Q?2: a) [5×1 marks] Suppose f : R ? Z where f (x) = é2x - 1ù. i. Is f one-to-one

Question

Q?2: a) [5×1 marks] Suppose f : R ? Z where f (x) = é2x - 1ù. i. Is f one-to-one? Explain. b) [5×1 marks] Prove or disprove each of the following: iii. If a ? b (mod 2m), then a ? b (mod m). v. For all integers a, b, c, if a | c and b | c, then ab | c2. Q?2: a) [5×1 marks] Suppose f : R ? Z where f (x) = é2x - 1ù. i. Is f one-to-one? Explain. b) [5×1 marks] Prove or disprove each of the following: iii. If a ? b (mod 2m), then a ? b (mod m). v. For all integers a, b, c, if a | c and b | c, then ab | c2. Q?2: a) [5×1 marks] Suppose f : R ? Z where f (x) = é2x - 1ù. i. Is f one-to-one? Explain. b) [5×1 marks] Prove or disprove each of the following: iii. If a ? b (mod 2m), then a ? b (mod m). v. For all integers a, b, c, if a | c and b | c, then ab | c2.

Explanation / Answer

I think the first question has too many typos to understand it.

So I'll answer the next 2 questions.

b) [5×1 marks] Prove or disprove each of the following:

iii. If a ? b (mod 2m), then a ? b (mod m).

Since a ? b (mod 2m), we know a = k · (2m) + b for some k.
Therefore,
a = (2k) · m + b
=> a = P . m + b ( where P = 2k)
=> a ? b (mod m).

v. For all integers a, b, c, if a | c and b | c, then ab | c2

Given a | c

=> a exactly divides c
=> c = a * k (where k is some integer)

Similarly,

b | c
=> b exactly divides c
=> c = b * l (where l is some integer)

Therefore,

c2
= c * c
= (a * k) * (b * l)
= (a*b) * (l*k)
= (a*b) * (P) (where P = l*k)
= ab . P

Hence ab | ab.P

=> ab | c2

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