Violation of Distributive law?

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NightOwl

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OK Folks: I ran into this while doing some Continuing Education in my field. I wouldn't have caught it but I used a calculator and then allowed the presenter to present his answer.

The presenter's contention was that 300 / 2.5 = 125 because 300 / 2 = 150 and 300 / 3 = 100. 150 - 100 = 50 and 0.5 * 50 = 25.

100 + 25 = 125

As I said, I used a calculator and just devided 300 by 2.5 and of course got 120.

Now the question is why? I know my math is rusty and though I suspect it is a violation of the distributive law I'm not sure how or why.

Sincerely, Craig
 
NightOwl said:
OK Folks: I ran into this while doing some Continuing Education in my field. I wouldn't have caught it but I used a calculator and then allowed the presenter to present his answer.

The presenter's contention was that 300 / 2.5 = 125 because 300 / 2 = 150 and 300 / 3 = 100. 150 - 100 = 50 and 0.5 * 50 = 25.

100 + 25 = 125

As I said, I used a calculator and just devided 300 by 2.5 and of course got 120.

Now the question is why? I know my math is rusty and though I suspect it is a violation of the distributive law I'm not sure how or why.
Click to expand...

\(\displaystyle 2.5=\dfrac{5}{2}\) so \(\displaystyle \dfrac{300}{2.5}=(300)\dfrac{2}{5}\)
 
NightOwl said:
OK Folks: I ran into this while doing some Continuing Education in my field. I wouldn't have caught it but I used a calculator and then allowed the presenter to present his answer.

The presenter's contention was that 300 / 2.5 = 125 because 300 / 2 = 150 and 300 / 3 = 100. 150 - 100 = 50 and 0.5 * 50 = 25.

100 + 25 = 125

As I said, I used a calculator and just devided 300 by 2.5 and of course got 120.

Now the question is why? I know my math is rusty and though I suspect it is a violation of the distributive law I'm not sure how or why.

Sincerely, Craig
Click to expand...
\(\displaystyle Given:\ a,\ b,\ c,\ and\ d \ne 0\ and\ d = \dfrac{b + c}{2}\)

\(\displaystyle \dfrac{a}{d} = \dfrac{a}{\dfrac{b + c}{2}} = \dfrac{a}{1} * \dfrac{2}{b + c} = \dfrac{2a}{b + c}.\)

So in your case \(\displaystyle \dfrac{300}{2.5} = \dfrac{300}{\dfrac{2 + 3}{2}} = \dfrac{2 * 300}{2 + 3} = \dfrac{600}{5} = 120.\)

Now what the presenter did (generically) was

\(\displaystyle \dfrac{a}{c} + 0.5\left(\dfrac{a}{b} - \dfrac{a}{c}\right)= \dfrac{a}{c} - \dfrac{0.5a}{c} + \dfrac{0.5a}{b} = \dfrac{ab}{bc} - \dfrac{0.5ab}{bc} + \dfrac{0.5ac}{bc} = \dfrac{0.5a(b + c)}{bc}.\)

So for your example he computed \(\displaystyle \dfrac{0.5 * 300(2 + 3)}{3 * 2} = \dfrac{150 * 5}{6} = \dfrac{750}{6} = 125.\)

Now that computation makes not an iota of sense. It is always wrong.
 
Thank you PKA.

I see your answer. And understand that the proper answer is 120. I guess the hard part for me, is visualizing why the presenter's argument though logical seeming, is invalid.

Sincerely, Craig
 
NightOwl said:
Thank you PKA.

I see your answer. And understand that the proper answer is 120. I guess the hard part for me, is visualizing why the presenter's argument though logical seeming, is invalid.

Sincerely, Craig
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I must admit that I do not see any logic anywhere in the presenter's computation. I can't figure out what the apparent merit is in it.
 
JeffM said:
\(\displaystyle Given:\ a,\ b,\ c,\ and\ d \ne 0\ and\ d = \dfrac{b + c}{2}\)

\(\displaystyle \dfrac{a}{d} = \dfrac{a}{\dfrac{b + c}{2}} = \dfrac{a}{1} * \dfrac{2}{b + c} = \dfrac{2a}{b + c}.\)

So in your case \(\displaystyle \dfrac{300}{2.5} = \dfrac{300}{\dfrac{2 + 3}{2}} = \dfrac{2 * 300}{2 + 3} = \dfrac{600}{5} = 120.\)

Now what the presenter did (generically) was

\(\displaystyle \dfrac{a}{c} + 0.5\left(\dfrac{a}{b} - \dfrac{a}{c}\right)= \dfrac{a}{c} - \dfrac{0.5a}{c} + \dfrac{0.5a}{b} = \dfrac{ab}{bc} - \dfrac{0.5ab}{bc} + \dfrac{0.5ac}{bc} = \dfrac{0.5a(b + c)}{bc}.\)

So for your example he computed \(\displaystyle \dfrac{0.5 * 300(2 + 3)}{3 * 2} = \dfrac{150 * 5}{6} = \dfrac{750}{6} = 125.\)

Now that computation makes not an iota of sense. It is always wrong.
Click to expand...

