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Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 11

October 5, 2014 etoix Leave a comment

Problem

Find all numbers $x$ for which

$|x-3|=8$
$|x -3| < 8$
$|x+4|< 2$
$|x-1|+|x-2| > 1$
$|x-1| + |x+1| < 2$
$|x-1|+|x+1| < 1$
$|x-1| \cdot |x+1| =0$
$|x-1| \cdot |x+2| = 3$

Solution

$|x-3|=8$

$x-3 =8 \to x = 11$

$3-x =8 \to x = -5$

$|x -3| < 8$

$x-3< 8 = x < 11$

$3-x< 8 = x > -5$

So $-5 < x < 11$ .

$|x+4|< 2$

$x+4 < 2 \to x < -2$

$-x-4< 2\to x > -6$

So $-6 < x < -2$ .

$|x-1|+|x-2| > 1$

$x-1 + x-2>1 \to 2x-3>1 \to 2x > 4 \to x > 2$ .

$1-x + 2-x>1 \to 3-2x>1 \to 2x < 2 \to x < 1$ .

$1-x+x-2>1 \to -1>1$ . This is nonsense.

$x-1+2-x > 1 \to 1 > 1$ . This is nonsense.

So $x < 1$ or $x > 2$ .

$|x-1| + |x+1| < 2$

$x-1 + x+1 < 2 \to 2x < 2 \to x < 1$

$1-x -x -1 < 2 \to -2x < 2 \to x > -1$

$1-x + x+1 < 2 \to 2 < 2$ . Nonsense.

$x-1 -x -1 < 2 \to -2 < 2$ . No use here.

So you might think that $-1 < x < 1$ . However, try letting $x = 0$ . You get $|0-1| + |0+1| < 2 \to |-1| + 1 < 2 \to 2 < 2$ . Doesn’t work.

Think about it this way, for any $x$ , the distance from $x-1$ to $x+1$ must be 2. So there is no number in $\mathbb{R}$ that satisfies this inequality.

$|x-1|+|x+1| < 1$

Following the logic from above. If the distance from $x-1$ to $x+1$ must be 2, it certainly cannot be less than 1.

$|x-1| \cdot |x+1| =0$

We need either $x-1 = 0$ or $x+1=0$ . The absolute value doesn’t even matter in this problem. So $x= 1$ or $x = -1$ .

$|x-1| \cdot |x+2| = 3$

3 is positive. For that to happens our two factors must be both positive or both negative. So $x > 1$ or $< -2$ . In this case $(x-1) (x+2) = 3$ . Expand to get $x^2 + x - 5 = 0$ . Using the quadratic formula $\frac{-1 \pm \sqrt{1-(4)(-5)}}{2} = \frac{-1 \pm \sqrt{21}}{2}$ . $\frac{-1 + \sqrt{21}}{2} > 1$ and $\frac{-1 - \sqrt{21}}{2} < -2$ so that fits what we’re looking for.

For $-2 < x < 1$ , we know that $(1-x)(x+2) = 3 \to -x^2 -x -1 = 0 \ to x^2 + x + 1 = 0$ . The solution here is $\frac{-1 \pm \sqrt{-3}}{2}$ , which is not $\mathbb{R}$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 10

October 4, 2014 etoix Leave a comment

Problem

Express each of the following without absolute value signs, treating various cases separately when necessary.

$|a+b|-|b|$ .
$|(|x|-1)|$ .
$|x| - |x^2|$ .
$a-|(a-|a|)|$ .

Solution

$|a+b|-|b|$

There are two possibilities for $|a+b|$ : $a+b$ and $-a-b$ .

There are two possibilities for $|b|$ : $b$ and $-b$ .

So there are four possibilities altogether.

$a+b-b = a$
$a+b+b = a+2b$
$-a-b+b = -a$
$-a-b-b = -a-2b$

$|(|x|-1)|$

This becomes $|x|-1$ and $1-|x|$ .

$|x|-1$ can be either $x-1$ or $-x-1$ .

$1-|x|$ can be either $1-x$ or $x+1$ .

Four possibilities.

$|x| - |x^2|$

$x^2$ will always be positive so this is really $|x| - x^2$ .

This gives two possibilities: $x-x^2$ and $-x-x^2$ .

