How do you rationalise a denominator that has another fraction within it?

>Here's what I've tried:
1×√2/1×√2-(1×√2/√2×√2) = √2/-√2-(√2/2) = √2/√2×(√2/2) = -√2/1 = -√2

I agree you want to multiply top and bottom of the 'main' fraction by √2

When you do this, you get...

1×√2 / ( 1-1/√2 )×√2     =     √2 / ( √2 - 1 )
So, I guess you fraction looks like this:

1/(1 - 1/√2)? Because 1/1 - (1/√2) would be 1 - 1/√2, which would just be a normal fraction.

As to how to solve it, treat the parts of the fraction individually.

If we now look at the denominator, we have 1 - 1/√2. We can bring this to the same denominator by multiplying 1 by √2 (because √2/√2 = 1), so we get (√2 - 1)/√2

Now we have a double fraction that we can rearrange. 1/((√2 - 1)/√2) can be rearranged to √2/(√2 - 1), as dividing by a value is the same as multiplying by the inverse (a/(b/c) = a \* c/b).

Now, a root in the denominator is always an ugly thing. Hence we use the third binomial (a + b)(a - b) = a² - b² and multiply what we currently have by (√2 + 1)/(√2 + 1) (as this can be cancelled to 1 and multiplying by 1 and adding 0 are always allowed.

Thus we arrive at (2 + √2)/(2 - 1) = 2 + √2

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LaTeX below:

$; \\frac{1}{1 - \\frac{1}{\\sqrt{2}}} = \\frac{1}{\\frac{\\sqrt{2} - 1}{\\sqrt{2}}} = \\frac{\\sqrt{2}}{\\sqrt{2} - 1} \* \\frac{\\sqrt{2} + 1}{\\sqrt{2} + 1} = \\frac{2 + \\sqrt{2}}{2 - 1} = 2 + \\sqrt{2} ;$
An approach to this could be taking the lcm.

In this case, 1/1-(1/√2) can be further simplified as:

1/[(√2-1)/√2]

Or √2/√2-1 which can be rationalised as 2+√2/2-1 which is 2 + √2

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