caveman caveman's cat

1. spivak's calculus chapter 1

For any numbers $a,b,c$:

Axiom 1. $a+(b+c) = (a+b)+c$.

Axiom 2. $a+0 = 0+a = a$.

Axiom 3. $a+(-a) = (-a)+a = 0$.

Axiom 4. $a+b=b+a$.

Axiom 5. $a(bc) = (ab)c$.

Axiom 6. $a 1 = 1 a = a$.

Axiom 7. $aa^{-1} = a^{-1}a = 1$, for all $a \ne 0$.

Axiom 8. $ab=ba$.

Axiom 9. $a(b+c) = ab+ac$.

1.1. problem 1

Prove these:

1.1.1. If $ax=a$ for some number $a\ne 0$, then $x=1$

Proof— by using Axiom 7: $$\begin{split} a^{-1}ax &= a^{-1}a\\ 1 x &= 1\\ \end{split}$$

By using Axiom 6 $$\begin{split} x &= 1\\ \end{split}$$

$\blacksquare$

1.1.2. $x^2 - y^2 = (x-y)(x+y)$

Proof— by using Axiom 9: $$\begin{split} (x-y)(x+y) &= x(x-y)+y(x-y) \\ &= (xx-xy)+(yx-yy) \\ &= x^2-xy+yx-y^2 \\ &= x^2-y^2 \\ \end{split}$$

$\blacksquare$

1.1.3. If $x^2 = y^2$, then $x=y$ or $x=-y$

Proof— by using the theorem we proved above: $$\begin{split} x^2 &= y^2 \\ x^2 - y^2 &= 0 \\ (x-y)(x+y) &= 0 \\ \end{split}$$

Then we use Axiom 7 and multiply both sides by $(x+y)^{-1}$ or $(x-y)^{-1}$. But since Axiom 7 is only defined for non-zero numbers, we need to state that those multiplications are for when $x+y\ne0$ or $x-y\ne0$, respectively. Hence we need to handle $x+y=0$ or $x-y=0$ separately as well.

Since the 2 cases above, $x+y\ne0$ and $x+y=0$, are exhaustive, then we can conclude that $x$ could only be either $y$ or $-y$.

$\blacksquare$

1.1.4. $x^3-y^3 = (x-y)(x^2+xy+y^2)$

Using Axiom 9 two times:

$$\begin{split} &(x-y)(x^2 + xy + y^2) \\ &= (x-y)x^2 + (x-y)xy + (x-y)y^2\\ &= (x-y)x^2 + (x-y)xy + (x-y)y^2\\ &= xx^2-yx^2 + xxy-yxy + xy^2-yy^2\\ &= x^3-yx^2 + x^2y-xy^2 + xy^2 - y^3\\ &= x^3 - y^3\\ \end{split}$$

$\blacksquare$

1.1.5. $x^n-y^n$ $=$ $(x-y)(x^{n-1}$ $+$ $x^{n-2}y$ $+$ $\ldots$ $+$ $xy^{n-2}$ $+$ $y^{n-1})$

I guess Spivak wants to say: $$x^n-y^n = (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)$$

1.1.5.1. proof by deduction

Using Axiom 9: $$\begin{split} & (x-y)\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(x^{n-n}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(x^{0}y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -y\left(y^{n-1} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i-1}\right)\\ &= x\left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x\left(x^{n-1}y^{1-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x\left(x^{n-1}y^{0} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x\left(x^{n-1} + \sum_{i=2}^{i=n} x^{n-i}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=2}^{i=n} x^{n-i+1}y^{i-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-(i+1)+1}y^{(i+1)-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i-1+1}y^{i+1-1}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= \left(x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right) -\left(y^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i} - y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\ \end{split}$$

Then using Axiom 1: $$\begin{split} & x^{n} + \sum_{i=1}^{i=n-1} x^{n-i}y^{i} - y^{n} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\\ &= x^{n} - y^{n} + \left(\sum_{i=1}^{i=n-1} x^{n-i}y^{i} - \sum_{i=1}^{i=n-1} x^{n-i}y^{i}\right)\\ &= x^{n} - y^{n} + \left(0\right)\\ &= x^{n} - y^{n} \end{split}$$

$\blacksquare$

1.1.5.2. proof by induction

Let: $$f(x,y,n) = \left(\sum_{i=1}^{i=n} x^{n-i}y^{i-1}\right)$$

Then we can say (is this sub-proof perfect?): $$\begin{split} x^{n+1}-y^{n+1} &= (x-y)\left(\sum_{i=1}^{i=n+1} x^{(n+1)-i}y^{i-1}\right)\\ &= (x-y)\left( \begin{split} &x^{(n+1)-(n+1)} y^{(n+1)-1}\\ &+ \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1} \end{split} \right)\\ &= (x-y)\left( x^{0} y^{n} + \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1} \right)\\ &= (x-y)\left( y^{n} + \sum_{i=1}^{i=n} x^{(n+1)-i}y^{i-1} \right)\\ &= (x-y)\left( y^{n} + x\sum_{i=1}^{i=n} x^{n-i}y^{i-1} \right)\\ &= (x-y)\left( y^{n} + xf(x,y,n) \right)\\ \end{split}$$

