Problem 1 · AMC 8 Stretch
Core
Arithmetic & Operations
Number Theory
recognizing-false-ruleslogical-reasoning
A common mistake is to 'break a square root apart' over a plus sign: \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\). Test it with \(a = 9, b = 16\): compute \(\sqrt{9}+\sqrt{16}\). (Compare it to the true value \(\sqrt{9+16}\) to see the rule is false.)
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Answer: 7 (while the true value is 5, so the rule is false)
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Hint 1 of 4
A 'rule' that is supposed to always work can be destroyed by a single example where it fails. Pick easy perfect squares so you can compute both sides in your head.
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Hint 2 of 4
Work out the RIGHT side: \(\sqrt{9}+\sqrt{16}=3+4\).
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Hint 3 of 4
Now the LEFT side: \(9+16=25\), and \(\sqrt{25}=5\). Are they equal?
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Approach: Disprove a false rule with one numerical counterexample
- Compute the right side of the supposed rule: \(\sqrt{9}+\sqrt{16}=3+4=7\).
- Compute the true left side: \(\sqrt{9+16}=\sqrt{25}=5\).
- Since \(7 \neq 5\), the rule \(\sqrt{a+b}=\sqrt{a}+\sqrt{b}\) is FALSE — one counterexample is enough to kill it.
- Same idea checks a claim like \(\sqrt{36}=\sqrt{30}+\sqrt{6}\): \(\sqrt{30} > 5\) and \(\sqrt{6} > 2\), so the right side exceeds \(7\), far more than \(\sqrt{36}=6\). A square root never splits over a plus sign.
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