0

In a proof like this: enter image description here

It just goes from a complicated looking expression into just $x$, is that allowed?
What are the rules for this?

Mauro ALLEGRANZA
  • 99,247
enoopreuse22
  • 187
  • 2
    There aren't rules for this sort of thing, unless you are in some context (like a classroom) where someone is imposing them. – lulu May 15 '23 at 13:59
  • "is that allowed?" There are no binding laws, it is rather the question, what is appropriate. I would like to see a step in between. Write $a=\sqrt{x^2-4}$ so that $a^2+4=x^2$. Then the fraction is $$ \frac{(x+a)^2+4}{2(x+a)}=\frac{x^2+2ax+(a^2+4)}{2(x+a)}=\frac{2(x+a)x}{2(x+a)}=x$$ – Dietrich Burde May 15 '23 at 14:04
  • 1
    Just a comment. What you have written is a proof that $x=\dots=x$. That's not a surprise. But of course it isn't really a proof since the first equality is not obvious. – ancient mathematician May 15 '23 at 14:08
  • To my knowledge, this question has no real answer. A proof is, more or less, just a completely convincing argument. Some proofs in combinatorics, for example, are just a paragraph containing no symbolic manipulations at all. – RTF May 15 '23 at 14:39
  • The proof looks fine, except that like in your previous question, you are begging the question by assuming the conclusion in the step "$x=$[complicated expression]". Deleting the fragment x = makes your proof valid. It is never valid to prove a result by assuming first assuming it. – ryang May 16 '23 at 14:58
  • @ryang Please correct me if im wrong but, the conclusion is "there exists number such that y+1/y=x", not "y+1/y=x". If i assumed the conclusion then i would be assuming that there exists y? Is it not okay to assume the equation is true? In this example I think the equation is being assumed to be true: https://math.stackexchange.com/questions/3605548/prove-that-y1-x-1y-x-iff-either-x-1-or-y-1 – enoopreuse22 May 18 '23 at 17:42
  • @enoopreuse22 You are correct, but I meant the conclusion of that section, not of the entire proof. Can you see that x = near the top is not merely redundant, but invalidates your proof? There, you aren't yet supposed to know that that equality is true; you only subsequently impicitly derive that equality using the transitive property of equality! $\quad$ If you (slowly) reread my linked answer to your previous question, hopefully you will become clearer about the moral of the story: no proof of an assertion begins by assuming that assertion (here the assertion is $1+\frac1y =x$). – ryang May 18 '23 at 18:06

0 Answers0