any confusion. | ||
Insanious
Canada1251 Posts
April 08 2011 18:47 GMT
#1608
On April 09 2011 03:35 Pufftrees wrote: Show nested quote + On April 09 2011 03:27 SharkSpider wrote: Some people are taught to approach these problems in a way that results in 2 Where do they teach Math incorrectly? I would like to know, seriously. In my university, implied multiplication goes before non-implied multiplcation... as such in my math tests 10/2(3) = 10/6 where 10/2*3 = 15 Although I am not taking puremath, I am taking finance... 99% of what I am tested on is math. This is simply how things are done here... might be different else where. But to be fair, many graphic calculators, as well as my own scientific calculator put implied > non-implied and give a result of 2 for the question given in the OP. | ||
rackdude
United States882 Posts
April 08 2011 18:48 GMT
#1609
On April 09 2011 03:38 mpupu wrote: Show nested quote + On April 09 2011 03:31 rackdude wrote: The operation of division is defined through the group which contains the set of the Integers and the operation division. In fact, the division on integers group is an Abelian group. Therefore the operator is a function from a pair integers to the set of integers that obeys the group axioms (which is really simple to prove and if you've ever taken math before you've already proved this so I'm letting this go as true). Not only is this true on the division operator, for each of the standard operators that you study in elementary school "math", the operator is part of an Abelian group. However, what this implies is that the input must be a pair of integers. 48÷2(9+3) does not give two integers, it gives 48 and some 2(9+3). However, if we denote that it is the value of 2(9+3), then we'd say 48÷(2(9+3)) and that is something division is defined on. However, we can also say it's (48÷2)(9+3) which is another way that division would be defined. While this doesn't have any relation to the original argument, I have to take you up on what you wrote because it is at least confusing and at worst, wrong. The integers with division as an operation are not an group, much less an Abelian group. The only operation for which this is true is addition. Check my edit. I wanted it to be integers to make it as simple as possible, but yeah, totally forgot that it had to be at least the rational numbers. And the relation to the original argument is pretty clear... it just says that by the way it is most commonly defined you better explicitly state what the two numbers we are doing the operation on are. It's not explicitly stated in the OP. | ||
Kazzoo
France368 Posts
April 08 2011 18:49 GMT
#1610
No, the thing that is concerning is that there are poeple who think it's 288 and say that 1/2x = 1/(2*x) | ||
dani_caliKorea
730 Posts
April 08 2011 18:50 GMT
#1611
http://forum.bodybuilding.com/showthread.php?t=133389973 | ||
mpupu
Argentina183 Posts
April 08 2011 18:51 GMT
#1612
On April 09 2011 03:48 rackdude wrote: Show nested quote + On April 09 2011 03:38 mpupu wrote: On April 09 2011 03:31 rackdude wrote: The operation of division is defined through the group which contains the set of the Integers and the operation division. In fact, the division on integers group is an Abelian group. Therefore the operator is a function from a pair integers to the set of integers that obeys the group axioms (which is really simple to prove and if you've ever taken math before you've already proved this so I'm letting this go as true). Not only is this true on the division operator, for each of the standard operators that you study in elementary school "math", the operator is part of an Abelian group. However, what this implies is that the input must be a pair of integers. 48÷2(9+3) does not give two integers, it gives 48 and some 2(9+3). However, if we denote that it is the value of 2(9+3), then we'd say 48÷(2(9+3)) and that is something division is defined on. However, we can also say it's (48÷2)(9+3) which is another way that division would be defined. While this doesn't have any relation to the original argument, I have to take you up on what you wrote because it is at least confusing and at worst, wrong. The integers with division as an operation are not an group, much less an Abelian group. The only operation for which this is true is addition. Check my edit. I wanted it to be integers to make it as simple as possible, but yeah, totally forgot that it had to be at least the rational numbers. It's still false. What's the (group) identity? Also, it cannot be Abelian because the operation is not commutative. | ||
Ace
United States16096 Posts
April 08 2011 18:53 GMT
#1613
On April 09 2011 03:49 levelnoobz wrote: The real problem isn't that there are people who voted what you didn't voted for, since all in all, it's a pretty dumb debate, the writting system being really confusing and theres a reason why we don't write divisions like this. No, the thing that is concerning is that there are poeple who think it's 288 and say that 1/2x = 1/2*x Variables and constants behave differently. 48/2(9+3) is not the same issue as 1/2x. | ||
MasterOfChaos
Germany2896 Posts
April 08 2011 18:53 GMT
#1614
On April 09 2011 03:49 levelnoobz wrote: The real problem isn't that there are people who voted what you didn't voted for, since all in all, it's a pretty dumb debate, the writting system being really confusing and theres a reason why we don't write divisions like this. No, the thing that is concerning is that there are poeple who think it's 288 and say that 1/2x = 1/2*x You can argue that implicit multiplication by a variable binds stronger than implicit multiplication with a bracketed expression. WolframAlpha takes this stance for example. I for one prefer interpreting both as binding stronger than normal multiplication/division. Simply because it's more useful in practice, and I'm convinced most of my fellow students would at my faculty would interpret it the same way. | ||
