What do you need to multiply ##\displaystyle \root 3 \of {2}## by to get ##2## ?barryj said:
Hi Barry,barryj said:
Well, you appear to know the trick already!barryj said:
Ideally we should say 'rationalise the denominator' (i.e. express with no surds in the denominator).gmax137 said:
That's what I think, but some mathematicians disagree:gmax137 said:
gmax137 said:
As well as virtually all high school algebra and college precalculus textbooks. As I said in the other thread that @PeroK linked to, the concept of "simplifying" fractions with radicals in the denominator means rewriting the fractions so that any radical now appears in the numerator. No doubt this topic precedes computers and calculators, but I'm sure that its basis is historical. For example, using a table to look up an approximation to ##\sqrt 2##, it's much quicker to calculate ##\frac{\sqrt 2}2 \approx \frac{1.4142}2## than it is to calculate ##\frac 1 {\sqrt 2} \approx \frac 1 {1.4142}##.PeroK said:
Yeah, mine does, too, but if you want more than 2 or so significant digits of precision, you'd do better by looking up ##\sqrt 2## in a table and then either dividing it by 2 or dividing it into 1.gmax137 said:
Mark44 said:
To simplify \( \frac{1}{\sqrt[3]{2}} \), you can multiply the numerator and the denominator by \( \sqrt[3]{4} \) to rationalize the denominator. This gives \( \frac{\sqrt[3]{4}}{\sqrt[3]{8}} \), which simplifies to \( \frac{\sqrt[3]{4}}{2} \).
We multiply by \( \sqrt[3]{4} \) because \( \sqrt[3]{4} \times \sqrt[3]{2} = \sqrt[3]{8} \), and \( \sqrt[3]{8} \) is a whole number (2). This process eliminates the cube root in the denominator, making it easier to work with.
Yes, for any \( \frac{1}{\sqrt[3]{a}} \), you can rationalize the denominator by multiplying by \( \sqrt[3]{a^2} \). This results in \( \frac{\sqrt[3]{a^2}}{\sqrt[3]{a^3}} = \frac{\sqrt[3]{a^2}}{a} \).
No, \( \frac{\sqrt[3]{4}}{2} \) is already in its simplest form. The cube root of 4 cannot be simplified further without using decimal or fractional approximations.
You can verify the simplification by calculating the numerical values of both \( \frac{1}{\sqrt[3]{2}} \) and \( \frac{\sqrt[3]{4}}{2} \). Both should yield the same decimal value, confirming that the simplification is correct.