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Set Theory

I am currently having trouble understanding these sets.

I am on letter a and I worked out that k is an element of P, where p is composed of positive integers, such that "k-1 is a multiple of 7." No clue what to do with this part.

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    Is $\Bbb P$ the set of all prime numbers? – 2017-01-10
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    @RSerrao In the book it says that P is a set of positive integers. – 2017-01-10
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    $k-1=n \times 7$ i.e. $k=(n \times 7) +1$. Then calculate for $n=1,2,3,4$ – 2017-01-10

3 Answers 3

3

$\mathbb{P}$ is weird notation for positive integers. (I would read it as the primes). Anyway, you just need to find four positive integers $k$ such that $k-1$ is a multiple of 7. This means $k$ is one more than a multiple of 7. 7 is a multiple of 7, so $k=8$ is in the set. Just think of a few other multiples of 7 and proceed accordingly.

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    So we are actually solving k?? – 2017-01-10
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    @keving A multiple of 7 is any integer times 7. So 0,7,-7,14,-14, etc. – 2017-01-10
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    @keving The set consists of all positive integers that are one more than a multiple of 7. The set notation says that for k to be in the set, k-1 must be a multiple of 7. So you can write $k-1 = 7n$ for some integer $n$ so that $k = 7n+1.$ So you can get different numbers in the set by plugging different integers $n$ into the right-hand-side of is equation. (Provided it comes out to a positive number.) – 2017-01-10
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    Still a bit confused as to why k would equal 8 as it is not a multiple of 7.. – 2017-01-10
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    @keving k does not equal a multiple of 7. k-1 equals a multiple of 7. So if k = 8, then k-1 =7. 7 is a multiple of 7. – 2017-01-10
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    So going off k=7n+1,{8, 15, 22, 29} ? – 2017-01-10
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    @keving Yes. That is right. – 2017-01-10
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51546/discussion-between-keving-and-spaceisdarkgreen). – 2017-01-10
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51548/discussion-between-keving-and-spaceisdarkgreen). – 2017-01-10
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Let us say that the set $A$ is as defined in a):

$$A = \{k \in \Bbb P | k-1\ \text{is a multiple of 7}\}$$

The part that is left of the vertical bar, $k \in \Bbb P$, reads: "elements that are in the set $A$ are the values $k$ that come from $\Bbb P$, the set of all positive numbers". The vertical bar can then be read as "such that". The part to the right of it reads "$k-1$ must be a multiple of $7$". Putting it all together, you have:

$$A = \{\text{set composed of all positive numbers $k$ such that $k-1$ happens to be a multiple of 7}\}$$

So for example, $2$ will not be in $A$ because even though $2$ is positive, $2-1 = 1$ is not a multiple of $7$. $8$ on the other hand will be there because $8$ is positive and $8-1 = 7$ is a multiple of $7$.

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This is my answer:

a)
$\mathbb{P}$ is a notation used for positive integers (and a lot of times a notation for prime numbers), I would take it as prime numbers. So you need to find $4$ prine numbers $p_i,~i\in\mathbb{N},1\le i\le 4$ such that $7\mid(p_i+1)$ so you need to $7q+1,~q\in\mathbb{N}$ be a prime number.

Obviously if $q$ is odd then $7q+1$ is even then it is prime if $7q+1=2\Rightarrow7q=1$ and this is not possible because $q\in\mathbb{N}$.

So $q$ is even lets find $p_i$:
$q=2\Rightarrow7k+1=7\cdot2+1=15$ and it is not prime $(3\mid15)$.
$q=4\Rightarrow7k+1=7\cdot4+1=29$ and it is prime, so $p_1=29$.
$q=6\Rightarrow7k+1=7\cdot6+1=43$ and it is prime, so $p_2$=43.
$q=8\Rightarrow7k+1=7\cdot8+1=57$ and it is not prime $(3\mid57)$.
$q=10\Rightarrow7k+1=7\cdot10+1=71$ and it is not prime $(3\mid57)$.
$q=12\Rightarrow7k+1=7\cdot12+1=85$ and it is not prime $(5\mid85)$.
$q=14\Rightarrow7k+1=7\cdot14+1=99$ and it is not prime $(3\mid99)$. $q=16\Rightarrow7k+1=7\cdot16+1=113$ and it is prime, so $p_3=113$. $q=18\Rightarrow7k+1=7\cdot18+1=127$ and it is prime, so $p_4=127$.

Then this are your four elements: $k\in\{29,43,113,127\}$.

b)
$x\in\{$apple, strawberry, pear, peach$\}$

c)
$\displaystyle x\in\left\{1,\frac{1}{2},\frac{1}{3},\frac{1}{4}\right\}$

d)
$n\in\{-1,-2,-3,-4\}$

e)
Again $\mathbb{P}$, lets take $n\in\{2,3,5,7\}$.