We can also rewrite what Spivak wants into: $$x^n-y^n = (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big)$$

We've already proven in Section 1.1.2: $$\begin{split} x^2-y^2 &= (x-y)(x+y)\\ &= (x-y)\left(\sum_{i=1}^{i=2} x^{2-i}y^{i-1}\right)\\ &= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\ \end{split}$$

Then, by induction, we prove that: $$\begin{split} x^2-y^2 &= (x-y)\Big(y^{2-1} + xf(x,y,2-1)\Big)\\ x^3-y^3 &= (x-y)\Big(y^{3-1} + xf(x,y,3-1)\Big)\\ \vdots\\ x^n-y^n &= (x-y)\Big(y^{n-1} + xf(x,y,n-1)\Big)\\ \end{split}$$

And since $(x-y)(y^{n-1} + xf(x,y,n-1))$ is only a rewrite of what Spivak wants, i.e. $(x-y)(x^{n-1}$ $+$ $x^{n-2}y$ $+$ $\ldots$ $+$ $xy^{n-2})$.

$\blacksquare$

1.1.6. $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$

1.1.6.1. boring proof

Using Axiom 9 and Axiom 8 $$\begin{split} &(x+y)(x^2 - xy + y^2)\\ &= (x+y)x^2 - (x+y)xy + (x+y)y^2\\ &= (x^3+x^2y) - (x^2y+xy^2) + (xy^2+y^3)\\ &= x^3 + x^2y - x^2y - xy^2 + xy^2 + y^3\\ &= x^3 + y^3\\ \end{split}$$

$\blacksquare$

1.1.6.2. fun proof

First, let's prove that $(-a) (-b) = ab$, by using Axiom 3, Axiom 5 and Axiom 9:

$$\begin{split} (-a) (-b) + (-ab) &=(-a) (-b) + (-a) b \\ &=(-a) \big((-b) + b\big) \\ &=-a \times 0 \\ &=0\\ (-a) (-b) + (-ab) + ab &= ab\\ (-a) (-b) &= ab\\ \end{split}$$

Now we know that $(-y)^2 = y^2$. and:

$$\begin{split} (-y)^3 &= (-y)^2 (-y)\\ &= y^2 (-y)\\ &= -y^3\\ \end{split}$$

Then by substituting $y$ in the proof in Section 1.1.4 by $(-y)$ we get:

$$\begin{split} x^3 - (-y)^3 &= (x-(-y)) (x^2 + x(-y) + (-y)^2)\\ x^3 + y^3 &= (x + y) (x^2 - xy + y^2)\\ \end{split}$$

$\blacksquare$

1.2. problem 2

What's wrong with this proof? Let $x=y$, then:

$$\begin{split} x^2 &= xy\\ x^2 - y^2 &= xy - y^2\\ (x+y)(x-y) &= y(x - y)\\ x+y &= y\\ 2y &= y\\ 2 &= 1\\ \end{split}$$

Answer:

The steps:

$$\begin{split} (x+y)(x-y) &= y(x - y)\\ x+y &= y\\ \end{split}$$

made use of multiplication of both sides by $(x-y)^{-1}$, which is undefined, since $x=y$ implies $(x-y)^{-1} = 0^{-1}$. Note Axiom 7 is not defined for when $a=0$.

1.3. problem 3

Prove the following.

1.3.1. $\frac{a}{b} = \frac{ac}{bc}$, if $b,c\ne 0$

IMO Spivak wants to say $ab^{-1} = acb^{-1}c^{-1}$.

Proof— Using Axiom 5:

$$\begin{split} acb^{-1}c^{-1} &= ab^{-1}(cc^{-1})\\ \end{split}$$

Using Axiom 7:

$$\begin{split} ab^{-1}(cc^{-1}) &= ab^{-1}(1)\\ \end{split}$$

Using Axiom 6:

$$\begin{split} ab^{-1}(1) &= ab^{-1}\\ \end{split}$$

$\blacksquare$

1.3.2. $\frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{bd}$, if $b,d\ne 0$

IMO Spivak wants to say:

$$ab^{-1} + cd^{-1} = (ad+bc)b^{-1}d^{-1}$$

Proof— Using Axiom 9:

$$\begin{split} (ad+bc)b^{-1}d^{-1} &= adb^{-1}d^{-1}+bcb^{-1}d^{-1}\\ \end{split}$$

Using Axiom 5:

$$\begin{split} adb^{-1}d^{-1}+bcb^{-1}d^{-1} &= ab^{-1}(dd^{-1}) + c(bb^{-1})d^{-1}\\ \end{split}$$

Using Axiom 7:

$$\begin{split} ab^{-1}(dd^{-1}) + c(bb^{-1})d^{-1} &= ab^{-1}(1) + c(1)d^{-1}\\ \end{split}$$

Using Axiom 6:

$$\begin{split} ab^{-1}(1) + c(1)d^{-1} &= ab^{-1} + cd^{-1}\\ \end{split}$$

$\blacksquare$

1.3.3. $(ab)^{-1} = a^{-1}b^{-1}$, if $a,b\ne 0$

Proof— If $a$ and $b$ are numbers, then their product, $(ab)$, is a number too.

