Pascal’s Triangle (Mod 2)
Reason: We are trying to prove that the number of ones in a given row of Pascal’s Triangle
(Mod 2) will be equal to 2n, where n is the number of 1’s in the binary expansion of that row number.
Pascal’s Triangle (Mod 2) is constructed by starting with row 0, which has one 1 in the row. To get
entries in the next row, you add the two previous entries that are diagonally above the entry you are
looking for. The entry is equal to these two entries added together (Mod 2). The four cases we will have
are:
This gives us:
for the first 8 rows of Pascal’s Triangle (Mod 2).
We can then break up the triangle into smaller triangles:
etc.
T1:
T2:
T3:
Pascal’s Triangle (Mod 2) will follow the form:
T1
T2 is the base case:
“
” = Tk is 3 Tk-1’s arranged so that Tk has 2 times the number of rows as Tk-1 and the remaining
entries are all zero.
Base Case: The number of 1’s in row 0 is 20=1.
The binary number is 0, so there are zero 1’s in the binary expansion, therefore n equals 0.
The number of 1’s in row 1 is 21=2.
The binary number is 1, so there is one 1 in the binary expansion, therefore n equals 1.
If we look at Pascal’s Triangle (Mod 2), we see that this matches.
We are doing induction on the number of digits in the binary expansion of row r.
Assume: The number of 1’s in a row r is equal to 2n, where n equals the number of 1’s in the binary
expansion of r, for D-1 digits. Assume r has D digits in its binary expansion.
-The leftmost digit must always be a 1.
Let r’=r with the leftmost digit removed. r’ has D-1 digits, therefore the number of 1’s in r’ is equal to
2n’, where n’ is the number of 1’s in the binary expansion of r’ by the induction assumption.
We need to show: The number of 1’s in r is equal to 2n’+1 which is double 2n’.
If we look at triangle TD that is composed to three copies of TD-1:
r will be in TD, because the leftmost 1 will double the number of rows from TD-1 to TD, putting r in
the second layer of TD-1’s in TD. r’ will be in the first layer of TD-1 in TD.
A “layer” can be defined as which TD-1 triangle we are looking at in triangle TD. TD will have two layers,
the first will have the original triangle TD-1, and the second layer will have two identical copies of TD-1.
The distance between r’ and r is 2D-1, because the number of rows in TD-1 is 2D-1. Since r is 2D-1 rows
below r’ it will be in the next layer of TD-1’s.
This means that r must be r’ rows below the first layer of TD-1’s, because TD-1 has the same number of
rows regardless of what layer it is in and the two TD-1’s in layer 2 are identical to TD-1 in layer 1, so r is in
the same relative position of TD-1 in the second layer as r’ in the first layer. The next layer will always
have two copies of TD-1 which will each have a copy of r’. There will also be some number of zeros
between the copies of TD-1. This means that the zeros will be between r’ in the first copy of TD-1 in the
second layer and r’ in the second cope of TD-1 in the second copy of TD-1 in the second layer. This means r
will have double the number of 1’s than r’ in the row of TD-1 in the first layer. Since r has double the
number of 1’s as r’, we can use this equation: 2(2n’)= the number of 1’s in row r where n’ is equal to the
number of 1’s in the binary expansion of r’. This equation can be written as: 2n’+1. n’ is the number of
1’s in the binary expansion of r’ and r’ is equal to r with the leftmost digit removed. This means that n’
has one less one than n, so n’+1 equals n. We can then write 2n’+1 as 2n. This proves inductively that the
number of ones in a row of Pascal’s Triangle (Mod 2) will be equal to 2n, where n is the number of 1’s in
the binary expansion of that row number.