# Two-way bounding problem 3

Here are the set definitions again for reference:

- \(A = \{(x,y) \in \mathbb{R}^2 : y = 3x + 7\} \)
- \( B = \{\lambda(-2,1) + (1-\lambda)(1,10) : \lambda \in \mathbb{R}\}\)

### Solution

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.

### Comments

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.