Question Video: Writing a Relation between Two Sets given Its Rule Mathematics • 8th Grade

π₯ = {3, 7, 5}, π¦ = {9, 6, 49, 25}, and π is a relation from π₯ to π¦, where πππ means that π = βπ for each π β π₯ and π β π¦. Determine the relation π.

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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 ππ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, weβve been given two sets. Weβre 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, weβre 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, weβre looking to find the relation, and weβre told that πππ means that π is equal to the square root of π. Letβs 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, weβll begin by looking at the values for π and then identifying possible corresponding values for π. So, letβs begin by looking at the first value of π, nine. If π is equal to nine, then, according to πππ, π must be equal to the square root of nine, and thatβs 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.

Weβll now look at the second element in set π¦, and weβre 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. Itβs roughly 2.4. We notice there are no elements in set π₯ that map onto this element in set π¦, and thatβs fine. It just means that there is no ordered pair for this value.

Letβs 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. Weβll 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.

Weβve 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.

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