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MATH1004
Summer School, 2017
Discrete Mathematics
Solutions to Assignment 1
1. Consider the sequence of triangular arrays of dots constructed as follows. Begin with the
first array consisting of one dot. Each subsequent array is obtained by adding a new row
of dots beneath the previous array so that each row contains exactly one more dot than
the one above it. The first four arrays are drawn below:
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The nth triangular number is defined to be the number of dots in the triangular array
consisting of n rows. For example, the first, second and third triangular numbers are 1, 3
and 6 respectively.
(a) [2 marks] Write down the next 5 triangular numbers.
Solution: The next 5 triangular numbers are 10, 15, 21, 28 and 35. Observe that the
nth triangular number can be obtained by adding n to the (n − 1)th triangular number.
(b) [1 mark] Guess a relationship between the sequence of triangular numbers and
Pascal’s triangle. (You do not need to provide any reasons.)
Solution: The triangular numbers coincide with the third diagonal of Pascal’s triangle.
2. [5 marks] Recall that E = {x | x = 2k, k ∈ Z} is the set of even integers. Let S =
{x | x = n2 , n ∈ Z} be the set of perfect squares and F = {x | x = 4n2 , n ∈ Z}. Prove
that
E ∩ S = F.
Solution: To prove E ∩ S = F we need to show both E ∩ S ⊆ F and F ⊆ E ∩ S.
For the first inclusion, given any x ∈ E ∩ S we need to show that x ∈ F . Now x ∈ E ∩ S
means that x ∈ E and x ∈ S. In other words, x is an even perfect square. Now x = n2 for some
n ∈ Z. Observe that n must be even, for otherwise if n was odd then x = n2 would also be odd,
which is a contradiction. Therefore n = 2k for some k ∈ Z and hence x = n2 = (2k)2 = 4k 2 ∈ F .
Hence E ∩ S ⊆ F .
For the reverse inclusion, let x ∈ F . Then x = 4n2 for some n ∈ Z. Observe that x must
be an even integer since x = 4n2 = 2(2n2 ) and 2n2 ∈ Z. Furthermore, x is a perfect sequare
since x = 4n2 = (2n)2 and 2n ∈ Z. Therefore x ∈ E and x ∈ S, which means x ∈ E ∩ S. Thus
F ⊆ E ∩ S.
In conclusion, we have E ∩ S = F .
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