Pliz help me, I have Set theory, Relations and Groups Problems?

A1. Determine whether the relation xRy where xRy means there exists an integer n such that x=2n y is an equivalence relation. Is the relation antisymmetric?

A2. Define an operation * on R by x*y=xy +1. Show that * is commutative but not associative. If we consider R\{0) does * have an identity on this set?

A3. Let a,b be fixed rel numbers and define θ:R→R by θ(x)=ax+b. Prove that θ is bijective if and only if a≠0 and give a formula for θ-1.

A4. Prove that (AxB)∩(BxA)=(A∩B)x(A∩B) for any sets A,B. Is the result true if we replace ∩by Ù?

A5. Let n be an integer. Prove that if n2 is even then n is even.

Hint: Assume that n is not even, ie. N is odd.

A6. Determine whether the relation “x divides y” is a total order on the set W={1,2,4}.

A7. Let M denote the set of all invertible 2x2 matrices with real entries. Prove that M is a commutative group under addition.

A8. Let S =P(A) (the set of all subsets of A). Do the semigroups (S, V ) and (S,^) have identities?

A9. Let (A,*) be a set closed under * with identity 1 and a,x be elements of A with inverses b and y respectively. What is the inverse of a*x in A?

A10. Let f:R→R and g:R→R be functions defined by f(x)= 2x+1 and g(x)+x2-2.

Determine which one is injective and which one is surjective.

Determine whether f and g are bijective.

Determine (g о f)(2).

A11. Determine whether f:R→R given by f(x)=√x is a mapping.

1 Answer

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  • 1 decade ago
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    i will help you with some.. . .

    2 .. . . x*y = xy + 1

    while y*x = yx + 1 = xy + 1 ... thus the relation is commutative

    now, consider (1 * (2 * 3)) = 1 * 7 = 8

    then ((1 * 2) * 3) = (3 * 3) = 10

    thus it is not associative.

    3. θ^(-1) = (x - b) / a ... thus it is necessary that a ≠ 0

    5. supposing n is odd, then n = (2k+1) for some integer k

    then n^2 = 4k^2 + 4k + 1 = odd

    thus we contradict the hypothesis.

    thus if n is odd, then n^2 is odd... by contrapositivity

    if n^2 is even, then n is even. (it also needs law of excluded middle, n is not odd means n is even)

    .. . .. .

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