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Another disjunction proof

  1. Jan 9, 2012 #1
    Could someone check another disjunction proof, please?

    1. The problem statement, all variables and given/known data

    I seem to think my proof of the following theorem has been made more complicated/lengthy than necessary. Any thoughts will be highly appreciated =)


    Part a) Prove that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.


    ADDENDUM! I went forward and completed part b) of this exercise before any responses were posted, so I figured I attach that as well (they are related).

    Prove that for any real number x, -|x|≤ x ≤ |x|.

    2. Relevant equations



    3. The attempt at a solution

    Proof:

    Let a and b be arbitrary real numbers.
    (-->) Suppose |a|≤ b.
    Case I. a ≥ 0. Then |a|= a and we have 0 ≤ |a| = a ≤ b. This also means that 0 ≤ b, and so -b ≤ 0. Thus, -b ≤ 0 ≤ |a| = a ≤ b. In particular, then, -b ≤ a ≤ b.
    Case II. a < 0. Then 0 < |a| = -a ≤ b. But since |a| = -a ≤ b, -|a| = a ≥ -b. Thus, -b ≤ a. Also, though, we have that 0 < -a and 0 < b. Adding these two inequalities yields 0 < b - a. Adding a to each side gives a < b. Clearly, then, a < b or a = b, so we may state that a ≤ b. Finally, together: -b ≤ a ≤ b.
    Since these cases are exhaustive and each results in -b ≤ a ≤ b, it follows that if|a|≤ b, then -b ≤ a ≤ b.
    (<--) Suppose -b ≤ a ≤ b. This means -b ≤ a and a ≤ b.
    Case I. a ≥ 0. Then |a| = a ≤ b. So in particular, |a|≤ b.
    Case II. a < 0. Then |a| = -a ≤ b. So in particular, |a| ≤ b.
    Since these cases are exhaustive and each results in |a| ≤ b, it follows that if -b ≤ a ≤ b, then |a| ≤ b.
    Lastly, since we have proven both directions of the biconditional for arbitrary real numbers a and b, we may conclude that for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b.

    Addendum proof:

    Let x be an arbitrary real number. We now break the proof into the following exhaustive cases, each of which results in the desired outcome.
    Case I. x ≥ 0. Then |x| = x, so clearly |x| = x or |x| ≤ x. Thus, we have that |x| ≤ x. With this, if we apply our theorem above (letting a = b = x), we may conclude that -x ≤ x ≤ x, which, implementing the equality established above, is the same as saying -|x|≤ x ≤ |x|.
    Case II. x < 0. Then |x| = -x. We may thus state that |x| = -x or |x| < -x, by which we rephrase to assert |x|≤ -x. Again applying the above theorem (this time with a = x and b = -x), we may conclude that -(-x) ≤ x ≤ -x. Applying the equality established above, we rexpress this as: -|x|≤ x ≤ |x|.
    Since x was chosen arbitrarily, we have shown that for any real number x, -|x|≤ x ≤ |x|.
     
    Last edited: Jan 9, 2012
  2. jcsd
  3. Jan 9, 2012 #2
    These two parts motivate a proof of the triangle inequality, which I finally solved:

    Theorem- For all real numbers x and y, |x + y| ≤ |x| + |y|.

    Proof- Let x and y be arbitrary real numbers. Note that x + y is also a real number, as is |x| + |y|. Using these notions, along with our first theorem above (for all real numbers a and b, |a|≤ b iff -b ≤ a ≤ b), letting a = x + y and b = |x| + |y|), we have |x + y| ≤ (|x| + |y|) iff -(|x| + |y|) ≤ x + y ≤ (|x| + |y|). Now, by our second theorem above (for any real number c, -|c|≤ c ≤ |c|, applying it once for c = x and again for c = y), we have both -|x|≤ x ≤ |x| and -|y|≤ y ≤ |y|. This means -|x|≤ x, -|y|≤ y, x ≤ |x|, and y ≤ |y|. Adding -|x|≤ x to -|y|≤ y yields -|x|-|y| ≤ x + y. Adding x ≤ |x|, to y ≤ |y| yields x + y ≤ |x|+|y|. If we then, in turn, add these resulting inequalities, we are left with -|x|-|y| ≤ x + y ≤ |x|+|y|, or otherwise, -(|x|+|y|) ≤ x + y ≤ |x|+|y|, as desired.
     
    Last edited: Jan 10, 2012
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