Week 33

Example 1. Prove that there is no positive integer in the interval I = (0, 1).

Example 2. Suppose that f(x) = ax4bx2 + x + 5 and that f(−3) = 2. Find the value of f(3).

Of course, you can find the solution just below, but it is highly recommended that, you first try to solve it on your own.

Just remember the words of Paul Halmos, who says

the only way to learn mathematics is to do mathematics.

Solution 1. If possible, assume that I is non-empty. Since I = (0, 1), it is the set of positive integers.

Being a set of positive integers, it must contain a least element, say m. Then 0 < m < 1.

Multiplying both sides of m < 1 by m gives m2 < m.

The square of a positive integer is a positive integer, so we have 0 < m2 < m < 1.  

But this is saying that I has a positive integer m2 which is smaller than its least positive integer m.

This is a contradiction. Thus, we have I = ϕ. Hence the result.

Solution 2. We first need to relate the value of f(−x) to the value of f(x). We then apply this relationship when x = 3. Notice that

(−x)n = xn when n is an even integer, and

(−x)n = −xn when n is an odd integer.

Applying this information implies that

f(−x) = a(−x)4b(−x)2 + (−x) + 5

= ax4bx2x + 5

= (ax4bx2 + x + 5) − 2x

= f(x) − 2x.

Therefore, f(3) = f[−(3)] − 2(−3) = 2 + 6 = 8.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: