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The following is from Elementary Differential Geometry by A.N. Pressley, page 102.

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Why is the author using the different choices of curves to prove $E_1=E_2$, $F_1=F_2$ and $G_1=G_2$, when it can be obtained by comparing the coefficients Eq(3)?

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2 Answers 2

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$A + 2B + C = D + 2E + F$ does not imply that $A = D$, $B =E$, and $C = F$!

Example:

$$1 + 2\cdot 2 + 1 = 6 = 2 + 2 \cdot 1 + 2$$

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  • $\begingroup$ Okay, my another question is that what is the meaning of doing this choice of curves? I feel like this is purposely done just to get the proof done. Why must this hold true? Why does the author make one parameter constant just to prove $E_1=E_2$? What if it cannot be done? $\endgroup$
    – danny
    Feb 14, 2020 at 15:25
  • $\begingroup$ I am not sure what the snippet of text you've shown is trying to accomplish (what is the overarching theorem/goal). And I am not sure how pedagogically the author arranged his/her definitions. So it is a bit difficult to address why that choice of curves work. // But with almost all constructive proofs in mathematics, you have to make a choice to "get the job done". In this case the author held one of the two values constant "because he could". In fact, from elementary linear algebra, any three generic curves passing through $u_0, v_0$ will do the job. The specific choice of holding ... $\endgroup$ Feb 14, 2020 at 15:46
  • $\begingroup$ ... $u_0$ and $v_0$ fixed merely simplifies the algebra. $\endgroup$ Feb 14, 2020 at 15:46
  • $\begingroup$ Yes but that ‘making the choice to get the job done’ part is causing me trouble. Makes me very uncomfortable. Is it not something like affirming the consequent? Making choice that would get the proof done? How can we be sure it is justified? $\endgroup$
    – danny
    Feb 14, 2020 at 15:51
  • $\begingroup$ Like I said: you need to show what the theorem statement is for me to know what you are objecting to then. $\endgroup$ Feb 14, 2020 at 16:14
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Because the first curve, with $v=v_0$ has $\dot v =0$, the second, with $u=u_0$ has $\dot u=0$ so are used to compare only the coefficients $E_i$ and $G_i$ . Than we need another curve to compare the other coefficients.

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  • $\begingroup$ Yes that’s true, but why in the first place such a curve as the first curve should exist? Even if so then we get $E_1=E_2$ similarly for $G$, then these results are used to prove $F_1=F_2$ but for that $E_1$ may not be equal to $E_2$ right? How can we use a value obtained from the first curve to find another value from the third curve? $\endgroup$
    – danny
    Feb 14, 2020 at 15:32
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    $\begingroup$ The two forms are equals iff they are the same for all the curves. $\endgroup$ Feb 14, 2020 at 15:35
  • $\begingroup$ But that’s what we have to prove right? Don’t know why, but this has been bothering me for two days. Unable to convince myself. $\endgroup$
    – danny
    Feb 14, 2020 at 15:44

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