Given:
Two numbers are [tex]\dfrac{9}{7}[/tex] and [tex]\dfrac{10}{7}[/tex].
To find:
A repeating decimal that is between [tex]\dfrac{9}{7}[/tex] and [tex]\dfrac{10}{7}[/tex].
Solution:
Using calculator, we get
[tex]\dfrac{9}{7}=1.285714285714...[/tex]
[tex]\dfrac{9}{7}=1.\overline{285714}[/tex]
and,
[tex]\dfrac{10}{7}=1.42857142857 1...[/tex]
[tex]\dfrac{10}{7}=1.\overline{428571}[/tex]
Now, the repeating decimal that is between [tex]\dfrac{9}{7}[/tex] and [tex]\dfrac{10}{7}[/tex] be x. So,
[tex]1.\overline{285714}<x<1.\overline{428571}[/tex]
On analyzing the numbers to hundredth places, we get 1.28 < 1.33 < 1.42, therefore
[tex]1.\overline{285714}<1.3333...<1.\overline{428571}[/tex]
[tex]1.\overline{285714}<1.\overline{3}<1.\overline{428571}[/tex]
And we know that [tex]1.\overline{3}[/tex] is the decimal form of [tex]\dfrac{4}{3}[/tex].
Therefore, the required number is [tex]1.\overline{3}[/tex] .
