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Hence the inequality is true when t is greater than —2 and less than or equal to — 1. We write this as —2 < r ^ — 1. Case (ii) r + 1 ^ 0 and so r > — 1. r + 2 < 0 and so t < —2. It is impossible to satisfy both r ^ — 1 and t < —2 and so this case yields no values of t. 36 Solve the following inequalities: (a) x2 > 4 Solution 2t + 3 ^ ^ ^ 1 when —2 < / ^ — 1. (b) x2 < 4 x2 > 4 (a) x2 — 4 > 0 (x — 2)(x + 2) > 0 For the product (x — 2)(a: + 2) to be positive requires either (i) x — 2 > 0 and x + 2 > 0 or (ii) x — 2 < 0 and x + 2 < 0.
Suppose we wish to multiply or divide an inequality by a quantity k. If k is positive the inequality remains the same; if k is negative then the inequality is reversed. If a > b then if k is positive if k is negative Note that when k is negative the inequality changes from > to c. Similar statements can be made for a ^ b, a < b and a ^ b. When asked to solve an inequality we need to state all the values of the variable for which the inequality is true. 34 Solve the following inequalities: (a) 3r + 1 > r + 7 Solution (b) 2 - 3z < 6 + z (a) 3r + 1 > r + 7 2r + 1 > 7 subtracting t from both sides 2t > 6 subtracting 1 from both sides ?
1 repeatedly we may write 200 = 128 + 72 = 128 + 64 + 8 = 27 + 26 + 23 = 1 (27) + 1 (26) + 0(25) + 0(24) + 1 (23) + 0(22) + 0(2') + 0(2°) = IIOOIOOO2 Another way to convert decimal numbers to binary numbers is to divide by 2 repeatedly and note the remainder. We rework the previous two examples using this method. 15 Solution Convert the following decimal numbers to binary: (a) 83 (b) 200 (a) We divide by 2 repeatedly and note the remainder. Remainder 83--2 = 41 -- 2 = 20--2= 10--2= 5--2= 2--2= 1 --2= 41 r 1 20 r 1 lOrO 5 r0 2r 1 1 r0 0r 1 1 1 0 0 1 0 1 To obtain the binary number we write out the remainder, working from the bottom one to the top one.