Each of the following is an attempted proof of the statement For each non zero r

Question

Each of the following is an attempted proof of the statement For each non zero real number x. there is a real number y so that xy = 1. proof with the correct analysis of its merits. Let x be a nonzero real number. Let y be a real number. Pick y = 1/x, which exists because x is not zero. Then by definition of y, xy = 1. Let x be a nonzero real number. Let y = 1/x, which exists because x is not zero. Then by definition of y, xy = 1. Let x be a nonzero real number. Let y be a real number such that xy = 1. We have shown that for each nonzero x, a y such that xy = 1 exists. Say x = 2. Then if we pick y = 1/2, we have xy = 1. It works the same way for any non-zero x. Only one proof is correct. The others are incorrect, for a variety of reasons. Match each (attempted) This proof is almost correct (but still wrong) because it instantiates y as an arbitrary real number that does not depend on x, and then redefines it. The writer of this proof is perhaps used to writing programs in a language where variables first need to be declared, and then assigned a value. This is inappropriate for an existence proof. By saying "Let y be a real number", we logically did much more than declare that y was going to be a real number- we introduced it as an arbitrary real number. To use the programming analogy, we didn't just declare y to be real, but we declared it to be real and assigned a random value to it. Once we have done that, we are stuck with that random value. It is improper to redefine it later to be a specific value that depends on x. This is correct. After assuming that an arbitrary nonzero x is given, we introduce the proper y as a function of x and then verify that it has the desired property. We're basically demonstrating how we can always win this challenge-response game. Someone challenges us with nonzero x. In response, we pick a y. By picking y to be 1/x, which is always possible because x is not zero, we guarantee xy = 1 and thereby win the game. This is wrong because it is a "proof by example". The proof writer picked an example for the x, and showed how y has to be chosen for that special x. What the attempted proof actually shows is that a real number y exists so that 2y = 1. It does not show that for any nonzero x. a y exists so that xy = 1. With the last sentence, the proof writer tried to salvage the situation by appealing to the reader's sense of pattern recognition: it works the same way for all nonzero x! But to explain just how it works "the same way" for any nonzero x was the center of the job the proof writer was supposed to do. This ''proof" is wrong because it shows nothing. It simply assumes the conclusion that the required y exists for each x. One could also call this an "attempted proof by affirmative declaration".

Explanation / Answer

The OPTION NUMBER 2 IS CORRECT.because for every non zero x its reciprocal will surely exist and product of a non zero number and its reciprocal is alwats 1

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