Rationalising the Denominator
Mathematicians are a tidy bunch, and one thing they quietly can't stand is a
surd sitting in the denominator of a fraction — the downstairs
of the fraction, the bit you divide by. Something like
\dfrac{1}{\sqrt{2}} is a perfectly good number, but it
feels untidy: how big is it, exactly? Dividing 1
by 1.41421356\ldots is horrible.
Rationalising the denominator is a clever little move that shoves
the surd upstairs, into the numerator, leaving a plain whole number on the
bottom. The messy \dfrac{1}{\sqrt{2}} becomes the tidy,
equal twin \dfrac{\sqrt{2}}{2} — the conventional
exact form you are expected to write:
\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{2}}{2}
Nothing has changed except the looks — both are
0.7071\ldots The trick works because
\sqrt{2}\times\sqrt{2} = 2: a root multiplied by itself
turns into a whole number, and the surd on the bottom vanishes.
The move: multiply by a disguised 1
Here is the whole idea in one sentence: multiply the top and the bottom by
the same surd. Because top and bottom get the same factor, you are really
multiplying the fraction by \frac{\sqrt{2}}{\sqrt{2}} = 1
— and multiplying by 1 never changes a number's value,
only its outfit.
To clear a surd from the bottom of a fraction:
-
for a single-surd denominator \sqrt{a},
multiply numerator and denominator by \sqrt{a};
-
\sqrt{a}\times\sqrt{a} = a, so the root disappears
from the bottom;
-
the fraction's value is unchanged, because you multiplied by
1;
-
for a two-term denominator like
(a + \sqrt{b}), multiply top and bottom by its
conjugate (a - \sqrt{b}) instead.
Worked example — \dfrac{6}{\sqrt{3}}
The bottom is \sqrt{3}, so multiply top and bottom by
\sqrt{3}:
\frac{6}{\sqrt{3}} = \frac{6}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}
The \sqrt{3} on the bottom became a plain
3, and \frac{6}{3} cancels to
2. Always simplify at the end — a rationalised answer is
expected in its lowest form, here 2\sqrt{3}.
The tricky case: two terms on the bottom
What about \dfrac{1}{1 + \sqrt{2}}? Multiplying by
\sqrt{2} is no help — you'd get
\sqrt{2} + 2 on the bottom, still carrying a
surd. The fix is the conjugate: the same two terms with the
middle sign flipped, here (1 - \sqrt{2}).
Why does that work? Because (a + b)(a - b) = a^2 - b^2 —
the difference of two squares — squares both terms and squaring a
surd kills it. Watch it worked through one line at a time:
\frac{1}{1 + \sqrt{2}} = \frac{1}{1+\sqrt{2}}\times\frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{1 - 2} = \frac{1-\sqrt{2}}{-1} = \sqrt{2} - 1
The denominator (1+\sqrt{2})(1-\sqrt{2}) = 1^2 - (\sqrt{2})^2 = 1 - 2 = -1
came out as a plain integer, exactly as promised, and the surd rode upstairs.
Two classic slips when rationalising:
-
You must multiply both the top and the bottom by the same thing.
Multiplying only the bottom would change the value of the fraction — that
is division, not tidying. Multiplying top and bottom is multiplying by a
disguised 1, which is safe.
-
For a two-term denominator, multiply by the conjugate (flip the
middle sign), not by the denominator itself. Multiplying
(1+\sqrt{2}) by (1+\sqrt{2})
gives 1 + 2\sqrt{2} + 2 — a surd survives. Only the
conjugate (1-\sqrt{2}) triggers the difference of two
squares that clears it.
The vanishing trick — (a+b)(a-b) = a^2 - b^2 — is the
very same difference-of-two-squares algebra hiding inside the
quadratic formula,
where a square root gets tamed in exactly this spirit. Learn to love conjugates now
and you'll meet them again.
And here's why anyone bothered in the first place: before calculators,
rationalising was genuinely useful, not just fussy. Working out
\frac{1}{\sqrt{2}} by hand meant dividing
1 by 1.41421356\ldots — an
agonising long division. But \frac{\sqrt{2}}{2} is just
1.41421356\ldots \div 2 — halving, which anyone can do in
their head. A surd on the bottom really was a problem worth removing.
See it explained