Dovall rational numbers under multiplication form a group.

In summary, the set of positive rationals forms a group under multiplication, while the set of negative irrationals does not due to the lack of an identity or inverse element. The question is whether the set of all rational numbers under multiplication forms a group, and the answer is yes. For any rational number a, there exists a unique element 1/a that serves as its inverse and the product of any two rationals is also rational, making it associative. However, the inverse of 0 would not exist, as there is no rational number that can be multiplied by 0 to equal 1. Therefore, the set of all rational numbers under multiplication does form a group, with the exception of
  • #1
buzzmath
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I know the set of positives rationals form a group under multiplication and that the negative irrationals do not form a group under multiplication because there is no identity or inverse. My question is does the set of all rational numbers under multiplication form a group.
 
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  • #2
buzzmath said:
I know the set of positives rationals form a group under multiplication and that the negative irrationals do not form a group under multiplication because there is no identity or inverse. My question is does the set of all rational numbers under multiplication form a group.

Hint: does there for every [tex]a \in \textbf{Q}[/tex] exist a unique element [tex]a^{-1}\in\textbf{Q}[/tex] such that [tex]a\cdot a^{-1}=1[/tex]?
 
  • #3
Yes, it would just be 1/a. That's what I put on my test and got it wrong. He wrote that the set of rational numbers under multiplication is not a group. 1 i rational, the product of any two rationals is rational, 1/a is the inverse, and it's associative. I don't understand.
 
  • #4
buzzmath said:
Yes, it would just be 1/a. That's what I put on my test and got it wrong. He wrote that the set of rational numbers under multiplication is not a group. 1 i rational, the product of any two rationals is rational, 1/a is the inverse, and it's associative. I don't understand.

Hint 2: does there exist 1/a for a = 0? :smile:
 
  • #5
buzzmath said:
Yes, it would just be 1/a. That's what I put on my test and got it wrong. He wrote that the set of rational numbers under multiplication is not a group. 1 i rational, the product of any two rationals is rational, 1/a is the inverse, and it's associative. I don't understand.

What's the inverse of 0?
 
  • #6
Wow, I can't believe I didn't think about that. Thanks
 

Related to Dovall rational numbers under multiplication form a group.

1. What are Dovall rational numbers?

Dovall rational numbers are a specific set of rational numbers that follow a specific pattern. They are defined as fractions where the numerator and denominator follow the sequence 1, 2, 3, 4, 5, and so on. For example, 1/1, 2/2, 3/4, 4/5 are all Dovall rational numbers.

2. How do rational numbers form a group under multiplication?

Rational numbers form a group under multiplication because they follow the four group axioms: closure, associativity, identity, and inverse. This means that when you multiply any two rational numbers, the result is always a rational number, the order in which you multiply does not matter, there is a multiplicative identity element (1), and every rational number has a multiplicative inverse.

3. What is the identity element of the group of Dovall rational numbers under multiplication?

The identity element of the group of Dovall rational numbers under multiplication is 1. This means that when you multiply any Dovall rational number by 1, the result is the original number. For example, 3/5 x 1 = 3/5.

4. Can Dovall rational numbers under multiplication form a subgroup?

Yes, Dovall rational numbers under multiplication can form a subgroup. This is because they follow the same group axioms as rational numbers under multiplication. Additionally, they are a subset of the larger group of rational numbers, so they inherit the group structure from the larger group.

5. What is the order of the group of Dovall rational numbers under multiplication?

The order of the group of Dovall rational numbers under multiplication is infinite. This is because there is no limit to the number of Dovall rational numbers that can be formed by following the pattern of 1, 2, 3, 4, 5, and so on. Therefore, the group contains an infinite number of elements.

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