Strangely enough for me, I think I see this. What it says using my example, is that division is NOT distributive. Whereas the presenter's assumption was that it is. yes?
 
NightOwl said:
OK Folks: I ran into this while doing some Continuing Education in my field. I wouldn't have caught it but I used a calculator and then allowed the presenter to present his answer.

The presenter's contention was that 300 / 2.5 = 125 because 300 / 2 = 150 and 300 / 3 = 100. 150 - 100 = 50 and 0.5 * 50 = 25.

100 + 25 = 125

This calculation so misguided - to quote Pauli - it is not even wrong. Calling it wrong would give it a unwanted shade of acceptance!!

As I said, I used a calculator and just devided 300 by 2.5 and of course got 120.

Now the question is why? I know my math is rusty and though I suspect it is a violation of the distributive law I'm not sure how or why.

Sincerely, Craig
Click to expand...

I don't think there is any use of distributive law here - just mere fantasy.
 
NightOwl said:
Strangely enough for me, I think I see this. What it says using my example, is that division is NOT distributive. Whereas the presenter's assumption was that it is. yes?
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Division is distributive

\(\displaystyle \dfrac{1}{5} * (30 + 45) = \dfrac{30 + 45}{5} = \dfrac{75}{5} = 15.\)

\(\displaystyle \dfrac{1}{5} * (30 + 45) = \dfrac{1}{5} * 30 + \dfrac{1}{5} * 45 = \dfrac{30}{5} + \dfrac{45}{5} = 6 + 9 = 15.\)

I repeat: the presenter's method makes no sense to me at all. So I cannot tell you where the flaw lies. Why do you think it does make some sense? Maybe then we could explain what the flaw is.
 
The presenter actually is a Dr in his field. The math was involving interpreting an analog graph of the rate of incidence of a real world occurance. The graphing speed is standardized such that if the distance between incidents is devided into 300, the rate of occurance can be determined. In this case a difference of 5 in accuracy wouldn't make a significant difference.

However, I tend to be detail oriented and needed to know why the estimate was off because at least on a NON-math person's level it seemed logical to assume that his verbal logic was correct. (Yes, I know what happens when you assume. :rolleyes:)

Sincerely, Craig
 
I fear to ask what is his field.

HOWEVER, this may be a method of approximation that he uses. He did get an answer that was approximately correct. I do not feel like determining its precision because calculators are now so ubiquitous that an approximation is seldom needed and that one does not seem particularly easy to use.
 
Subhotosh Khan said:
I don't think there is any use of distributive law here - just mere fantasy.
Click to expand...

Thank you Subhotosh. Perhaps so.

I freely admit that my math is not the best and I know that in itself is aggravating to those with better memories for their math training and probably inexcusable others. To me it is quite aggravating to KNOW that something is incorrect and not understand why. What rule it violates. If I understand this then I can avoid a similar mistake later that perhaps could cost a life.



What all of this proves to me is that I need to go back and break out the books for yet another project :(.
 
NightOwl said:
The presenter's contention was that 300 / 2.5 = 125 because 300 / 2 = 150 and 300 / 3 = 100. 150 - 100 = 50 and 0.5 * 50 = 25.
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Wow. I've seen beginning-algebra students try to "average" rates of change (such as speeds), but never an instructor. What a newb move! :shock:
 
:p:p:p


They might not have been the presenter but they may have had the same teacher. :wink:
 
JeffM said:
\(\displaystyle Given:\ a,\ b,\ c,\ and\ d \ne 0\ and\ d = \dfrac{b + c}{2}\)

\(\displaystyle \dfrac{a}{d} = \dfrac{a}{\dfrac{b + c}{2}} = \dfrac{a}{1} * \dfrac{2}{b + c} = \dfrac{2a}{b + c}.\)

So in your case \(\displaystyle \dfrac{300}{2.5} = \dfrac{300}{\dfrac{2 + 3}{2}} = \dfrac{2 * 300}{2 + 3} = \dfrac{600}{5} = 120.\)

Now what the presenter did (generically) was

\(\displaystyle \dfrac{a}{c} + 0.5\left(\dfrac{a}{b} - \dfrac{a}{c}\right)= \dfrac{a}{c} - \dfrac{0.5a}{c} + \dfrac{0.5a}{b} = \dfrac{ab}{bc} - \dfrac{0.5ab}{bc} + \dfrac{0.5ac}{bc} = \dfrac{0.5a(b + c)}{bc}.\)

So for your example he computed \(\displaystyle \dfrac{0.5 * 300(2 + 3)}{3 * 2} = \dfrac{150 * 5}{6} = \dfrac {750}{6} = 125.\)

