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secondary 4 | E Maths
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For the complete set, does the value if x go up to infinity? If so, how are we supposed to draw the values?
Date Posted: 5 years ago
Views: 730
P is a prime number
S is a perfect square
T is an integer ending in “2”
Since epsilon contains integers from 2 upwards and there are no limits, the set goes to infinity.
Unfortunately the sets P, S and T go to infinity as well, so we are unable to actually list the entire set out. There is no hard and fast rule on how to write out the elements in the set (which is why the question does not require you to list out all the elements in epsilon and each set).
If I have to list down the elements, what I would personally do is to list down perhaps the first 10 qualifying numbers in each set and then write “and other integers ending in 2” etc.
S is a perfect square
T is an integer ending in “2”
Since epsilon contains integers from 2 upwards and there are no limits, the set goes to infinity.
Unfortunately the sets P, S and T go to infinity as well, so we are unable to actually list the entire set out. There is no hard and fast rule on how to write out the elements in the set (which is why the question does not require you to list out all the elements in epsilon and each set).
If I have to list down the elements, what I would personally do is to list down perhaps the first 10 qualifying numbers in each set and then write “and other integers ending in 2” etc.
Thanks Eric! Just asking, is there a chance that there will be a number matching the criteria going further down the list of numbers?
Thanks!!
Thanks!!
The fact that O Level examiners can confidently post such questions shows that there must be some sort of pattern going on.
Let’s consider the respective sets.
The list of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and so on.
The list of perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400 and so on. Observe that the final digits are never going to end in 2, 3, 7 and 8 even if the numbers go down the list (I believe the reason is covered in university).
The list of numbers ending in 2 are 2, 12, 22, 32, 42, 52, 62, 72, 82, 92, 102 and so on.
Obviously P n T = {2} since the common element is 2 (do not write 2 on its own without the brackets since we are talking about sets and not elements when we say P n T; we only write 2 without the brackets when the question is phrased such that you are to list down the common elements in P and T). This follows that n (P n T) = 1.
Similarly, there is nothing common in S and T, so S n T = o/ (you can also put { } to denote that the set of solutions contains nothing at all, but do not write the null set o/ inside the brackets { } at the same time!) and therefore n (S n T) = 0.
Let’s consider the respective sets.
The list of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and so on.
The list of perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400 and so on. Observe that the final digits are never going to end in 2, 3, 7 and 8 even if the numbers go down the list (I believe the reason is covered in university).
The list of numbers ending in 2 are 2, 12, 22, 32, 42, 52, 62, 72, 82, 92, 102 and so on.
Obviously P n T = {2} since the common element is 2 (do not write 2 on its own without the brackets since we are talking about sets and not elements when we say P n T; we only write 2 without the brackets when the question is phrased such that you are to list down the common elements in P and T). This follows that n (P n T) = 1.
Similarly, there is nothing common in S and T, so S n T = o/ (you can also put { } to denote that the set of solutions contains nothing at all, but do not write the null set o/ inside the brackets { } at the same time!) and therefore n (S n T) = 0.
They won't give you a question asking you to find the number of terms to infinity. For this question it's quite simple.
i)
n(P∩T) = 1 as the only integer ending with 2 and is a prime number is the number 2.
All other integers ending with 2 would be divisible by 2 since ending with a 2 means they are even. Which makes them non-prime.
ii)
n(S∩T) = 0
The perfect squares are squares of integers. There are only 10 possibe digits that the integers end with :
0,1,2,3,4,5,6,7,8,9
When the integers are squared, the digit the product ends with (i.e in the ones place) will be the last digit of the square of those digits.
0x0 =0
1x1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
For example,
44² = 1936
4² = 16
The digit ends with a 4.
So the last digit can only be 0,1,4,5,6,9. It can never be 2.
i)
n(P∩T) = 1 as the only integer ending with 2 and is a prime number is the number 2.
All other integers ending with 2 would be divisible by 2 since ending with a 2 means they are even. Which makes them non-prime.
ii)
n(S∩T) = 0
The perfect squares are squares of integers. There are only 10 possibe digits that the integers end with :
0,1,2,3,4,5,6,7,8,9
When the integers are squared, the digit the product ends with (i.e in the ones place) will be the last digit of the square of those digits.
0x0 =0
1x1 = 1
2 x 2 = 4
3 x 3 = 9
4 x 4 = 16
5 x 5 = 25
6 x 6 = 36
7 x 7 = 49
8 x 8 = 64
9 x 9 = 81
For example,
44² = 1936
4² = 16
The digit ends with a 4.
So the last digit can only be 0,1,4,5,6,9. It can never be 2.
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I've included some side notes in red (see pic). Hope that helps! :)
Date Posted:
5 years ago
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