Is Spivak's Proof Methodology Essential for Understanding Calculus?

  • Thread starter Thread starter driscol
  • Start date Start date
  • Tags Tags
    Spivak
driscol
Messages
12
Reaction score
0

Homework Statement


Spivak's "Calculus" Chapter 1, Problem 4v

Find all numbers x for which
x^2-2x+2>0


Homework Equations





The Attempt at a Solution



<br /> x^2-2x+2&gt;0<br />

<br /> x^2-2x&gt;-2<br />

<br /> x^2-2x+1&gt;-2+1<br />

<br /> (x-1)^2&gt;-1<br />

<br /> x\in R<br />

Would that be an adequate proof? Anything squared is always going to be positive...
Also, for these Spivak proofs, do I have to keep showing every single simple step?

<br /> \frac{a}{b}=\frac{ac}{bc}\ \ \ \ \ \mbox{b,c\neq0}<br />

<br /> \frac{a}{b}=(\frac{a}{b})(\frac{c}{c})<br />

<br /> \frac{a}{b}=(\frac{a}{b})(1)<br />

<br /> \frac{a}{b}=\frac{a}{b}<br />

I understand that doing this in the beginning will help me develop good habits/skills, but is it really necessary for every single problem? Could I just have canceled the c's in that equation without doing all the extra work?

Thanks.
 
Physics news on Phys.org
driscol said:

Homework Statement


Spivak's "Calculus" Chapter 1, Problem 4v

Find all numbers x for which
x^2-2x+2&gt;0


Homework Equations





The Attempt at a Solution



<br /> x^2-2x+2&gt;0<br />

<br /> x^2-2x&gt;-2<br />

<br /> x^2-2x+1&gt;-2+1<br />

<br /> (x-1)^2&gt;-1<br />

<br /> x\in R<br />

Would that be an adequate proof? Anything squared is always going to be positive...

Adequate except for a technicality. At the beginning of your proof, you assume what you are trying to prove, which renders the rest of the calculations pointless. They should be written in reverse order: Begin with the true statement that for all x in R, (x-1)^2 > -1, and work from there.

Also, for these Spivak proofs, do I have to keep showing every single simple step?

<br /> \frac{a}{b}=\frac{ac}{bc}\ \ \ \ \ \mbox{b,c\neq0}<br />

<br /> \frac{a}{b}=(\frac{a}{b})(\frac{c}{c})<br />

<br /> \frac{a}{b}=(\frac{a}{b})(1)<br />

<br /> \frac{a}{b}=\frac{a}{b}<br />

I understand that doing this in the beginning will help me develop good habits/skills, but is it really necessary for every single problem? Could I just have canceled the c's in that equation without doing all the extra work?

Thanks.

But then the problem would be pointless, right?
 
So it would it go:
x in R
x^2-2x+2x>0
.
.
.

or

x in R
(x-1)^2>-1
.
.
.

If the second one, then wouldn't I still have to go through the first one to get (x-1)^2>-1?
 
driscol said:
Would that be an adequate proof? Anything squared is always going to be positive...
Two things:
(1) Since you're doing these calculations with a lot of detail, you should prove that squares are nonnegative, rather than just assume.
(2) You omitted a description of what you're doing -- I can't be sure if you did it right, or if you are suffering from a common misunderstanding.

Typically, if you don't make any comments, it's presumed that for each line of your proof, you implicitly mean something like "this line is a consequence of what we've done up to this point". And with ths interpretation, what you have proven is the converse of what you wanted -- you proved the (utterly trivial statement):
If (the real number) x satisfies the inequation x^2 - 2x + 2 > 0, then x is a real number​
when instead what you wanted to prove was
If (the real number) x is a real number, then it satisfies the inequation x^2 - 2x + 2 > 0​

However, it turns out that each line in your proof is logically equivalent to the line before it (i.e. an "if and only if") -- and if you indicated that in your proof, then it would be correct. (Ignoring the possible hole of assuming that squares are nonnegative)



Could I just have canceled the c's in that equation without doing all the extra work?
If you have already proven that you can cancel, then sure. :-p


I understand that doing this in the beginning will help me develop good habits/skills, but is it really necessary for every single problem?
If you've been asked to, then yes. :-p

Ignoring issues of grade, one of the things you learn is how the basic algebraic properties of the real numbers combine to produce all of the useful arithmetic tricks you like to use. This is important for two reasons:

(1) In time, you will be doing arithmetic with other kinds of objects (e.g. vectors, matrices, operators, functions, etc) that do not share all of the basic algebraic properties of the real numbers -- and consequently, many of your favorite arithmetic tricks don't work at all. Understanding why they work for real numbers will help you comprehend why they don't work in these other situations.

(2) At some point, you may be faced with situations where you need to be able to derive interesting arithmetic facts yourself, rather than just being told. For example, maybe you think some useful thing should be true but you can't find it in your book... or maybe you are faced with disbelief of some fact you are told, and need to convince yourself it's really true. And if you're going to learn to prove things, it's much easier to start with proving "easy" things that you already believe should work out.

(3) This doesn't apply in this particular case, but as an extension of (1), working through the proofs will help you appreciate the cases in which things don't work. For example, proving the theorem
If ac = bc and c is nonzero, then a = b​
will help you understand why it's important for c to be nonzero, and will reduce your temptation to blindly cancel things in the future when you don't know if they are nonzero.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top