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Question about dividing this fraction

  1. Jul 27, 2015 #1

    JR Sauerland

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    I encountered this question recently, and didn't really understand what they did in the solution. This time I posted the entire question to assist with knowing what is going on. Basically, I multiplied both sides by 2pi/3, but the question wanted me to multiply both sides by 3/2pi, the reciprocal. Does this have anything to do with the division rules were you flip the fraction before multiplying it against a new one?
     
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  3. Jul 27, 2015 #2
    Which part are you having problems with in the solution specifically? It multiplies by 3/2pi in order to cancel out the 2pi/3 term to get r by itself.
     
  4. Jul 27, 2015 #3

    JR Sauerland

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    here is another example. When I stack them like this, I have absolutely no idea how to finish out the problem. I guess I don't know how to simplify S/pi/4
     
  5. Jul 27, 2015 #4

    JR Sauerland

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    Again, no idea how they got from 7/4/2 to 7/8.....
     
  6. Jul 27, 2015 #5
    There's a rule to memorize that, when you divide by fractions, you flip the fraction and multiply. It would be a good idea to memorize that, but, you don't even have to think of it that way. After all, division is just the inverse of multiplication, so we can think of it in terms of multiplication.

    You know that [tex]S = r \frac{\pi}{4}.[/tex]
    Except, you want to know what [itex]r[/itex] is in terms of [itex]S[/itex], which means you have to get [itex]r[/itex] alone on one side. So instead of thinking of it as dividing by [itex]\frac{\pi}{4}[/itex], think of it as multiplying both sides by the inverse.

    So what's the inverse of a number, you might ask? The inverse of a number [itex]a[/itex] is the number [itex]b[/itex] such that [itex]a b = 1[/itex]. We usually denote [itex]b[/itex] as [itex]a^{-1}[/itex] or [itex]\frac{1}{a}[/itex].

    So what's the inverse of [itex]\frac{\pi}{4}[/itex]? Well, we just look at what number we can multiply with it to get [itex]1[/itex]. Just by looking at it, we see that the inverse of [itex]\frac{\pi}{4}[/itex] is [itex]\frac{4}{\pi}[/itex], because [itex]\frac{\pi}{4} \frac{4}{\pi} = \frac{4 \pi}{4 \pi} = 1[/itex]. So, if we simply multiply this inverse by both sides, then it'll deal with that term and we'll get [itex]r[/itex] by itself: [tex]S = r \frac{\pi}{4} \implies \frac{4}{\pi} S = r \frac{\pi}{4} \frac{4}{\pi} \implies \frac{4S}{\pi} = r.[/tex]

    So that's really what's going on when you perform that operation. Now, the easy way to do all that is to just know that [tex]\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{a}{b} \frac{d}{c}[/tex]
     
  7. Jul 27, 2015 #6

    JR Sauerland

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    Thank you so much!!<3
     
  8. Jul 27, 2015 #7

    Mentallic

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    Also notice that a fraction of the form

    [tex]\frac{a}{\left(\frac{b}{c}\right)}[/tex]

    is directly related to the reciprocal of fractions that I mentioned to you in your previous thread.

    [tex]\frac{a}{\left(\frac{b}{c}\right)}=a\left(\frac{1}{\left(\frac{b}{c}\right)}\right)=a\left(\frac{c}{b}\right)=\frac{ac}{b}[/tex]

    Whenever you have a fraction in the denominator (let's call this the little fraction in the big fraction), you take the denominator of the little fraction and put it into the numerator of the big fraction.
     
  9. Jul 28, 2015 #8

    JR Sauerland

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    In this problem I attached below, he said they could be grouped with a common denominator (of 4). However, I don't see what the guy did or how. He somehow took [itex] {\pi} - \frac{\pi}{2} - \frac{\pi}{4} [/itex] and turned it all into [itex]\frac{\ 4 pi - 2 pi - pi}{4}[/itex].... However, I simply don't understand how he did it. I put it into my TI-84 verbatim of [itex]{\pi} - \frac{\pi}{2} - \frac{\pi}{4}[/itex] and even used the math button to command the calculator to report the answer as a fraction, although it just came as a string of numbers like 0.988997. I believe the calculator solved and converted pi to 3.14... rather than leaving it as pi, as we wanted to do of course.

    My best attempt to guess how/why he used 4 was to set an entirely new fraction up with the denominator of 4, and see how many times each of the three fractions went into 4. For the first one, [itex]\frac{\pi}{pi}[/itex] is 1, so it goes into 4 four times, which is why they would have used [itex]{4pi}[/itex] for the first portion. for the second portion [itex]\frac{\pi}{2}[/itex], I suppose they used the logic that 2 goes into 4 two times so it 'becomes' (I use quotations because I don't know how it becomes, but it does) [itex]{2pi}[/itex]. Am I on the right track?
     

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    Last edited: Jul 28, 2015
  10. Jul 28, 2015 #9
    You can only add fractions if the denominator is the same. He looks at it and notices that you can easily get a denominator of 4 in all of the terms. You just have to multiply the first two terms by 1 (since that doesn't change their value), but you disguise the 1 in a very clever way:
    [tex]\pi - \frac{\pi}{2} - \frac{\pi}{4} = 1 \cdot \pi - 1 \cdot \frac{\pi}{2} - \frac{\pi}{4} = \frac{4}{4} \cdot \pi - \frac{2}{2} \cdot \frac{\pi}{2} - \frac {\pi}{4}[/tex]

    Then you multiply that out, and you add the numerators together and keep the denominator as 4.

    Why does adding fractions work this way? Remember the distributive property:
    [tex]a(b +c) = ab + ac[/tex]
    So, we can distribute the product over the sum like that. Now say you want to add [itex]\frac{1}{4} + \frac{2}{4}[/itex], then you can write:
    [tex]\frac{1}{4} + \frac{2}{4} = \frac{1}{4}(1 + 2) = \frac{3}{4}[/tex]

    (Note that I have worked backwards from the distrubutive property). So you can see, we can add numbers with the same denominator together, because we can factor the denominator out and then multiply it back after we've added. We can't do that, however, if the denominators are different, because we can't factor it out. What would we factor out of we wanted [itex]\frac{1}{3} + \frac{1}{2}[/itex]? Answer: we can't factor anything out of there. We just have to manipulate it until the denominators are the same.
     
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