1. ## Rationalise Denominator

Hi,

In order to rationalise the fraction $\frac{1}{1+\sqrt{3}}$ I must multiply the numerator and denominator by $1-\sqrt{3}$. I do not understand why this is so. Why does the sign only change for the surd and not the rational number (1) which is added to it?

Thanks for any help

2. Originally Posted by webguy
Hi,

In order to rationalise the fraction $\frac{1}{1+\sqrt{3}}$ I must multiply the numerator and denominator by $1-\sqrt{3}$. I do not understand why this is so. Why does the sign only change for the surd and not the rational number (1) which is added to it?

Thanks for any help
(a+b)(a-b) = a^2 - b^2

3. Hello,

Recall the identity $(a+b)(a-b)=a^2-b^2$.

In order to use this identity, you only need one among a or b to change its sign.

It's not a formal explanation but I hope you will understand

4. Would anyone mind going into deeper detail? I would like to understand this logic in its completeness.

5. Originally Posted by webguy
Would anyone mind going into deeper detail? I would like to understand this logic in its completeness.
I'm not sure what you're asking. You mean this?

$(a+b)(a-b) = a^2 -ab + ab -b^2 = a^2-b^2$

6. So why does that identity help us rationalise that fraction?

7. Code:
a + b
a - b
=====
a^2 + ab
- ab - b^2
==============
a^2      - b^2
Just a multiplication:
a times (a + b) = a^2 + ab
-b times (a + b) = -ab - b^2

8. Originally Posted by webguy
So why does that identity help us rationalise that fraction?
For integers a and b with b non-negative, if you have denominator $a + \sqrt{b}$ and you multiply it by $a - \sqrt{b}$ you get $a^2 - b$ which is an integer. Thus we have made the denominator rational.

9. Thank you! Now it makes sense!