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    Foundation of maths questions

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    1: Use the Euclidean Algorithm to find the Highest Common Factor of 1176 and
    2: Use the rules of natural deduction to prove ((A --> (B -->C)) -->((A B)--> C)).
    [12 marks]
    3: Use the rules of natural deduction to prove (A (B U C)) ((A B) C)
    [14 marks]
    4:Determine which of the following functions are one-one and which are onto
    i) f:R?>R f(s) =3x+1
    ii) g: N?>N, f(s) =4x+1
    iii) h:R?>R. h(s) =x2-x
    [14 marks]
    5: Given two functions f B ?> C and g.A ?> B such that f°g:A ?> C is onto prove that f is onto.
    Give an example in which fog is onto but g is not.
    [16 marks]
    6: Determine which of the following relations,[xpyl on the set N of natural numbers are (i) Reflexive, (ii) symmetric (iii transitive.(iv) antisymmetric, justifying your answers.
    (a) x is an integer multiple of y
    (b) x<y
    (c) y>x2
    [18 marks]
    7: Define a relation p on the set Z by xpy iff x-y is a multiple of 5.
    Prove p that is an equivalence relation and describe the equivalence classes of p.
    [14 marks]

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    Solution Preview

    1. 2 | 1176 1960
    2 |588 980
    299 490
    Since there are no factors other than 1 between 299 and 490,
    the highest common factor of 1176 and 1960 is 2*2=4.

    2. Since A=>(B=>C) and B=>C,

    3. Let us denote the union of A and B by AUB. Then A(BUC) means all elements belong to A but not to BUC, Of course not belong to either B or C. So every element in A(BUC) belongs to A but not to B meaning belonging to AB, at the same time it does not belong to C, so every element in A(BUC) belongs to (AB)C.

    4.(i) This function is one-to-one and onto. Since, let y1=f(x1) and y2=f(x2), if y1=y2, then we have 3x1+1=3x2+1, so x1=x2, which means this function f(x)=3x+1 is ...

    Solution Summary

    A selection of foundation of math problems are solved.