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Common invalid steps include division by zero, assuming the converse, taking only one square root, multiplying an inequality without checking sign, and treating a non-injective function as injective. A correct-looking conclusion does not make an invalid derivation a proof. Identify the exact condition under which the questionable operation would be legitimate.
Esempio svolto
A claimed proof begins with a = b and concludes a + b = b. Locate the first invalid step.
Step 1: multiplying a = b by a gives a² = ab. This is valid.
Step 2: subtract b² from both sides to obtain a² − b² = ab − b². This is also valid.
Step 3: factor both sides: (a − b)(a + b) = b(a − b). Still valid.
Step 4: cancel a − b to obtain a + b = b. This is invalid, because the original condition a = b means a − b = 0. Cancellation divides both sides by that quantity; division by zero is undefined.
The valid factored equation only states 0 = 0. It does not constrain a + b. For a concrete check, take a = b = 1: all the equations through Step 3 hold, but the final claim becomes 2 = 1. Diagnose the exact operation rather than saying vaguely that “the proof looks wrong”.
Accesso a vita a tutte le 4 lezioni di Reasoning, mappe concettuali interattive, quesiti risolti passo per passo e simulazioni d'esame a tempo per TMUA.