This IS a TEST; Do you know the ACTUAL reason division by zero is not permitted?
Moderators: Calilasseia, ADParker
scherado wrote:OlivierK wrote:scherado wrote:Another thing I remember from college-days is that multiplication is addition by a different name: 7 x 3 = 7 + 7 + 7.
For division, we have subtraction by a different name: For 6/2, we subtract 2 from 6, take the reduced value and repeat UNTIL we either reach 0 or there's not enough remaining; count the number of iterations:
6 - 2 = 4
4 - 2 = 2
2 - 2 = 0
3 times we subtracted 2: 6/2 = 3
Now, do the same for 3/0:
3 - 0 = 3
3 - 0 = 3 ...
Infinite iterations.
There's that pesky infinity, again. (Not whistling Dixie.)
You think you reach 0 after subtracting 0 from 6 an infinite number of times?
How precious.
Do you want to reconsider...when you've sobered up?
OlivierK wrote:
No need to reconsider.
You're counting interations taken to reach zero.
In your first example you come up with the (correct) count of 3 iterations to reach zero.
In your second example you come up with the (incorrect) count of infinity iterations to reach zero.
The fact that "infinity" is a braindead answer to the second part is not my problem.
scott1328 wrote:How to tell if a troll has invaded your online forum:
Issues challenges to which he cannot recognize correct answers and fails to even make the attempt to answer challenges given back.
check
scherado wrote:OlivierK wrote:
No need to reconsider.
You're counting interations taken to reach zero.
In your first example you come up with the (correct) count of 3 iterations to reach zero.
In your second example you come up with the (incorrect) count of infinity iterations to reach zero.
The fact that "infinity" is a braindead answer to the second part is not my problem.
You most certainly to have a problem: Your problem is that, STILL, you won't accept the entire point of this thread. To wit, (you): "you come up with the (incorrect) count of infinity"
INFINITY IS NOT A VALUE, but I repeat myself.
And this: "you come up with the (incorrect) count of infinity iterations to reach zero."
The example NEVER REACHES ZERO, which is the point of that example.
Do you want to reconsider, sober or not?
OlivierK wrote:... Infinity doesn't have anything to do with it, unless someone claims that's the answer to division by zero. ...
theropod wrote:Isn’t there a math symbol for infinity?
Well, yes there is, and it has a term called “lemniscate“. I think is it a two dimensional representation of a möbius loop, but it’s been a “few” years since my last math class.
Does an ♾ represent a value? If it didn’t it wouldn’t need be represented in formal math.
RS
felltoearth wrote:The problem is that infinity isn’t a number. It’s math, there has to be a number, right?
Must have blown people’s minds when zero was invented.
scott1328 wrote:scherado wrote:Can anyone tell me--after it has been revealed already--the reason we can't divide by zero? In my 5 words or Matt_B's 6 words.
This thread has been over for a few days now. I thought it would take much longer to get a CORRECT answer: Give Matt_B the keys to the forum.
I'll have to start a new one.
Your 5 word simplistic answer is wrong. And your presumption that there is but one answer is wrong.
Why don't you answer my question? why is it incorrect to use L'Hopital's rule to evaluate the limit as x approaches 0 of sin(x)/x ?
scott1328 wrote:L'Hopital's rule states that if the limit as x approaches some value c of f(x) = 0 and g(x)=0, then the limit as x approaches c of the quotient of those two functions [ f(x)/g(x) ] can be found finding the limit as x approaches c of f'(x)/g'(x).
So it would seem that applying L'Hopital's rule to the quotient of sin(x)/x would be useful in evaluating the limit: since sin(0)=0 so we have the case 0/0. The rule says that we should be able to evaluate the limit by finding the derivative of sin(x) which turns out to be cos(x) and the derivative of x which turns out to be the constant 1. Now, evaluating the limit : the limit as x approaches 0 of cos(x)/1 is 1.
However, where the the trouble comes in, is in how you actually prove that the derivative of sin(x) does in fact equal cos(x). To find the derivative of sin(x) relies on knowing how to evaluate the limit of sin(x)/x. And therein lies the circularity.
scott1328 wrote:Absolutely you can define the trig functions(and their hyperbolic counterparts) in terms of their MacLauren series expansions. In which case finding the derivatives of sin and cos are trivial matters. Motivating such a definition is problematic. This introduces a pedagogical difficulty however. Such definitions of sine and cosine are going to seem entirely arbitrary to first and second year calculus students. And would likely impede understanding rather than aid it.
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