SOLUTION: show the conjecture is false by finding a counterexample.
the square of the sum of two numbers is equal to the sum of the squares of the two numbers. that is, (a+b)^2 = a^2 +
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Question 471852: show the conjecture is false by finding a counterexample.
the square of the sum of two numbers is equal to the sum of the squares of the two numbers. that is, (a+b)^2 = a^2 + b^2
Found 2 solutions by bucky, richard1234:
Answer by bucky(2189) (Show Source): You can put this solution on YOUR website!
There are many possibilities for counter-examples.
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Here's one:
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Suppose we let a = 2 and b = 3. In this case the term on the left side of the equation that is given in the problem becomes:
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(a + b)^2 = (2 + 3)^2 = 5^2 = 25
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But with a = 2 and b = 3 the right side of the equation proposed by the problem is:
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a^2 + b^2 = 2^2 + 3^2 = 4 + 9 = 13
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So the equation proposed by the problem ... namely (a + b)^2 = a^2 + b^2 cannot be true because substituting 2 for a and 3 for b results in this equation becoming 25 = 13 and this is obviously NOT true.
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This one counter-example shows that you cannot trust this equation to be true in all cases.
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Hope this helps you to understand the problem. You should be able to try other values for a and b and by following the above method find other counter-examples that result in the left side of the equation NOT being equal to the right side and therefore enabling the same conclusion to be reached ... the equation cannot be trusted to be true in all cases.
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Expanding the left hand side, we get a^2 + 2ab + b^2, which is supposedly equal to a^2 + b^2. This implies that 2ab = 0 for all a,b. To obtain a counterexample to this statement, simply choose some nonzero a,b e.g. a = 5, b = -2 and it will contradict the claim.
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