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$X_1$ is a set of numbers which are neither prime nor composite. $X_2$ is a set of numbers from 1 to 40 that are multiples of 10.

Find $X_1 \cup X_2$).


When I tried I got the answer as $\{1,10,20,30,40\}$
There is only one number which is neither prime nor composite, i.e. $1$ [I didn't know it included $0$]
$\Rightarrow X_1 = \{1\}, X_2 = \{10,20,30,40\}$
$X_1 \cup X_2 = \{1,10,20,30,40\}$.

But in the book it is written:
There is only one number which is neither prime nor composite, i.e. $0$
$\Rightarrow X_1 = \{0\}, X_2 = \{10,20,30,40\}$
$X_1 \cup X_2 = \{0,10,20,30,40\}$.

So, can you help me with this? Thanks.

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Could you make sure you're talking of integer or even natural numbers? –  DonAntonio Feb 10 '13 at 10:19
    
Not even natural numbers –  chndn Feb 10 '13 at 10:20
    
The question just says 'numbers' so i think it is non negative numbers –  chndn Feb 10 '13 at 10:21
2  
$\sqrt 2$ is neither orime nor composite ... –  Hagen von Eitzen Feb 10 '13 at 10:21
1  
The book, if it says what you say it says, is wrong. $1$ is neither prime nor composite. –  Gerry Myerson Feb 10 '13 at 10:53

1 Answer 1

up vote 1 down vote accepted

I believe that this is simply mistake - someone printed $0$ instead of $1$ in the answers. And then your solution is correct. (But this explanation becomes more improbable if the answer mentions $0$ more than once. Then I believe it's even bigger author's mistake - in thinking instead of typing.)

Number $1$ is usually excluded from primes because of uniqueness of natural number factorization ("Every number can be expressed as product of primes in only one way, if order is unimportant") - because number $1$ as prime would allow many different factorizations like $2=1\cdot 2=1\cdot 1\cdot 2=\ldots$

Also the context - primes and compounds - usually doesn't include number $0$ (maybe only as reminder).

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You may be correct. thnx –  chndn Feb 10 '13 at 12:21

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