basicman wrote:
The problem seems to be that there is multiplication and division but the parenthesis are always first.
What is *INSIDE* the parenthesis are always first. The implicit multiplication is *NOT* always first, it is just multiplication.
Quote:
I do not think that you are right and if there is some proof of this I would like to see it. 6/2(1+2) is the same as 6/2x. Solve for X=(1+2)=3. So 6 would be divided by 6. I do not see how you could look at this any other way.
Uh, no, you botched that.. 6/2x = 6/2*x = 3*x. Implicit multiplication does not automatically get higher precedence.
Multiplying by x is dividing by 1/x and adding x is subtracting -x; those who stand behind the linearity of PEMDAS would have me rearrange my interpretation between two of either M/D or A/S to compensate for the form in which one of them arrives in the expression.
For example: 3–4+5 and A>S. If I asked you to substitute the minus-four with a plus-negative-four, you would need to show me 3+(-4–5). I'm not so keen on incorporating adjacent terms into alterations on one of them. This, to me, defeats the purpose of term decomposition.
As for explicit versus implicit, I can understand 2*5log(100) ≠ 2 * 5log(100), but I shall go no further than that. (And please define this intended behavior in the abstract.)
Weregoose wrote:
As for explicit versus implicit, I can understand 2*5log(100) ≠ 2 * 5log(100), but I shall go no further than that. (And please define this intended behavior in the abstract.)
All you did was add some spaces in there...
But there is zero difference between explicit vs. implicit multiplication. None, zilch, nada - the two are identical.
I didn't preview hard enough... 2/5log(100) ≠ 2 / 5log(100)
Where the numerator and denominator are separated by a space.
Picture unrelated.
KermMartian wrote:
Since when do multiplication and division have the same precedence? It's PEMDAS, not PE(MD)(AS)...
They're the same operation. You're living in a strange world if you want 6/5*(5/6) to be different from 6/5/(6/5)
Seeing this thread helps me to realize just how terrible American -m-a-t-h- arithmetic education is. There's no excuse to actually have to think about order of operations after, say, 4th grade unless you're actually writing your own compiler and have to worry about more than just the usual arithmetic operators.
KermMartian wrote:
Since when do multiplication and division have the same precedence? It's PEMDAS, not PE(MD)(AS)...
For the record, it's usually referred to as "BODMAS" in the UK (Brackets Orders Division Multiplication Addition Subtraction) which exchanges the order of division and multiplication in comparison to the mnemonic that appears to be popular in the USA.
That said, I'd have thought you'd be aware that division and multiplication have the same precedence from your experience with programming languages.
Well sure, I know the ranks of operator precedence in programming languages, but for no adequately-explained reason it didn't occur to me to apply what I know there to arithmetic precedence.
souvik1997 wrote:
Basicman: put 6/2(3) in your calculator. What does it say?

6/2(1+2)=1
6/2*(1+2)=9
just as I said () acts as x. 2(x) must be resolved before you can continue. These are the answers that my calculator gave me and that is exactly what I thought the answers would be. I really don't understand the confusion or the difference in views or beliefs by others. I'm not an expert, it's just my belief as it seems everyone has one on this subject.
basicman wrote:
just as I said () acts as x. 2(x) must be resolved before you can continue.
x=1+2
6/2x
That yields the same answer: 9 (try it on your calculator if you don't believe me). I guess you're interpreting it as 6/(2x) rather than (6/2)x.
basicman wrote:
6/2(1+2)=1
6/2*(1+2)=9
What calculator are you using? Or did you even bother to type that into a calculator?
Quote:
just as I said () acts as x.
Correct, which is why it has the *SAME* precedence as the division.
Quote:
2(x) must be resolved before you can continue.
False, the 6/2 must be resolved first because it is the leftmost operation in an equal-precedence situation.
Oddly enough you came up with the perfect example of why this is 9, you just screwed it up. I'll do it step by step:
6/2(1+2)
substitute x for (1+2)
6/2x, x = (2+1)
3x; x = 3
3*3
9
Quote:
These are the answers that my calculator gave me and that is exactly what I thought the answers would be. I really don't understand the confusion or the difference in views or beliefs by others. I'm not an expert, it's just my belief as it seems everyone has one on this subject.
There is no beliefs, you and everyone who thinks it is 1 are simply wrong. The *correct* answer is 9 for *both* 6/2(1+2) and 6/2*(1+2). The two are literally identical in meaning. Like I said, there is no ambiguity here - the rules are clear cut here.
Kllrnohj wrote:
Quote:
What calculator are you using? Or did you even bother to type that into a calculator?
basicman wrote:
Quote:
6/2(1+2)=1
6/2*(1+2)=9"]
I'm using a Casio fx-9750GA. I am sure that my 84+ would have given me the same answers.
6/2(1+2)=1
6/2*(1+2)=9
basicman wrote:
I'm using a Casio fx-9750GA. I am sure that my 84+ would have given me the same answers.
6/2(1+2)=1
6/2*(1+2)=9
Well quite simply your Casio is wrong as the right answer for both of those is 9. I'm curious now, though, what does the 84+ (or even better an 89) come up with? I'm curious how many calculators screw that up.
The last TI calc that would screw this up was the TI-85, and the manual explicitly states that it doesn't follow OOP.
My TI-84+SE gets it right (says they are both 9). My Casio Prizm gets it wrong. More importantly,
Wolfram|Alpha thinks they're the same.
My TI-83+SE says 9 for both. As Merth said, the Prizm makes a mistake. My new TI-86, reading between the missing LCD rows, says 9 for both.
It's 10.
Wow, that is even worse than the original one...
comicIDIOT wrote:
http://a4.sphotos.ak.fbcdn.net/hphotos-ak-snc6/250424_10150205689963108_685698107_7124165_4332047_n.jpg
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