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3. Prove: if $c > 0$ Then $a > b$ implies $ac > bc$ 4. Prove that the equation $x^2 + 1 = 0$ has no integer solution 5. Prove that $||a| - |b|| \le |a - b|$ 6. Prove $|a - b| \le |a - c| + |c - b|$ 7. Prove $||x| - |y|| \le |x - y|$ 8.(* ) Prove $a < b$ implies $a^3 < b^3$

          3. Prove: if $c > 0$ Then $a > b$ implies $ac > bc$
4. Prove that the equation $x^2 + 1 = 0$ 
has no integer solution
5. Prove that $||a| - |b|| \le |a - b|$
6. Prove $|a - b| \le |a - c| + |c - b|$
7. Prove $||x| - |y|| \le |x - y|$
8.(* ) Prove $a < b$ implies $a^3 < b^3$

3. Prove: if c > 0 Then a > b implies ac > bc
4. Prove that the equation x^2 + 1 = 0
has no integer solution
5. Prove that ||a| - |b|| ≤ |a - b|
6. Prove |a - b| ≤ |a - c| + |c - b|
7. Prove ||x| - |y|| ≤ |x - y|
8.(* ) Prove a < b implies a^3 < b^3

Added by Lauren O.

Computer Science and Information Technology

Trishna Knowledge Systems 2018 Edition

Instant Answer

Solved by Expert Tarandeep Singh

Step 1

Prove: if c > 0, then a > 5 implies ac > bc. To prove this statement, we can use the transitive property of inequalities. Given: c > 0 and a > 5 We want to prove: ac > bc Since a > 5, we can multiply both sides of the inequality by c: ac > 5c Since c > 0, we Show more…

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3. Prove: if c > 0, then a > 5 implies ac > bc. 4. Prove that the equation x + 1 = 0 has no integer solution. 5. Prove that |a - b| < |a - b|. 6. Prove |a - b| < |c - a| + |c - 5|. 7. Prove: |x - 1| - |y - 1| < |x - y|. 8. Prove that a < 5 implies a^3 < 53.