$a-|(a-|a|)|$

There is just one variable here. There is a case where $a \geq 0$ and a case where $\leq 0$ .

$a \geq 0$ . In this case we just drop the bars and get $a - ( a -a ) = a - 0 = a$ .

$a \leq 0$ . In this case we negative all the bars and get $a - - (a - - a) = a - - 2a = 3a$ .

So $a \geq 0 \to a-|(a-|a|)| = a$ and $a \leq 0 \to a-|(a-|a|)| = 3a$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 9

October 3, 2014 etoix Leave a comment

Problem

Express each of the following with at least one less pair of absolute value signs.

$|\sqrt{2} + \sqrt{3} - \sqrt{5} + \sqrt{7}|$ .
$|(|a+b|-|a|-|b|)|$ .
$|(|a+b|+|c|-|a+b+c|)|$ .
$|x^2-2xy+y^2|$ .
$|(|\sqrt{2}+\sqrt{3}|-|\sqrt{5}-\sqrt{7}|)|$ .

Solution

$|\sqrt{2} + \sqrt{3} - \sqrt{5} + \sqrt{7}|$

$\sqrt{7} > \sqrt{5}$ so the whole expression will be positive without absolute value, so we can simply drop the bars and write $\sqrt{2} + \sqrt{3} - \sqrt{5} + \sqrt{7}$ .

$|(|a+b|-|a|-|b|)|$

We can rewrite the above as $|(|a+b|- ( |a|+|b|))|$ .

In the text we learned that $|a+b| \leq |a|+|b|$ .

So we know that the difference between $|a+b|$ and $|a|+|b|$ will be negative. So we can swap them to make the difference positive and drop the bars.

$|a| + |b| - |a+b|$

$|(|a+b|+|c|-|a+b+c|)|$

The this similar to the previous part. Let $d = a + b$ . Now you’re dealing with $|(|d|+|c|-|d+c|)|$ . Since $|d+c| \leq |d| + |c|$ , we know that that subtraction will always be positive, as is. So we can simply drop the bars.

$|a+b|+|c|-|a+b+c|$

$|x^2-2xy+y^2|$

This is simply $|(x-y)^2|$ . Since square is always positive, we can simply drop the bars.

$x^2 - 2xy + y^2$

$|(|\sqrt{2}+\sqrt{3}|-|\sqrt{5}-\sqrt{7}|)|$

It should be clear that $\sqrt{2} + sqrt{3} > 0$ and $\sqrt{5} - \sqrt{7} < 0$ . So the subtraction above will be positive, as is. We can simply drop the bars.

$|\sqrt{2}+\sqrt{3}|-|\sqrt{5}-\sqrt{7}|$

We can also drop the bars around $\sqrt{2} + \sqrt{3}$ since that’s always positive.

$\sqrt{2}+\sqrt{3} -|\sqrt{5}-\sqrt{7}|$

For $\sqrt{5} - \sqrt{7}$ , we can simply swap and still get the same positive number.

$\sqrt{2}+\sqrt{3} - (\sqrt{7}-\sqrt{5})$

$\sqrt{2}+\sqrt{3} + \sqrt{5} - \sqrt{7}$

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 8

October 2, 2014 etoix Leave a comment

Problem

Although the basic properties of inequalities were stated in terms of the collection P of all positive numbers, and < was defined in terms of P, this procedure can be reversed. Suppose that P10-P12 are replaced by

(P’10) For any numbers $a$ and $b$ one, and only one, of the following holds:

$a = b$ ,
$a < b$ ,
$a > b$ .

(P’11) For any numbers a, b, and c, if $a < b$ and $b < c$ , then $a < c$ .

(P’12) For any numbers a, b, and c, if $a < b$ , then $a+c < b+c$ .

(P’13) For any numbers a, b, and c, if $a < b$ and $0 < c$ , then $ac < bc$ .

Show that P10-P12 can then be deduced as theorems.

Solution

(P10) (Trichotomy law) For every number a, one and only one of the following holds:

$a = 0$ ,
$a$ is in the collection P,
$-a$ is in the collection P.

Let a be some arbitrary number. Let P be the set of all positive numbers. By P’10, assuming $b = 0$ , one of the following is true about a:

$a = 0$
$a < 0$
$0 < a$

In the first case, $a = 0$ .

