Question

\(\large \begin{align} \color{black}{ \text{prove that if a and b are positive integers } \hspace{.33em}\\~\\
\text{then } \sqrt 2\hspace{.33em}\\~\\
\text{lies between} \hspace{.33em}\\~\\
\dfrac{a}{b}~~~\text{and}~~~\dfrac{a+2b}{a+b} \hspace{.33em}\\~\\
}\end{align}\)

Answer 1

do two cases...
case 1) \(\dfrac{a}{b}>\sqrt{2}\)

case 2) \(\dfrac{a}{b}<\sqrt{2}\)

and use the fact that

\[\frac{a+2b}{a+b}=\frac{a+b+b}{a+b}=\frac{a+b}{a+b}+\frac{b}{a+b}=1+\frac{b}{a+b}\]

Answer 2

still confused on how to use this fact and proceed

Answer 3

please note that if a/b<sqrt(2), the we have:
\[\frac{ a }{ b }+1<\sqrt{2}+1\]
and:
\[\frac{ 1 }{ \frac{ a }{ b }+1 }>\frac{ 1 }{ \sqrt{2}+1 }=\sqrt{2}-1\]
so:
\[1+\frac{ b }{ a+b }=1+\frac{ 1 }{ \frac{ a }{ b }+1 }>1+\sqrt{2}-1\]
or:
\[1+\frac{ b }{ a+b }=1+\frac{ 1 }{ \frac{ a }{ b }+1 }>\sqrt{2}\]

Answer 4

Assume \(\dfrac{a}{b}>\sqrt{2}\)



\[\frac{a+2b}{a+b}=1+\frac{b}{a+b}=1+\frac{1}{\frac{a}{b}+1}<1+\frac{1}{\sqrt{2}+1}=\sqrt{2}\]

Answer 5

Here is a complete proof using same argument as above

Let \(p = \frac{a}{b}\) and \(q = \dfrac{p+2}{p+1}\), then we need to show \(\sqrt{2}\) lies between \(p\) and \(q\).

Let \(A\) be the set of all positive rationals \(p\) such that \(p^2 \lt 2\)
and let \(B\) be the set of all positive rationals \(p\) such that \(p^2\gt
2\).

If \(p\) is in \(A\) then \(p^2 \lt 2 \).
\(q^2-2 = \left(\dfrac{p+2}{p+1}\right)^2 - 2 = \dfrac{2-p^2}{(p+1)^2} \gt 0\tag{1}\)


If \(p\) is in \(B\) then \(p^2 \gt 2 \).
\(q^2-2 = \left(\dfrac{p+2}{p+1}\right)^2 - 2 = \dfrac{2-p^2}{(p+1)^2} \lt
0\tag{2}\)

the result follows from above two inequalities. \(\blacksquare\)

Answer 6

thanks all !!