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I need help... I did most of my hw, but these just can't? IDK. but...

Among all fractions $x$ that have a positive integer numerator and denominator and satisfy $$\frac{9}{11} \le x \le \frac{11}{13},$$ which fraction has the smallest denominator?

AND

Evaluate: $\dfrac{(x+y)^2 - (x-y)^2}{y}$ for $x=6$, $y \not= 0$.

Dec 2, 2018

#1
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1. First, try to change the x in the middle, to a fraction. Let's say x/y. We then multiply both sides by b, to get 9b/11 and 11b/13. Now, the trick here is to find a positive integer that satisfies and is less than 1, since 11/13 is less than 1.

Dec 2, 2018
#2
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deleted

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Dec 2, 2018
edited by Rom  Dec 3, 2018
edited by Rom  Dec 3, 2018
#3
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Not quite, Rom. Continuing with my method and plugging the values of b, we get b=6, so the answer is $$\boxed{\frac{5}{6}}.$$ Can someone verify this?

tertre  Dec 3, 2018
#4
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9/11 = .8181.....

5/6 = .8333....

11/13 = .846....

Mmmmm.....it looks like you could  be correct, tertre.....

CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
edited by CPhill  Dec 3, 2018
#5
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Here might be another way to look at this

Notice that the fractions seem to have the form  n / [ n + 2]    where n is an integer

Suppose that there exists a fraction such that

9/11 < n / [n + 2]  < 11/13

We can split this into two inequalities

Looking at the inequality on the left....we have

9[n + 2] < 11n

9n + 18 < 11n

18 < 2n

9 < n    or

n > 9

Looking at the inequality on the right.... we have

13(n) < 11[n + 2]

13n < 11n + 22

2n < 22

n < 11

This implies that

9 < n < 11

So...n = 10 is the only integer that satisfies this

And n + 2 = 12

So....the fraction is 10/12  = 5/6  ......just as tertre found!!!

Good job, tertre   !!!!

Dec 3, 2018
edited by CPhill  Dec 3, 2018
#6
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Thank you, CPhill! Great solution!

tertre  Dec 3, 2018