Gregory from Magnus C of E School in
Nottingham and Luke from St Patrick's School reasoned correctly.
Here is Luke's explanation:
For this question I found a pattern in it:
For example
has length of 1 and
has length of
and there
is a pattern in this because 1 x 2 = 2, so there is your numerator,
and to get
your denominator you multiply 1 by 3, so there you have your
.
So to put this into easier words, it simply means multiply the numerator by 2
the whole way and multiply the denominator by 3.
So there is my strategy.
So The answer to
is
and the answer to
is
and
finally
is
.
is
and when n
infinity,
0
David from Gordonstoun School got the same result and added:
In a geometric progression,
=
term
= constant factor
= number of terms
Any value in a geometric progression is
In this case
= 1
=
=
(which is less than 1),
so the higher its power,
the closer the result is to zero.
(Any positive number smaller than 1, to the power of infinity, tends to zero.
Therefore
tends to zero)
So as
tends to infinity,
tends to
So the Cantor Set's length is zero.
Liam from Wilbarston School reasoned in a similar way:
The length of
is simply two thirds of the length of
,
as
is purely
with the middle thirds removed. Now taking
to be the
length of
:
=
,
=
,
=
etc. etc.
It's obvious that
=
.
So as n tends to
infinity,
gets increasingly smaller, i.e. tends to zero.
Therefore the length of the Cantor set is zero. In fact, the Cantor set is
a set of points, because endpoints of line segments will never be removed,
only middle thirds.
And as Euclid said,
'A point is that which has no
part'
i.e. a point has zero length, zero width and zero height.
Well done to you all.