PEDMAS left to right solving parentheses: (9+3)=12 which leaves us with a string of operations of the the same rank 48/2*12 solve that left to right 48/2 = 24 24*12 = 288
/ should really be ÷ in my post above or, as bchad pointed out, the fractional line will act as a grouping symbol.
In the original equation does the fact that it reads 2(9+3) and not 2*(9+3) make a difference? idk. I think the answer is 2 48÷2(9+3) --> 48÷24=2 Where as 48÷2*(9+3) --> 24*12=288
I too think the answer is 288, but… Substitute x for (9+3), so x = 12 48 ÷ 2x = 2
So it is controversial after all, Geeez.
FOIL?
Will I get in trouble if I burn this math problem?
Yah… it’s funny to see that such a simple question can draw so much of controversy. You guys probably have seen a bunch of questions like 48 ÷ 2a… then “a” is multiplied to 2 not 24… but the problem about this quesion is that “a” is an integer (9+3). I believe it’s mathematically wrong to exclude * between integers just like 2*3 is not equal to 23. But there is also a bracket between the integers as well… 2(3) = 2*3 so I think the answer is 2.
Mobius Striptease Wrote: ------------------------------------------------------- > 48/(2*(9+3))=2 > 48/2*(9+3)=288 > > the answer to the original question is obviously > 288 and nothing else - if you assume parenthesis > in places where there arent any, of course you’ll > get a wrong answer Thanks Mobius. This was insightful
So since I’m no expert at math here, you guys are saying that using a ÷ will run left to right, but using a / also groups everything to the right? I never knew that?
Well, when you are doing it in text, using a “/” also goes left to right. The issue is really about how you translate “text formulas” into what you would write on a piece of paper. 48/2(9+3) could be 48/2 * (9+3) = 24 * 12 = 288 or 48 — * (9+3) = 24 * 12 =288 2 But it is not hard to see why someone (like me, earlier) might read it is 48 --------- = 48/24 = 2 2(9+3) If we have a piece of paper, there’s no ambiguity, but in text-ish, it’s easy to mess up. Mobius is right that the error involves mentally putting parentheses where they aren’t there, but, maybe because I learned algebra before computers were widespread in elementary schools, I visually associate “/” with the ratio bar. When I do excel work, I often use more parentheses than I technically need to resolve these kinds of ambiguities.
I saw it as 288 right away, i don’t see how you can possible mistake it for anything else.
If you looked at it quickly, then yes you can make the mistake of grouping: 48÷2(9+3) -> 48 times 12 ___ 2 You need to add brackets if you don’t want division first. That is the whole point of adding parentheses/brackets. The calculator is always right. The only math controversy I’ve come across (well the most common anyway) is: 1/3 = .33333 repeating .33333 repeating * 3 = .9999 repeating 3(1/3) = 1 .9999 repeating = 1?
0.999 repeating does equal 1 Proof Let x=0.9999 repeating Then 10x - x = 9.9999repeat - 0.9999repeat = 9.0000 Solving for x: 10x - x = 9x = 9 x=1 Alternately… Ask yourself what number you must subtract from 1 to get 0.9999repeat You eventually convince yourself it has to be zero. Thus 0.99999repeat is another representation of 1, just like 5/5ths.
Bchad, I don’t think that’s theoretically correct. All of your answers are subject to some level of rounding. Thus, I think the answer to parts of your proof would be unsolvable with exact precision (no rounding) using discrete mathematics. Instead, wouldn’t this become a basic calc limit problem where “0.99999repeating” is replaced with "LIM as x->0 : f(x) = 1-x, x<>0. The result may be the same, but I think the process may be different to get there. Maybe I’m being nitpicky. Also, I could be wrong. I’m retaking my undergrad calc courses now in the evenings, so maybe to the man with the hammer everything’s a nail in my case.
Well, the proof with equations is your standard “rationalization of repeating decimals” procedure and works with discrete math. The proof without an equation is a heuristic argument that says that you can show that, 0.999repeat = 1 to any precision required. This is a more physics-y style of proof, sort of like the dirac delta function. I’m not sure if it holds up in the pure math world, but I think it can be done with limits. More fun: 0.1111repeat = 1 in base 2.
Approaches != Equals, come one guys this math!
So comp-sci… prove that (1 - 0.99999repeat) != 0 for me.
bchadwick Wrote: ------------------------------------------------------- > So comp-sci… prove that > > (1 - 0.99999repeat) != 0 for me. Easy. What is a definition of 0? 0*a = 0. now, lets say we have 9*10^-n. Let’s multiple this by a = 0^n we get 9*10^(n-n) = 9. 9 != 0
It’s definitely been well debated: http://en.wikipedia.org/wiki/0.999… Like the fractional equation, the algebraic equation is still a manipulation of an equation. It’s not necessarily an explanation since the fractional equation alone could be regarded as a proof. In calculus, we go into explaining this phenomenon through the use of limits and note that the repeating number is a process. In order to understand that process, one needs to understand the concept of infinity. Unlike the originally posted PEMDAS equation, there are still mathematicians and critics out there who challenge this concept of the repeating 9’s. For logic would tell us that a repeating decimal would not equal a real number; we are taught that point the repeating number is that it should never lead to a fixed number – again, it goes against logic. (even if our calculations tell us otherwise) It isn’t the same same as x/x = 1 because that is easily understood. Ask people to put .9999 repeating on a number line compared to 1 and how many would say .9999 repeating < 1? As a fifth grader, this problem blew my mind.