I've been dealing with some equations and I've got a bit of a problem. The equation as follows. $$t=\frac S{(V_f+V_i)/2}$$ I couldn't quite simplify it because on one hand I've got multiplication by the reciprocal rule which states that $$t=\frac S1×\frac2{V_f+V_i}=\frac{2S}{V_f+V_i}$$ and on the other hand I've got the order of operations stating that $$a/b/c =\frac a{bc}$$ which leads me to $$t=\frac S{2(V_f+V_i)}$$ I'm sure I'm missing something here, but I can't quite get my head around it. Help would be appreciated. Cheers!
Does order matter in chained divisions?
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algebra-precalculus
2 Answers
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The first interpretation is correct: $$t=\frac{2S}{V_f+V_i}$$ When dealing with chained multiplications and divisions the order does matter; without parentheses an expression is evaluated left to right. So $a/b/c$ evaluates as $(a/b)/c$, which can be simplified to $\frac a{bc}$ as you have pointed out in the second interpretation. However, the given expression is of the form $a/(b/c)$, which is not equivalent to $(a/b)/c$ and simplifies to the different result of $\frac{ac}b$.
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0Thanks, I understand now, but I wonder how come I've never heard of it in a class or somewhere else... Maybe I just wasn't paying attention. The division tends to be trickier than multiplication and it's kinda annoying... Oh, well... The intuition for that is important. – 2017-02-18
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Order matters. For example, $2/(2/3)=3$ while $(2/2)/3=1/3$. In your rule, $a/b/c$ is a short notation for $$(a/b)/c=\frac{\frac{a}{b}}{c}$$ which differs from $$a/(b/c)=\frac{a}{\frac{b}{c}}$$ which is the form of your original expression. Hence you cannot apply $a/b/c=\frac{a}{bc}$ to the original formula, because it is not the form of the original formula.