Have the following system of equations: 2 + x 2 − y 2 = 0 x 2 − y 2 − 2 = 0And if I substitute y by a function of x and vice versa I get: 2 + x 2 − x 2 + 2 = 0 y 2 − y 2 − 4 = 0I therefore get: 4 = 0 , − 4 = 0 Therefore I don't have any solutions for that system of equationsIn theory, am I allowed to get to this conclusion?

Question

Have the following system of equations:  2  +      x    2    −      y    2    =  0      x    2    −      y    2    −  2  =  0And if I substitute y by a function of x and vice versa I get:  2  +      x    2    −      x    2    +  2  =  0      y    2    −      y    2    −  4  =  0I therefore get:  4  =  0  ,  −  4  =  0 Therefore I don&#039;t have any solutions for that system of equationsIn theory, am I allowed to get to this conclusion?

lodosr · Accepted Answer

You have       y    2    =      x    2    +  2 and       y    2    =      x    2    −  2 as your initial equations. Think about that for a secod.

watch5826c · Accepted Answer

1. Let&#039;s suppose one and only one condition: that there exists at least one   (  x  ,  y  ) in       R    2   solution of the system of equations. Let   (  x  ,  y  ) be that solution.2. In consequence, the two equations are verified by   (  x  ,  y  ).3. Therefore,   [  .  .  .  ]  4  =  0. Since   x and   y are both in the   (  R  ,  +  ,  ∗  ) ring, at least one of our supposed conditions is false.4. Conclude.This is a typical Reductio ad absurdum. As long as you don&#039;t violate the Zermelo–Fraenkel set theory and the properties of the objects you&#039;re using, there&#039;s no reason for your reasoning to be incorrect.

Have the following system of equations: 2 + x 2 </msup> −

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