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    H denotes the set of numbers of the form a + b 5, where a and b are rational. The numbers are
    combined under multiplication.
    (i) Show that the product of any two members of H is a member of H .
    [2]
    It is now given that, for a and b not both zero, H forms a group under multiplication.
    (ii) State the identity element of the group.
    [1]

    (iii) Find the inverse of a + b 5.
    [2]
    (iv) With reference to your answer to part (iii), state a property of the number 5 which ensures that
    every number in the group has an inverse.
    [1]
    I'm stuck on part iv). The answer says that it is because 5 is prime or sqrt5 is irrational, however I'm not sure why this is relevant.

    (The answer to part (iii) was a/(a^2-5b^2) - (b/(a^2-5b^2))sqrt5 )

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    Notice that there is always an inverse if the denominators in the expression  \dfrac{a}{a^2-5b^2} - \dfrac{b}{a^2-5b^2} \sqrt{5} are not zero; but why can't  a^2-5b^2 equal zero?, with  a, b \in \mathbb{Q}.
 
 
 
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Updated: May 23, 2010

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