Deduce the following statement from the properties shown below. For all real numbers a and b,...

 

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Deduce the following statement from the properties shown below. For all real numbers a and b, llal - |bl|s|a - bl. 1. Property 1. For all real numbers x and y, x + y x + lyl. (The triangle inequality-Theorem 4.5.6.) 2. Property 2. For every positive real number c, if -c srs c, then Irl sc. (Section 4.5, exercise 42.) Proof: Suppose a and b are any real numbers. By applying the triangle inequality (Property 1) with x = ab and y = b, then (a - b) + bl s have, lal Hence, by subtracting [b] from both sides, we get, lal (*) and so, because (a - b) + b = a, we Similarly, by applying the triangle inequality (Property 1) with x = ba and y = a, we have |(ba) +as and so, because (ba) + ab, we have |b| Hence, by subtracting |al from both sides, we get [b]- From this we get the following. b-2 -(101- ? lal ? Deduce the following statement from the properties shown below. For all real numbers a and b, llal - |bl|s|a - bl. 1. Property 1. For all real numbers x and y, x + y x + lyl. (The triangle inequality-Theorem 4.5.6.) 2. Property 2. For every positive real number c, if -c srs c, then Irl sc. (Section 4.5, exercise 42.) Proof: Suppose a and b are any real numbers. By applying the triangle inequality (Property 1) with x = ab and y = b, then (a - b) + bl s have, lal Hence, by subtracting [b] from both sides, we get, lal (*) and so, because (a - b) + b = a, we Similarly, by applying the triangle inequality (Property 1) with x = ba and y = a, we have |(ba) +as and so, because (ba) + ab, we have |b| Hence, by subtracting |al from both sides, we get [b]- From this we get the following. b-2 -(101- ? lal ?