In the second case, $a < 0$ , so $0 < 0 - a = -a$ , so $-a \in P$ .

The solution book uses a proof by contradiction to prove $-a \in P$ . I think mine is just as valid.

In the third case, it’s clear that $a \in P$ .

By P’10, these are the only three possible cases, therefore, for all a, $a = 0$ , $a \in P$ , or $-a \in P$ .

(P11) (Closure under addition) If a and b are in P, then $a+b$ is in P.

Let a and b be arbitrary numbers in P. It follows that $0 < a$ . By P’12, we know that $0 + b = b < a + b$ (b in P’12 is 0. c in P’12 is b here). Then, by P’11, we know that since $0 < b < a + b$ , it must be that $0 < a + b$ , which means $a+b \in P$ .

Therefore, $a, b \in P \to a+b \in P$ .

(P12) (Closure under multiplication) If a and b are in P, then $a \cdot b$ is in P.

Let a and b be arbitrary numbers in P. It follows that $0 < a$ . By P’13, we know that $0 \cdot b = 0 < ab$ (b in P’13 is 0. c in P’13 is b here). So $0 < ab$ , which means $ab \in P$ .

Therefore, $a, b \in P \to ab \in P$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 7

October 1, 2014 etoix Leave a comment

Problem

Prove that if $0 < a < b$ , then $a < \sqrt{ab} < \frac{a+b}{2} < b$ .

Notice that the inequality $\sqrt{ab} \leq \frac{a+b}{2}$ holds for all $a, b \geq 0$ . A generalization of this fact occurs in Problem 2-22.

Solution

Since $0 < a < b$ , we know that $a^2 < ab$ and consequently $a < \sqrt{ab}$ .

Similarly $a+b < 2b$ and consequently $\frac{a+b}{2} < b$ .

Since $a, b > 0$ , we know that $(a - b)^2 > 0$ . It follows that $a^2 - 2ab + b^2 > 0$ and $a^2 + 2ab + b^2 > 4ab$ . Further, $(a+b)^2 > 4ab$ , so $\frac{(a+b)^2}{4} > ab$ and $\frac{a+b}{2} > \sqrt{ab}$ .

By the transitive property $a < \sqrt{ab} < \frac{a+b}{2} < b$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 6

September 30, 2014 etoix Leave a comment

Problem

Prove that if $0 \leq x < y$ , then $x^n < y^n, n = 1,2,3, ...$
Prove that if $x < y$ and n is odd, then $x^n < y^n$ .
Prove that if $x^n = y^n$ and n is odd, then $x = y$ .
Prove that if $x^n = y^n$ and n is even, then $x = y$ or $x = -y$ .

Solution

Prove that if $0 \leq x < y$ , then $x^n < y^n, n = 1,2,3, ...$

In Problem 5 we showed that if $0 \leq a < b$ and $0 \leq c < d$ , then $ac < bd$ . Let $a = c = x$ and $b = d = x$ , which implies $x^2 < y^2$ .

Now consider if $a = x^2$ , $c = x$ , $b = y^2$ , and $d = y$ . By the same logic, we see that $x^3 < y^3$ . Now, we can repeat the process again by setting $a = x^3$ and $b = y^3$ . In fact, we can do this again and again ad infinitum.

Therefore, if if $0 \leq x < y$ and $n = 1,2,3, ...$ , then , then $x^n < y^n$ .

Prove that if $x < y$ and n is odd, then $x^n < y^n$ .

Consider the case where $x \geq 0$ . This is the same as the previous part.

Consider the case where $x < y \leq 0$ . From the text we learn that when $a < b$ , $b -a$ is positive. If we apply this to $y \leq 0$ , we see that $0 - y = -y > 0$ . We also see that $-x > 0$ and $-x < -y$ . Using what we proved in the previous part, we know that $(-x)^n < (-y)^n$ .

By closure under multiplication $(-a) (-b) > 0$ . Let $a = b = x$ , so $(-x) (-x) > 0$ . In fact, any pair, $(-x) (-x)$ , $(-x) (-x) (-x) (-x)$ , $(-x) (-x) (-x) (-x) (-x) (-x)$ , …, will always be positive. So we know that $x^{2n} > 0$ . Similarly, we know that $y^{2n} > 0$ . By extension, $x^{2n} \cdot x = x^{2n+1} < 0$ . So for any odd exponent $x^n < 0$ . Where n is odd, we do not change signs. So we know that $(-x)^n < (-y)^n$ where n is odd.

