Prove that ![]()
Wednesday, 13 March 2019
11:09 AM
Prove that for all ![]()
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Discovery
Case when :![]()
ie ![]()
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Case when :![]()
ie ![]()
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[complete the square]![]()
Proof
Case 1: Suppose that Then ![]()
Ie ![]()
Adding to both sides shows that ![]()
Since we have ![]()
Therefore ![]()
Case 2: Now suppose that then ![]()
Ie ![]()
Adding to both sides give ![]()
But in this case so![]()
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