Question

Find a number that is between 2/7 and 4/11 .

Answer 1

Make denominators equal first..

Answer 2

ok so what is it

Answer 3

@naranjja you were writing something.

Carry on buddy..

Answer 4

So @Hero ?

Answer 5

\[\frac{2}{7} and \frac{4}{11}\]

After cross multiplying notice that

22 < 28

So

\[\frac{2}{7} < \frac{4}{11}\]

Now \(\dfrac{3}{7} > \dfrac{2}{7}\)

So test if \(\dfrac{3}{7} < \dfrac{4}{11}\)

If so, then you have found a fraction between \[\frac{2}{7} and \frac{4}{11}\]

Answer 6

My browser crashed. Sorry, I'll start over again.

Answer 7

Okay, so that didn't work. But now you can test \(\dfrac{3}{11}\)

We know that \(\dfrac{3}{11} < \dfrac{4}{11}\)

So test to see if \(\dfrac{3}{11} > \dfrac{2}{7}\)

That doesn't work either. But you can try adding both fractions together then dividing by 2 to find a fraction between both:

\[\left(\frac{2}{7} + \frac{4}{11}\right) \div 2\]

Answer 8

Establishing the common denominator 77 (multiplying the first by 11 and the second by 7),
\[\frac{ 2 }{ 7 }=\frac{ 22 }{ 77 }\]\[\frac{ 4 }{ 11 }=\frac{ 28 }{ 77 }\]
So say you choose, \[\frac{ 24 }{ 77 }\]
That fraction is irreducible, and it is approximately 0.312

Answer 9

That's probaby the easiest way to do it.

Answer 10

\[\frac{ 22 }{ 77 }\approx0.28\]\[\frac{ 28 }{ 77 }\approx0.36\]
You can choose 0.3 for example, which is 1/3

Answer 11

so is it 22/77

Answer 12

@naranjja, there is no need to reduce further. If you have two fractions

\(\dfrac{a}{c}\) and \(\dfrac{b}{c}\) and a < b, then

\(\dfrac{a}{c}\) < \(\dfrac{b}{c}\)

Answer 13

and if you find a third fraction \(\dfrac{d}{c}\) such that a < d < c, then
\(\dfrac{a}{c} < \dfrac{d}{c} < \dfrac{b}{c}\)

Answer 14

soooo

Answer 15

is it 22/77?

Answer 16

No, it isn't. I guess you haven't understood anything me or @naranjja has written.

Answer 17

It can be any number between 22/77 and 28/77. Which is the same as saying between 0.28 and 0.36. I think 0.333 is a good number to choose because it is 1/3.

Answer 18

oh ok and yes I did just making sure @Hero