Finally, we have the case where $x < 0 \leq y$ . We have shown earlier that where n is odd, $x^n < 0$ . On the other hand, $y^n \geq 0$ . So again $x^n < y^n$ .

Prove that if $x^n = y^n$ and n is odd, then $x = y$ .

Assume $x \neq y$ . So either $x < y$ or $x > y$ .

If $x < y$ , by part 2 above, we know that it must be that $x^n < y^n$ , since n is odd. This contradicts our given that $x^n = y^n$ .

If $y < x$ , by part 2 above, we know that it must be that $y^n < x^n$ , since n is odd. This contradicts our given that $x^n = y^n$ .

Since our assumption leads to contradictions, it must be that when $x^n = y^n$ and n is odd that $x = y$ .

Prove that if $x^n = y^n$ and n is even, then $x = y$ or $x = -y$ .

Consider the case where $x, y \geq 0$ . Assume $x \neq y$ . Applying the proof in part 1, if $x > y$ then $x^n > y^n$ . Similarly, if $y > x$ then $y^n > x^n$ . In either case, we contradict our given that $x^n = y^n$ . Therefore, it must be that $x = y$ .

Consider the case where $x, y \leq 0$ . Assume $x \neq y$ . In this case $(-x), (-y) \geq 0$ . Applying the proof in part 1, if $(-x) > (-y)$ then $(-x)^n > (-y)^n$ . Similarly, if $(-y) > (-x)$ then $(-y)^n > (-x)^n$ . In either case, we contradict our given that $x^n = y^n$ . Therefore, it must be that $-x = -y$ .

Finally, consider the case where $x \leq 0 \leq y$ . Assume $x \neq (-y)$ . In this case $x, (-y) \geq 0$ . Applying the proof in part 1, if $x > (-y)$ then $x^n > (-y)^n$ . Similarly, if $(-y) > x$ then $(-y)^n > x^n$ . In either case, we contradict our given that $x^n = y^n$ . Therefore, it must be that $x = -y$ .

Therefore, if $x^n = y^n$ and n is even, then $x = y$ or $x = -y$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 5

September 29, 2014 etoix 2 Comments

Problem

Prove the following:

If $a < b$ and $c < d$ , then $a+c < b+d$ .
If $a < b$ , then $-b < -a$ .
If $a < b$ and $c > d$ , then $a - c < b - d$ .
If $a < b$ and $c > 0$ , then $ac < bc$ .
If $a < b$ and $c < 0$ , then $ac > bc$ .
If $a > 1$ , then $a^2 > a$ .
If $0 < a < 1$ , then $a^2 < a$ .
If $0 \leq a < b$ and $0 \leq c < d$ , then ac < bd.
If $0 \leq a < b$ , then $a^2 < b^2$ . (Use 8.)
If $a, b \geq 0$ and $a^2 < b^2$ , then $a < b$ . (Use 9 backwards.)

Solution

If $a < b$ and $c < d$ , then $a+c < b+d$ .

$a < b \to b - a > 0$ . $c < d \to d - c > 0$ . Since $b -a$ and $d - c$ are both positive, $b+d - (c+a) > 0$ , by P11 in the text, closure under addition. Therefore $a + c < b + d$ .

If $a < b$ , then $-b < -a$ .

Since $a < b$ , $b - a > 0$ , Since $b-a$ is positive, we know that $-(b-a)$ , which is $a - b$ , is negative. Since $a - b < 0$ , we conclude $-b < -a$ .

If $a < b$ and $c > d$ , then $a - c < b - d$ .

Here $b - a > 0$ and $c - d > 0$ . By closure under addition for inequalities, $-a + c + b - d > 0$ . Therefore, $b - d > - (- a + c)$ and $a - c < b - d$ .

If $a < b$ and $c > 0$ , then $ac < bc$ .

We know that $b - a > 0$ . By closure under multiplication for inequalities, since $c > 0$ , $c (b-a) > 0$ . We distribute and see $bc - ac > 0$ , so $ac < bc$ .

