# Foundation of maths questions

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1: Use the Euclidean Algorithm to find the Highest Common Factor of 1176 and

1960.

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 Summary

A selection of foundation of math problems are solved.

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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,

A^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 ...

###### Education

- BSc , Wuhan Univ. China
- MA, Shandong Univ.

###### Recent Feedback

- "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
- "excellent work"
- "Thank you so much for all of your help!!! I will be posting another assignment. Please let me know (once posted), if the credits I'm offering is enough or you ! Thanks again!"
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- "Thank you very much for your valuable time and assistance!"

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