Solutions to Problem Set 12

Math 310-01/02

10-6-2017

1. Prove that if n is an integer, then $2 n^2 + n + 1$ is not divisible by 3.

If n is an integer, then by the Division Algorithm one of the following 3 cases holds:

$$n = 3 q, \quad n = 3 q + 1, \quad n = 3 q + 2 \quad\hbox{for}\quad q \in \integer.$$

If $n = 3 q$ , then

$$2 n^2 + n + 1 = 2(3 q)^2 + (3 q) + 1 = 18 q^2 + 3 q + 1 = 3(6 q^2 + q) + 1.$$

Hence, $2 n^2 + n + 1$ is not divisible by 3.

If $n = 3 q + 1$ , then

$$2 n^2 + n + 1 = 2(3 q + 1)^2 + (3 q + 1) + 1 = (18 q^2 + 12 q + 2) + (3 q + 1) + 1 = 18 q^2 + 15 q + 4 = 3(6 q^2 + 5 q + 1) + 1.$$

Hence, $2 n^2 + n + 1$ is not divisible by 3.

If $n = 3 q + 2$ , then

$$2 n^2 + n + 1 = 2(3 q + 2)^2 + (3 q + 2) + 1 = (18 q^2 + 24 q + 8) + (3 q + 2) + 1 = 18 q^2 + 27 q + 11 = 3(6 q^2 + 9 q + 3) + 2.$$

Hence, $2 n^2 + n + 1$ is not divisible by 3.

Therefore, if n is an integer, then $2 n^2 + n + 1$ is not divisible by 3.


2. Prove that if x is a real number, then

$$-4 \le |x - 1| - |x - 5| \le 4.$$

Consider the cases $x \le 1$ , $1 < x \le 5$ , and $x > 5$ .

Case 1. $x \le 1$ .

In this case, $x - 1 \le 0$ , so $|x - 1| = -(x - 1)$ . Also, $x \le 1$ certainly implies $x \le 5$ , so $x - 5 \le
   0$ , and $|x - 5| = -(x - 5)$ . Therefore,

$$|x - 1| - |x - 5| = -(x - 1) - [-(x - 5)] = -4.$$

Thus, in this case, $-4 \le |x -
   1| - |x - 5| \le 4$ is true.

Case 2. $1 < x \le 5$ .

In this case, $x - 1 > 0$ , so $|x - 1| = x - 1$ . Also, $x \le 5$ , so $x - 5 \le 0$ , and $|x - 5| =
   -(x - 5)$ . Therefore,

$$|x - 1| - |x - 5| = (x - 1) - [-(x - 5)] = 2 x - 6.$$

Now $1 < x \le 5$ gives $2 < 2 x \le 10$ , so $-4 < 2 x - 6 \le 4$ .

Thus, in this case, $-4 \le |x -
   1| - |x - 5| \le 4$ is true.

Case 3. $x > 5$ .

In this case, $x - 5 > 0$ , so $|x - 5| = x - 5$ . Also, $x > 5$ certainly implies $x > 1$ , so $x - 1 >
   0$ , and $|x - 1| = x - 1$ . Therefore,

$$|x - 1| - |x - 5| = (x - 1) - (x - 5) = 4.$$

Thus, in this case, $-4 \le |x -
   1| - |x - 5| \le 4$ is true.

Since the inequality holds in all cases, it is true.


I don't know that I have ever found any satisfactory answers of my own. But every time I ask it, the question is refined. ... questioning as exploration, rather than the search for certainty. - Ta-Nehisi Coates


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