SOLUTION: If 4 < a < 7 < b < 9, then which of the following best defines a/b? I tried to look at it like a triangle, with (a) being 6 and (b) being 8, but I'm stuck after that. a. 4/9

Algebra ->  Inequalities -> SOLUTION: If 4 < a < 7 < b < 9, then which of the following best defines a/b? I tried to look at it like a triangle, with (a) being 6 and (b) being 8, but I'm stuck after that. a. 4/9      Log On


   



Question 697295: If 4 < a < 7 < b < 9, then which of the following best defines a/b?
I tried to look at it like a triangle, with (a) being 6 and (b) being 8, but I'm stuck after that.
a. 4/9 < a/b < 1
b. 4/9 < a/b < 7/9
c. 4/7 < a/b < 7/9
d. 4/7 < a/b < 1
e. 4/7 < a/b < 9/7

Found 4 solutions by MathLover1, Edwin McCravy, AnlytcPhil, Edwin Parker:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
If 4+%3C+a+%3C+7+%3C+b+%3C9+, then 4+%3C+a+%3C+7 and 7+%3C+b+%3C+9
so
4 7 ---------a%2Fb will be
4/7
c. 4%2F7+%3C+a%2Fb+%3C+7%2F9+ best defines a%2Fb

Answer by
Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!
The other tutor's answer is wrong, because, for example
suppose a = 4.5 and b = 8
4 < a < 7 < b < 9 because 4 < 4.5 < 7 < 8 < 9
However
a/b = 4.5/8 = .5625
and
4/7 = 0.5714 (rounded to 4 decimal places)
and so 4/7 is GREATER THAN a/b, not LESS THAN a/b!!!
I'll show you how to do it later.
Edwin

Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
Here is the correct solution:

First we must prove another theorem:

The reciprocals of two given unequal positive numbers are
positive numbers unequal in the opposite order from the 
two given positive numbers.

[Maybe you've already proved this, but in case you haven't,
here is the proof of that]:

Proof:

Suppose we are given:

0 < x < y  and we are to prove that 0 < 1%2Fy < 1%2Fx

Starting with

0 < x < y

divide all three sides by y:

0%2Fy < x%2Fy < y%2Fy

Simplify:

0 < x%2Fy < 1

Divide all three sides by x:

0%2Fx < expr%28x%2Fy%29÷x < 1%2Fx
 
0 < expr%28x%2Fy%29÷x%2F1 < 1%2Fx 

0 < x%2Fy·1%2Fx < 1%2Fx

0 < cross%28x%29%2Fy·1%2Fcross%28x%29 < 1%2Fx

0 < 1%2Fy < 1%2Fx

----------------------------------

Now that we have proved that, we can turn to your
problem:

4 < a < 7 < b < 9

We are given 

      4 < a  

Divide both sides by 9

(1)   4%2F9 < a%2F9 

We are given 

      b < 9  

By the theorem above:

     1%2F9 < 1%2Fb

Multiply both sides by "a":

(2)   a%2F9 < a%2Fb

 
We are given 

      7 < b  

By the theorem above:

     1%2Fb < 1%2F7

Multiply both sides by "a":

(3)   a%2Fb < a%2F7

We are given 

      a < 7  

Divide both sides by 7

(4)   a%2F7 < 7%2F7 

      a%2F7 < 1

Putting (1), (2), (3) and (4) together:

      4%2F9 < a%2Fb < a%2F7 < 1

So omit a%2F7 and we have

      4%2F9 < a%2Fb < 1

as the only one of the inequalities given that 
we can be sure of.

Edwin

Answer by Edwin Parker(36) About Me  (Show Source):
You can put this solution on YOUR website!
Here is a counterexample to show
that
b. 4%2F9 < a%2Fb < 7%2F9
is not always true:
Suppose a = 6 and b = 7.5
4 < a < 7 < b < 9 because 4 < 6 < 7 < 7.5 < 9
However
a/b = 6/7.5 = .8
and
7/9 = 0.7778 (rounded to 4 decimal places)
So a%2Fb > 7%2F9,
and thus this is a case when choice b.
is false.
-------------------------------
The second response above gives a
counterexample of choice c.
-------------------------------
Here is a counterexample to show
that
d. 4%2F7 < a%2Fb < 1
is not always true:
Suppose a = 4.2 and b = 8
4 < a < 7 < b < 9 because 4 < 4.2 < 7 < 8 < 9
However
a/b = 4.2/8 = .525
and
4/7 = 0.5714 (rounded to 4 decimal places)
So a%2Fb > 4%2F7,
and thus this is a case when choice d.
is false.
This is also a counterexample for choice e.
Edwin