By Axiom 7, we know that: $$(ab)(ab)^{-1}=1$$

We also know that the equality holds if we multiply both sides by the same numbers1:

$$(ab)(ab)^{-1} a^{-1} b^{-1}=(1) a^{-1} b^{-1}$$

By applying Axiom 5 we get2: $$(aa^{-1}) (bb^{-1}) (ab)^{-1} =(1) a^{-1} b^{-1}$$

By applying Axiom 7 we get:

$$(1) (1) (ab)^{-1} =(1) a^{-1} b^{-1}$$

By applying Axiom 6 we get:

$$(ab)^{-1} = a^{-1} b^{-1}$$

$\blacksquare$

1.3.4. $\frac{a}{b}\cdot \frac{c}{d} = \frac{ac}{db}$, if $b,d\ne 0$

IMO Spivak wants to say $(ab^{1})(cd^{-1}) = (ac)(db)^{-1}$.

Proof— By proof in Section 1.3.3, the right hand becomes:

$$(ac)(db)^{-1} = (ac)(d^{-1}b^{-1})$$

By Axiom 5:

$$(ac)(d^{-1}b^{-1}) = (ab^{-1})(cd^{-1}) $$

$\blacksquare$

1.3.5. $\frac{a}{b}/\frac{c}{d} = \frac{ad}{bc}$, if $b,c,d\ne 0$

IMO Spivak wants to say prove this:

$$(ab^{-1})(cd^{-1})^{-1} = adb^{-1}c^{-1}$$

1.3.5.1. intermediate proof to massage things a bit

Proof— First let's prove that $(x^{-1})^{-1} = x$.

By Axiom 7 we know that:

$$\begin{split} x^{-1} x &= 1\\ \end{split}$$

Since $x^{-1}$ is, itself, yet another number, then we can also apply Axiom 7 to get:

$$\begin{split} (x^{-1})^{-1} x^{-1} &= 1\\ \end{split}$$

Since both of the expressions equal $1$, we can put them in an equation:

$$\begin{split} (x^{-1})^{-1} x^{-1} &= x^{-1}x\\ \end{split}$$

By multiplying both sides by $x$:

$$\begin{split} (x^{-1})^{-1} x^{-1} x &= x^{-1}xx\\ \end{split}$$

By Axiom 5:

$$\begin{split} (x^{-1})^{-1} (x^{-1} x) &= (x^{-1}x)x\\ \end{split}$$

By Axiom 7:

$$\begin{split} (x^{-1})^{-1} (1) &= (1)x\\ \end{split}$$

By Axiom 6:

$$\begin{split} (x^{-1})^{-1} &= x\\ \end{split}$$

$\blacksquare$

1.3.5.2. actual proof

Proof— By proof in Section 1.3.3, we can spread the exponent of $(cd^{-1})^{-1}$ as follows:

$$\begin{split} (ab^{-1})(cd^{-1})^{-1} = (ab^{-1})c^{-1}(d^{-1})^{-1} \end{split}$$

By proof in Section 1.3.5.1:

$$\begin{split} (ab^{-1})c^{-1}(d^{-1})^{-1} = (ab^{-1})c^{-1}d \end{split}$$

By Axiom 5:

$$\begin{split} (ab^{-1})c^{-1}d &= (ad)(b^{-1}c^{-1}) \\ &= adb^{-1}c^{-1} \\ \end{split}$$

$\blacksquare$

1.3.6. if $b,d\ne 0$, then $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$. also determine when $\frac{a}{b} = \frac{b}{a}$

IMO Spivak wants me to prove $ab^{-1} = cd^{-1}$ if and only if $ad = bc$ (of course, $b,d\ne 0$), and then show all cases where $ab^{-1} = ba^{-1}$ is possible.

Proof— By multiplying both sides by $bd$:

$$ab^{-1}bd = cd^{-1}bd$$

By Axiom 5

$$a(b^{-1}b)d = c(d^{-1}d)b$$

By Axiom 7:

$$a(1)d = c(1)b$$

By Axiom 6:

$$ad = cb$$

$\blacksquare$

1.4. find all numbers $x$ for which

1.4.1. $4-x < 3-2x$


  • 1. I think we need axioms to define how equality works. Maybe Spivak will define equality axioms later on? Dunno.
  • 2. note we cannot pop $(ab)^{-1}$ and spread the $-1$ exponent because there is no axiom for it, so we cannot apply Axiom 5 on $(ab)^{-1}$. The reason there is no axiom for it is because of the fact that the exponent $-1$ has a special meaning used in Axiom 7, and —so far— is not just a normal number