### Video Transcript

π₯ is the set containing elements
three, seven, five, π¦ is the set containing elements nine, six, 49, 25, and π
is a
relation from π₯ to π¦, where ππ
π means that π is equal to root π for π which
is an element of π₯ and π which is an element of π¦. Determine the relation π
.

So, weβve been given two sets. Weβre told that π is an element of
set π₯. Since set π₯ contains the elements
three, seven, five, π could be equal to three, it could be equal to seven, or it
could be equal to five. Similarly, weβre told that π is an
element of set π¦. Since π¦ contains the elements
nine, six, 49, 25, then π could be any of these four values. Now, weβre looking to find the
relation, and weβre told that ππ
π means that π is equal to the square root of
π. Letβs define the relation as the
set of ordered pairs, where π, π being, remember, means π is related to π, in
other words, a pair of numbers such that the first number is equal to the square
root of the second. We might say then that π
is the
set containing the elements of the ordered pair π, π.

But we need to identify the exact
values of π and π that satisfy π is equal to the square root of π. So, weβll begin by looking at the
values for π and then identifying possible corresponding values for π. So, letβs begin by looking at the
first value of π, nine. If π is equal to nine, then,
according to ππ
π, π must be equal to the square root of nine, and thatβs equal
to three. And so, we can observe that the
element three in set π₯ must map onto the element nine in set π¦. And so, the ordered pair three,
nine satisfy our relation.

Weβll now look at the second
element in set π¦, and weβre going to let π be equal to six. In this case, π will be equal to
the square root of six. Well, root six is an irrational
number. It falls somewhere between the
square root of four, two and the square root of nine, three. Itβs roughly 2.4. We notice there are no elements in
set π₯ that map onto this element in set π¦, and thatβs fine. It just means that there is no
ordered pair for this value.

Letβs continue, now calculating
what happens when π is equal to 49. When π is 49, π is equal to the
square root of 49, which is equal to seven. So, the element seven in set π₯
maps onto the element 49 in set π¦. And so, our next ordered pair must
be seven, 49. Weβll do this one more time,
finally looking at the element 25. When π is 25, π is equal to the
square root of 25, which is five. And so, our ordered pair is five,
25.

Weβve now identified all of the
mappings from set π₯ to π¦. And so, π
, our relation, is the
set containing the ordered pairs three, nine; seven, 49; five, 25.