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Chapter 2: Problem 79

Show that $$ (a+b)^{2}=a^{2}+b^{2} $$ if and only if \(a=0\) or \(b=0\).

Short Answer

Expert verified
In summary, we have proven that \((a + b)^2 = a^2 + b^2\) if and only if \(a = 0\) or \(b = 0\). We first demonstrated this by directly substituting \(a = 0\) or \(b = 0\), and then we proved the converse by expanding \((a + b)^2\) and manipulating the resulting algebraic equation.

Step by step solution

01

Prove the forward direction (if)

Assume either \(a = 0\) or \(b = 0\). - If \(a = 0\), then \((a + b)^2 = (0 + b)^2 = b^2\), and \(a^2 + b^2 = 0^2 + b^2 = b^2\). Therefore, \((a + b)^2 = a^2 + b^2\). - If \(b = 0\), then \((a + b)^2 = (a + 0)^2 = a^2\), and \(a^2 + b^2 = a^2 + 0^2 = a^2\). Therefore, \((a + b)^2 = a^2 + b^2\). Thus, we have shown that if \(a = 0\) or \(b = 0\), then \((a + b)^2 = a^2 + b^2\).
02

Prove the reverse direction (only if)

Now, assume that \((a + b)^2 = a^2 + b^2\). We want to show that this implies either \(a = 0\) or \(b = 0\). First, let's expand the expression \((a + b)^2\): \((a + b)^2 = a^2 + 2ab + b^2\) Now, since we've assumed that \((a + b)^2 = a^2 + b^2\), we have: \(a^2 + 2ab + b^2 = a^2 + b^2\) Subtracting \(a^2\) and \(b^2\) from both sides, we get: \(2ab = 0\) Since 2 is nonzero, the only way this equation can hold is if either \(a = 0\) or \(b = 0\). Therefore, we have shown that if \((a + b)^2 = a^2 + b^2\), then either \(a = 0\) or \(b = 0\).
03

Conclusion

We have proven that \((a + b)^2 = a^2 + b^2\) if and only if \(a = 0\) or \(b = 0\). The forward direction was shown by direct substitution of \(a = 0\) or \(b = 0\), and the reverse direction was proven by manipulating the algebraic equation under the assumption that \((a + b)^2 = a^2 + b^2\).

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