Now that computation makes not an iota of sense. > > It is always wrong. < <
Click to expand...
No, it is not always wrong. It's just not true in general. \(\displaystyle \ \ \ \ \)As long as b = c and they're not 0, then it holds. Let a = 300 and \(\displaystyle \ \ \)b = c = 2.5 \(\displaystyle \ \ \ \ \ \ \ \ \dfrac{300}{2.5} \ = \ \dfrac{300}{\bigg(\dfrac{5}{2}\bigg)} \ = \ \dfrac{300}{\bigg(\dfrac{2.5 + 2.5}{2}\bigg)} \ = \ \dfrac{2(300)}{5} \ = \ \dfrac{600}{5} \ = \ \boxed{120}.\) \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)Compare that to using the professor's way: \(\displaystyle \ \ \ \ \ \ \)\(\displaystyle \dfrac{0.5a(b + c)}{bc} \ = \ \dfrac{0.5(300)(2.5 + 2.5)}{(2.5)(2.5)} \ = \ \dfrac{(150)(5)}{6.25} \ = \ \dfrac{750}{6.25} \ = \boxed{120}. \)
 
lookagain said:
No, it is not always wrong. It's just not true in general. \(\displaystyle \ \ \ \ \)As long as b = c and they're not 0, then it holds. Let a = 300 and \(\displaystyle \ \ \)b = c = 2.5 \(\displaystyle \ \ \ \ \ \ \ \ \dfrac{300}{2.5} \ = \ \dfrac{300}{\bigg(\dfrac{5}{2}\bigg)} \ = \ \dfrac{300}{\bigg(\dfrac{2.5 + 2.5}{2}\bigg)} \ = \ \dfrac{2(300)}{5} \ = \ \dfrac{600}{5} \ = \ \boxed{120}.\) \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)Compare that to using the professor's way: \(\displaystyle \ \ \ \ \ \ \)\(\displaystyle \dfrac{0.5a(b + c)}{bc} \ = \ \dfrac{0.5(300)(2.5 + 2.5)}{(2.5)(2.5)} \ = \ \dfrac{(150)(5)}{6.25} \ = \ \dfrac{750}{6.25} \ = \boxed{120}. \)
Click to expand...

[FONT=MathJax_Main]0.5[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Main]300[/FONT][FONT=MathJax_Main]([/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main])/([/FONT][FONT=MathJax_Main]3[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Main]2)[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]150[/FONT][FONT=MathJax_Main]∗[/FONT][FONT=MathJax_Main]5[/FONT][FONT=MathJax_Main]/6[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main]750[/FONT][FONT=MathJax_Main]/6 [/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Main] 125.[/FONT] for 300/2.5

Now that computation makes not an iota of sense. It is always wrong.
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However, that computation is ALWAYS wrong - as JeffM stated.
 
Subhotosh Khan said:
However, that computation is ALWAYS wrong - as JeffM stated.
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What is the significance of the inclusion of "ALWAYS" here? If it's a specific arithmetical computation that is false, then "wrong" should suffice to describe it, else the inclusion of "always" is redundant. But if it's a method/general formula/algorithm that never produces a correct answer, then I could see describing it as "always wrong" instead of just "wrong." \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)For instance, I would state that the equation 1 = 2 is "false," but not "always false," as there are no variables in it and it never has the possibility to be true.
 
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lookagain said:
What is the significance of the inclusion of "ALWAYS" here? If it's a specific arithmetical computation that is false, then "wrong" should suffice to describe it, else the inclusion of "always" is redundant. But if it's a method/general formula/algorithm that never produces a correct answer, then I could see describing it as "always wrong" instead of just "wrong." \(\displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \)For instance, I would state that the equation 1 = 2 is "false," but not "always false," as there are no variables in it and it never has the possibility to be true.
Click to expand...
Lookagain

You seem to be assuming (incorrectly) that I did not test to see under what circumstances an algorithm along the lines proposed would work and so did not realize that

\(\displaystyle Given\ a, b, c, d \ne 0\ and\ d = \dfrac{b + c}{2}\,\ then\ \dfrac{0.5a(b + c)}{bc} = \dfrac{a}{d} \iff b = c = d.\)

Even though I knew that such an algorithm would work in a special case, discussing the special case would hardly have helped the OP, who wanted to know why it did not work generally. (You may have missed that the post you are criticizing deliberately used the word "generically.")

Furthermore, I believe that the algorithm actually intended excluded the special case. As far as I can see, the whole point of the algorithm was to avoid dividing by a number that was not an integer. As the algorithm was explained by the OP,

NightOwl said:
The presenter's contention was that 300 / 2.5 = 125 because 300 / 2 = 150 and 300 / 3 = 100. 150 - 100 = 50 and 0.5 * 50 = 25.

100 + 25 = 125
Click to expand...


I did not make the inane assumption that the presenter's contention was to solve 300 / 2.5 thusly

300 / 2.5 = 120 because 300 / 2.5 = 120 and 300 / 2.5 = 120. 120 - 120 = 0 and 0 * 0 = 0.

120 + 0 = 120

The algorithm (as I interpreted it) has a purpose only if b and c are integers, one greater than d and one less than d. That algorithm is ALWAYS wrong.

Of course I agree that your algorithm of dividing the numerator by the denominator twice, subtracting the two quotients, multiplying that difference by zero, and adding the product to the first quotient is ALWAYS right. I congratulate you on your lovely algorithm.
 
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