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MTH 231, Winter 2021
Week 5, Recitation
Remember to show all work in the space provided to obtain full credit. Put your answer in the box or the
space provided under each problem.
1. Show that at least ten of any 64 days chosen must fall on the same day of the week.
We can show this via contradiction. That is, we want to assume that 9 or fewer days out of 64 fall on each
day of the week. Now since there are 7 days in a week and each one has at most 9 days that fall on it, this
gives us a maximum total of
7 · 9 = 63 days
This is a contradiction to our assumption of having 64 days. Therefore our assumption must be false. In
other words it is true that at least ten of any 64 days chosen must fall on the same day of the week.
2. Show that the two statements are equivalent.
3x + 2 is even
x + 5 is odd
To show these are equivalent we must show that 3x + 2 is even, if and only if x + 5 is even.
(→) Assume that 3x + 2 is even. By definition there exist an integer k such that 3x + 2 = 2k. This shows
that 2 is a factor of 3x + 2. Since 3 is not divisible by 2, this means that x must be divisible by 2. This
tells us that x is an even number. By definition there exist an integer m such that x = 2m. Turning our
attention to x + 5 we have that
x + 5 = 2m + 5
= 2m + 4 + 1
= 2(m + 2) + 1 factor a 2
Since m is an integer, we know that m + 2 is also an integer. So by definition x + 5 is even.
(←) Assume that x + 5 is odd. By definition there exist an integer k such that x + 5 = 2k + 1. Subtracting
1 from both sides of this we get that x + 4 = 2k. By definition then x + 4 is an even number. This implies
that x is divisible by 2, hence x is an even number. So there exist an integer m such that x = 2m. Turning
our attention to 3x + 2 we have that
3x + 2 = 3(2m) + 2
= 2(3m + 1) factor out a 2
Since m is an integer, then so is 3x + 1. So by definition 3x + 2 is even.
Since we have proven both directions (→) and (←), we know the two statements are equivalent.
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3. Find a counterexample to the statement that every positive integer can be written as the sum of the
squares of three integers.
A counter example to this would be the number 7. Included in the explanation you should say why it can
not be written as the sum of squares of three integers.
For example we know that any time we square an integer it will be a positive number. So if it is possible, we
would want the sum of three positive numbers to equal 7. Here are all the possible sums of three positive
numbers. (note changing the order does not count as a separate case)
7=0+0+7
=0+1+6
=0+2+5
=0+3+4
=1+1+5
=1+2+4
=1+3+3
=2+2+3
In each case we have a number that is not a square number. Since this exhaust all the possibilities, we
know that 7 is a counterexample.
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4. Prove the triangle inequality, which states that if x and y are real numbers, then
|x| + |y| ≥ |x + y|
where |x| represents the absolute value of x. Note that |x| = x if x is positive and |x| = −x if x is negative.
Case 1 (x = 0) In this case we have that
|x| + |y| = 0 + |y|
= |y|
= |0 + y|
= |x + y|
So in this case the triangle inequality holds. WLOG we know that this will also be true when y = 0.
Case 2 (x > 0 and y > 0) In this case we know that indivisually x and y will be positive. |x| = x and
|y| = y. We also know the sum of x and y will also be positive. So |x + y| = x + y. This gives us that
|x| + |y| = x + y and |x + y| = x + y
So in this case the triangle inequality holds. WLOG we know that this will be true when x < 0 and y < 0.
Case 3 x > 0 and y < 0 In this case we have that |x| = x and |y| = −y. So we have that
|x| + |y| = x + (−y)
From here we consider three sub-cases
(a) x + y = 0 In this case we have |x + y| = |0| = 0. Since y < 0 we know that −y > 0, and so
x + (−y) > 0 + (−y) = −y > 0. So it must be that
|x| + |y| = x + (−y)
>0
= |x + y|
(b) x + y > 0 In this case we have that that |x + y| = x + y. Since y < 0 it must be that y < 0 < −y, and
so x + (−y) > x + y. This gives us
|x| + |y| = x + (−y)
>x+y
= |x + y|
(c) x + y < 0 In this case we have that |x + y| = −(x + y) = (−x) + (−y). Since x > 0 it must be that
−x < 0 < x, and so x + (−y) > (−x) + (−y). This gives us
|x| + |y| = x + (−y)
> (−x) + (−y)
= −(x + y)
= |x + y|
So in this case (and all sub-cases) the triangle inequality holds. WLOG we also know it will hold when
x > 0 and y > 0.
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5. Prove or disprove that you can use dominoes to tile the standard checkerboard with two adjacent corners
removed (that is, corners that are not opposite).
Since this is a prove involving the existence of such a tiling, one valid proof would be to draw a standard
8 × 8 checker board and show that it can be tiled with dominoes.
6. Find a formula for the sum of the first n even positive integers. Then prove the formula you came up
with is true using mathematical induction.
To begin searching for a formula, compute the first few sums and look for a pattern.
n
1
2
3
4
5
Sum
2
2+4
2+4+6
2+4+6+8
2 + 4 + · · · + 10
Result
2
6
12
20
30
From the table it appears the sum for the first n integers will be n(n + 1). In other words
2 + 4 + 6 + · · · + 2n = n(n + 1)
We will prove this via induction.
Base Case The sum of the first even integer is 2. Looking at the formula for n = 1 we have that 1(1+1) = 2.
So in this case the formula holds true.
Inductive Step Assume that the formula holds true for the sum of the first k even integers, and show
that it holds for the sum of the first k + 1 even integers.
Starting with the sum we have
2 + 4 + · · · + 2k + 2(k + 1) = k(k + 1) + 2(k + 1) by assumption
= k 2 + k + 2k + 2
= k 2 + 3k + 2
= (k + 1)(k + 2)
This show the formula holds for k + 1.
From our base case and the inductive step, we know that the sum of the first n even integers is given by
n(n + 1), where n ≥ 1.
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