Question

{3,7,11,15,19}=write the set in set builder form A

Answer 1

Answer:

A={(x,y) where y=2x+1 :2≤x≤9}

Answer 2

Answer:

A = {x : x = 4m + 3, and 0 ≤ m ≤ 4, where m ∈ N}

Step-by-step Explanation:

We have,

A = {3, 7, 11, 15, 19}

Now, when we check the pattern,

we can say that,

All these numbers are odd, so we can directly say it is of the form 2n + 1

So,

In set builder form we can write,

x = 2n + 1

But there is a mistake here,

We can't put n = 0, then x = 1

and 1 is not in the set

Also,

For n = 2, x = 5, which is again not in the set.

Similarly,

For n = 4, x = 9 which is again not in the set.

But,

For n = 1, x = 3

For n = 3, x = 7

For n = 5, x = 11

You see,

n also must be an odd number then only we get just the numbers in the set,

So,

Let n = 2m + 1

Then,

x = 2n + 1

From our observations, n = 2m + 1

x = 2(2m + 1) + 1

x = 4m + 2 + 1

x = 4m + 3

Here if we substitute m = 0, we get x = 3

For m = 1, x = 7

For m = 2, x = 11

For m = 3, x = 15

For m = 4, x = 19

So, We can say that,

0 ≤ m ≤ 4, where m ∈ N

Then, set builder form of set A is

A = {x : x = 4m + 3, and 0 ≤ m ≤ 4, where m ∈ N}

Hope it helped you and believing you understood it...All the best