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Sorry, I don't really know how to improve this question, this quiz has been given me today and I think there is no a true answer, because in the different cases of $a > 0, a < 0, b > 0, b < 0$(and nested cases) you have that every answer can be false.

Taken $a, b \in \mathbb{R}$, given $a^2 - b^2 = 0$, which one is true?

  1. $ab > 0$
  2. $a + b = 1$
  3. $ab < -1$
  4. $a > b$
  5. $a + b = 0$

These are answers I gave myself(all the examples respect the given condition of $a^2 - b^2 = 0$:

  1. If $a = 1 \land b = -1$ then $ab = -1$
  2. If $a = 1 \land b = -1$ then $a + b = 0$
  3. If $a = 0 \land b = 0$ then $ab = 0$
  4. If $a = -1 \land b = 1$ then $a < b$
  5. If $a = 1 \land b = 1$ then $a + b = 2$

Any suggestions? Thanks!

user1729
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Niccolò Campolungo
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2 Answers2

4

All of the options are false, your couter-examples are okay. Some statements you can get from $a^2 - b^2 = 0, a,b \in \mathbb R$ are: $$\begin{align*}a^2 & = b^2\\|a|^2 & = |b|^2\\|a| &= |b|\\a&=\pm b\\b&=\pm a\end{align*}$$

AlexR
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  • These are statements I already found myself, thanks anyway! I'll ask the teacher to make it clear. – Niccolò Campolungo Sep 23 '13 at 14:07
  • @LightStyle I guess you answered the question by yourself anyways ;-) multiple choice does not imply that at least one answer is correct ^^ – AlexR Sep 23 '13 at 14:12
  • Yep I only wanted to be totally sure before making a dumb question to the teacher :P however this specific question had to have an answer, since it was inside a test :P – Niccolò Campolungo Sep 23 '13 at 14:57
  • @LightStyle Proof by counter-example :P My Profs (if they have MC, it's not very usual here) explicitly say that any and all or none may also be a correct answer. – AlexR Sep 23 '13 at 15:11
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since a2 - b2 = 0 by factorization (a+b)(a-b) = 0

a + b = 0 and a - b = 0

Thus , among all options last one a + b = 0 is correct

Monarch
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    Not necessarily - EITHER a + b =0 OR a - b = 0. So it is not a necessity that a + b = 0 – Mufasa Sep 23 '13 at 12:45
  • Look at the counter example provided by the OP. The last one is not guaranteed to be true just because $a^2-b^2=0$ – Mufasa Sep 23 '13 at 12:50
  • $ab = 0 \Leftrightarrow a = 0 \vee b = 0$ note the big OR ($\vee$). This is wrong, the OP even gave a counter-example. – AlexR Sep 23 '13 at 14:09