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Let (y=x^2): (y^2-5y+4=(y-1)(y-4)=(x^2-1)(x^2-4)=(x-1)(x+1)(x-2)(x+2)).
[ A\cup B = 1,2,3,4,\quad A\cap B = 2,3 ] [ A\setminus B = 1,\quad B\setminus A = 4 ] Remark : Set difference removes elements of the second set from the first.
Let remainder be (ax+b). Write (x^100 = (x^2-1)Q(x) + ax+b). Set (x=1): (1 = a+b). Set (x=-1): (1 = -a+b). Solve: adding → (2=2b \Rightarrow b=1,\ a=0). Remainder = 1. Chapter 7 – Relations and Functions Exercise 7.2 Define relation (R) on (\mathbbZ) by (aRb) if (a-b) is even. Prove (R) is an equivalence relation.
Multiply numerator and denominator by conjugate (1+i): [ \frac(2+3i)(1+i)(1-i)(1+i) = \frac2+2i+3i+3i^21+1 = \frac2+5i-32 = \frac-1+5i2 = -\frac12 + \frac52i ]
Work mod 7: (2^1\equiv 2,\ 2^2\equiv 4,\ 2^3\equiv 1 \pmod7) (since (8\equiv 1)). Thus (2^3k\equiv 1). Write (100 = 3\cdot 33 + 1). (2^100 = (2^3)^33\cdot 2^1 \equiv 1^33\cdot 2 \equiv 2 \pmod7). Remainder = 2.
(x^2 < 1 \Rightarrow x^2 -1 < 0 \Rightarrow (x-1)(x+1) < 0). Product negative iff one factor positive, the other negative. Case 1: (x-1<0) and (x+1>0) → (x<1) and (x>-1) → (-1<x<1). Case 2: (x-1>0) and (x+1<0) impossible (would require (x>1) and (x<-1)). Thus (-1<x<1).
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