Thread: 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..

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?

Hello, ice_syncer!

An interesting problem . . .

Prove that is always between and
for any positive whole numbers and

Let be any positive rational number less than

We have: . .[1]

. . Add 1 to both sides: .

. . Take reciprocals: .

. . Add 1 to both sides: . .[2]


The left side is: .

The right side is: .

. . Hence, [2] becomes: . .[3]


Combining [1] and [3], we have: .

Elegant solution, Soroban!

Grandad

Originally Posted by Soroban

Hello, ice_syncer!

An interesting problem . . .


Let be any positive rational number less than

We have: . .[1]

. . Add 1 to both sides: .

. . Take reciprocals: .

. . Add 1 to both sides: . .[2]


The left side is: .

The right side is: .

. . Hence, [2] becomes: . .[3]


Combining [1] and [3], we have: .

Thanks mate! well done.