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Prove that for all real numbers r and c where c ≥ 0, −c ≤ r ≤ c if

and only if |r| ≤ c. (Recall that one way to prove a statement of
the form P if a

User TommyT
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Final answer:

By considering the definition of absolute value, we demonstrated that for any non-negative number c and any real number r, the inequality −c ≤ r ≤ c is equivalent to |r| ≤ c by examining the relationship in both cases where r is non-negative and where r is negative.

Step-by-step explanation:

To prove that for all real numbers r and c where c ≥ 0, −c ≤ r ≤ c if and only if |r| ≤ c, we can consider two scenarios based on the definition of absolute value:

  1. If r ≥ 0, then |r| = r. In this case, saying |r| ≤ c is the same as saying r ≤ c, which is already half of the inequality −c ≤ r ≤ c. Because r ≥ 0, it automatically satisfies the other half −c ≤ r.
  2. If r < 0, then |r| = −r. In this scenario, saying |r| ≤ c means −r ≤ c, which can be rewritten as r ≥ −c. This provides one half of the inequality −c ≤ r, and since r is less than zero, r ≤ c is also satisfied.

Thus, we can conclude that −c ≤ r ≤ c is equivalent to |r| ≤ c for any non-negative c and any real number r.

User Okcoker
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