Two-way bounding problem 3

Here are the set definitions again for reference:


Lemma: \(\lambda(-2,1) + (1-\lambda)(1,10) = (-2\lambda + (1 - \lambda), \lambda + (10 - 10\lambda)) = (1 - 3\lambda, 10 - 9\lambda)\)

\(A \subseteq B\)

Let (x,y) be an element of A. Then \(y = 3x + 7\) by the definition of A.

Consider \(\lambda = \frac{1-x}{3}\).

Then, using our lemma above, we can calculate:

\(\lambda(-2,1) + (1-\lambda)(1,10) = (1 - 3\lambda, 10 - 9\lambda) = (1 - 3\frac{1-x}{3}, 10 - 9\frac{1-x}{3}) = (1 - (1-x), 10 - 3(1-x)) = (x, 3x+7) = (x,y) \)

So we've shown that (x,y) is an element of B.

Since every element of A is also an element of B, \(A \subseteq B\).

\(B \subseteq A\)

Let (x,y) be an element of B. By the definition of B, we know that \((x,y) = \lambda(-2,1) + (1-\lambda)(1,10)\) for some real number \(\lambda\).

So, using our lemma above, \((x,y) = (1 - 3\lambda, 10 - 9\lambda)\).

Then \(3x+7 = 3(1 - 3\lambda) + 7 = 3 - 9\lambda + 7 = 10 - 9\lambda = y \). So 3x+7 = y, which means that (x,y) is an element of A.

Since every element of B is also an element of A, \(B \subseteq A\).

Since \(A \subseteq B\) and \(B \subseteq A\), A=B, which is what we needed to show.


Notice that the proof for the backward direction was closely related to the proof for the forward direction, but not an exact reversal of the steps. This two-part technique is powerful because you can use slightly different approaches to handle the two halves.

The lemma wasn't strictly necessary. It barely merits the name "lemma." I just didn't feel like writing out that piece of arithmetic twice.

In upper-level classes, you might not see the summary line at the end of each section and at the end of the proof. Because this is a very standard way to show that two sets are equal, authors may just expect you to know that this is what they are doing.