jacen
Austria3644 Posts
April 08 2011 18:57 GMT
#1615
On April 09 2011 00:48 Ceril wrote: Math, the language that cannot lie. Obviously depending on school, local notation habits and intepreation rules. The universal language is not universal at all =( It is universal. But if you start leaving out stuff or simplifying notation/syntax, you have to state your conventions. Everyone really trying to get across a message would not leave out parenthesis that make the statement ambiguous. | ||
rackdude
United States882 Posts
April 08 2011 18:58 GMT
#1616
On April 09 2011 03:51 mpupu wrote: Show nested quote + On April 09 2011 03:48 rackdude wrote: On April 09 2011 03:38 mpupu wrote: On April 09 2011 03:31 rackdude wrote: The operation of division is defined through the group which contains the set of the Integers and the operation division. In fact, the division on integers group is an Abelian group. Therefore the operator is a function from a pair integers to the set of integers that obeys the group axioms (which is really simple to prove and if you've ever taken math before you've already proved this so I'm letting this go as true). Not only is this true on the division operator, for each of the standard operators that you study in elementary school "math", the operator is part of an Abelian group. However, what this implies is that the input must be a pair of integers. 48÷2(9+3) does not give two integers, it gives 48 and some 2(9+3). However, if we denote that it is the value of 2(9+3), then we'd say 48÷(2(9+3)) and that is something division is defined on. However, we can also say it's (48÷2)(9+3) which is another way that division would be defined. While this doesn't have any relation to the original argument, I have to take you up on what you wrote because it is at least confusing and at worst, wrong. The integers with division as an operation are not an group, much less an Abelian group. The only operation for which this is true is addition. Check my edit. I wanted it to be integers to make it as simple as possible, but yeah, totally forgot that it had to be at least the rational numbers. It's still false. What's the (group) identity? Also, it cannot be Abelian because the operation is not commutative. Ehh, didn't think it all the way through. I was just thinking the inverse of multiplication so it must be a group. Multiplication is a group on the rationals so I was just jumping ahead to say division was too so that way it would seem less confusing than talking about multiplication then defining division as a way of dealing with rational numbers and then getting to the conclusion that it needs to be explicitly stated for the definition. However, yeah, that assumption was wrong and I couldn't just quickly jump to it being a group. Post deleted, thank you for pointing that out. | ||
Kazzoo
France368 Posts
April 08 2011 18:58 GMT
#1617
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shadowy
Bulgaria305 Posts
April 08 2011 18:58 GMT
#1618
On April 09 2011 03:24 Severedevil wrote: Instead, it's a clash between people who're aware of differing conventions for implicit multiplication versus explicit multiplication, and people who aren't + refuse to learn. Thank you - so, well said. Just for the sake of arguing - some basic algebra: 48 / 2(9+3) = 48 / (2*9 + 2 *3) = 2 Did I do this wrong? And if not, how it's my interpretation any different than yours (for those keep repeating it's 288). There is no right or wrong answer here - just BADLY, POORLY, WRONGLY written equation, which can not give correct output, since the input is wrong. Duhhhh! | ||
SharkSpider
Canada606 Posts
April 08 2011 18:58 GMT
#1619
On April 09 2011 03:57 jacen wrote: Show nested quote + On April 09 2011 00:48 Ceril wrote: Math, the language that cannot lie. Obviously depending on school, local notation habits and intepreation rules. The universal language is not universal at all =( It is universal. But if you start leaving out stuff or simplifying notation/syntax, you have to state your conventions. Everyone really trying to get across a message would not leave out parenthesis that make the statement ambiguous. This. Write it out in C, write it out in LISP (or any programming language), you get zero ambiguity. LaTeX it up and use a real division line, zero ambiguity. Use notation reserved for middle schools and "Math Essentials," and you get ambiguity. Not surprising. | ||
Ace
United States16096 Posts
April 08 2011 19:03 GMT
#1620
On April 09 2011 03:58 shadowy wrote: Show nested quote + On April 09 2011 03:24 Severedevil wrote: Instead, it's a clash between people who're aware of differing conventions for implicit multiplication versus explicit multiplication, and people who aren't + refuse to learn. Thank you - so, well said. Just for the sake of arguing - some basic algebra: [b]48 / 2(9+3) = 48 / (2*9 + 2 *3) = 2 Did I do this wrong? And if not, how it's my interpretation any different than yours (for those keep repeating it's 288). There is no right or wrong answer here - just BADLY, POORLY, WRONGLY written equation, which can not give correct output, since the input is wrong. Duhhhh! why are you distributing 2 to (9+3)? These aren't variables these are constants. | ||
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