# Thread: prove

1. ## prove

Prove that root2 is always between a/b and (a+2b)/(a+b) for any positive whole number a and b

Which means we have to prove that a/b is lesser than root2 and root2 is lesser than (a+2b)/(a+b)...how

I htink you need to use contradictory method and euclid's divison lemma..

2. Hi -

Root2 is irrational. Therefore a/b is not equal to root2 for any values of a and b.

Assume first that a/b is greater than root2. Then, squaring:

a^2/b^2 > 2

So a^2 > 2b^2

Then square (a + b)/(a + 2b).

The result can be written as 1 + 3b^2/(a^2 + 2ab + b^2)

and this is less than

1 + 3b^2/(3b^2 + 2ab) since we have replaced a^2 by 2b^2 in the denominator (and a^2 > 2b^2, from above).

This in turn is less than 2, since 2ab is positive.

So (a + b)/(a + 2b) < root2.

QED

Next, assume a/b is less than root2. And so on ... as before with the inequalities reversed.

OK?

3. Hello, ice_syncer!

An interesting problem . . .

Prove that $\displaystyle \sqrt{2}$ is always between $\displaystyle \frac{a}{b}$ and $\displaystyle \frac{a+2b}{a+b}$
for any positive whole numbers $\displaystyle a$ and $\displaystyle b .$

Let $\displaystyle \frac{a}{b}$ be any positive rational number less than $\displaystyle \sqrt{2}$

We have: .$\displaystyle \frac{a}{b} \:<\:\sqrt{2}$ .[1]

. . Add 1 to both sides: .$\displaystyle \frac{a}{b} + 1 \:<\:\sqrt{2} + 1$

. . Take reciprocals: .$\displaystyle \frac{1}{\frac{a}{b} + 1} \:>\:\frac{1}{\sqrt{2}+1}$

. . Add 1 to both sides: .$\displaystyle 1 + \frac{1}{\frac{a}{b}+1} \:>\:1 + \frac{1}{\sqrt{2}+1}$ .[2]

The left side is: .$\displaystyle 1 + \frac{1}{\frac{a}{b}+1} \;=\;1 + \frac{1}{\frac{a+b}{b}} \;=\;1 + \frac{b}{a+b} \;=\;\frac{a + 2b}{a+b}$

The right side is: .$\displaystyle 1 +\frac{1}{\sqrt{2}+1}\!\cdot\!{\color{red}\frac{\s qrt{2}-1}{\sqrt{2}-1}} \;=\;1 + \frac{\sqrt{2} - 1}{1} \;=\;\sqrt{2}$

. . Hence, [2] becomes: .$\displaystyle \frac{a+2b}{a+b} \:> \:\sqrt{2}$ .[3]

Combining [1] and [3], we have: .$\displaystyle \frac{a}{b} \:<\:\sqrt{2} \:<\:\frac{a+2b}{a+b}$

4. Elegant solution, Soroban!

Grandad

5. Originally Posted by Soroban
Hello, ice_syncer!

An interesting problem . . .

Let $\displaystyle \frac{a}{b}$ be any positive rational number less than $\displaystyle \sqrt{2}$

We have: .$\displaystyle \frac{a}{b} \:<\:\sqrt{2}$ .[1]

. . Add 1 to both sides: .$\displaystyle \frac{a}{b} + 1 \:<\:\sqrt{2} + 1$

. . Take reciprocals: .$\displaystyle \frac{1}{\frac{a}{b} + 1} \:>\:\frac{1}{\sqrt{2}+1}$

. . Add 1 to both sides: .$\displaystyle 1 + \frac{1}{\frac{a}{b}+1} \:>\:1 + \frac{1}{\sqrt{2}+1}$ .[2]

The left side is: .$\displaystyle 1 + \frac{1}{\frac{a}{b}+1} \;=\;1 + \frac{1}{\frac{a+b}{b}} \;=\;1 + \frac{b}{a+b} \;=\;\frac{a + 2b}{a+b}$

The right side is: .$\displaystyle 1 +\frac{1}{\sqrt{2}+1}\!\cdot\!{\color{red}\frac{\s qrt{2}-1}{\sqrt{2}-1}} \;=\;1 + \frac{\sqrt{2} - 1}{1} \;=\;\sqrt{2}$

. . Hence, [2] becomes: .$\displaystyle \frac{a+2b}{a+b} \:> \:\sqrt{2}$ .[3]

Combining [1] and [3], we have: .$\displaystyle \frac{a}{b} \:<\:\sqrt{2} \:<\:\frac{a+2b}{a+b}$

Thanks mate! well done.