If $a < b$ and $c < 0$ , then $ac > bc$ .

We know that $b - a > 0$ . Since $c < 0$ , we know that $-c > 0$ . By closure under multiplication for inequalities $-c ( b - a ) > 0$ . We distribute and see that $ac - bc > 0$ , so $ac > bc$ .

If $a > 1$ , then $a^2 > a$ .

Start with $a > 1$ . Multiply both sides by a to get $a \cdot a = a^2 > a$ . Since $a > 1$ , then $a > 0$ . Part 4 tells us that if $c > b$ and $d > 0$ , then $cd > bd$ . Let $b = c = a$ . Let $b = 1$ . Substitute and we see that $a^2 > a$ .

If $0 < a < 1$ , then $a^2 < a$ .

Similar to the previous problem, we know that $a^2$ will be positive. Since $0 < a < 1$ , $a \cdot a$ will produce a number smaller than a.

If $0 \leq a < b$ and $0 \leq c < d$ , then $ac < bd$ .

Consider the case where $a, c \neq 0$ . Here $c > 0$ . By closure under multiplication for inequalities $ac < bc$ (multiply $a < b$ by c on both sides). Similarly, $bc < bd$ (multiply $c < d$ by b on both sides). So we have $ac < bc$ and $bc < bd$ . By the transitive property $ac < bd$ .

Consider the case where either $a = 0$ or $c = 0$ or both. In this case, $ac = 0$ . Since $0 \leq a < b$ and $0 \leq c < d$ , $b > 0$ and $d > 0$ . By closure under multiplication of inequalities $bd > 0 = ac$ . So $ac < bd$ .

If $0 \leq a < b$ , then $a^2 < b^2$ . (Use 8.)

Part 8 says if $0 \leq a < b$ and $0 \leq c < d$ , then $ac < bd$ . Let $c = a$ and $d = b$ . Then $a^2 < b^2$ .

If $a, b \geq 0$ and $a^2 < b^2$ , then $a < b$ . (Use 9 backwards.)

Assume $a < b$ were false, so $a = b$ or $a > b$ .

If $a = b$ , then $a^2 = b^2$ , which contradicts our given that $a^2 < b^2$ .

If $a > b$ and $b \geq 0$ , then by part 9, we know that $a^2 > b^2$ , which contradicts our given that $a^2 < b^2$ .

Since our assumption that $a < b$ is false can only lead to contradictions, we can assert that $a< b$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 4

September 28, 2014 etoix Leave a comment

Problem

Find all numbers $x$ for which

$4 -x < 3-2x$
$5-x^2<8$
$5-x^2<-2$
$(x-1)(x-3) >0$ (When is a product of two numbers positive?)
$x^2-2x+2>0$
$x^2+x+1>2$
$x^2-x+10 >16$
$x^2+x+1>0$
$(x- \pi) (x+5)(x-3) >0$
$(x- \sqrt[3]{2} ) (x- \sqrt{2} ) > 0$
$\sqrt{x}{2} < 8$
$x + 3^x < 4$
$\frac{1}{x} + \frac{1}{1-x} > 0$
$\frac{x-1}{x+1} > 0$

Solution

$4 -x < 3-3x$

Collect the terms.

$x < -1$

$5-x^2<8$

Move terms around and combine.

$-3 < x^2$

$x^2$ will always be positive so any $\mathbb{R}$ will work.

$5-x^2<-2$

Move terms around and combine.

$7 < x^2$

If this were $x^2 = 7$ , we see that $x = \pm \sqrt{7}$ .

We looking for when $x^2$ is greater than 7, so the solution is $x < -\sqrt{7}$ or $x > \sqrt{7}$ .

$(x-1)(x-3) >0$

Our critical points are $x= 1$ and $x = 3$ .

If $x = 0$ , then we would have $- * - = + > 0$ .

If $x = 2$ , then we would have $+ * - = - < 0$ .

If $x = 4$ , then we would have $+ * + = + > 0$ .

So the inequality works when $x < 1$ or $x > 3$ .

$x^2-2x+2>0$

If this were $x^2 - 2x + 2 = 0$ , the solution would be $\frac{2 \pm \sqrt{-4}}{2}$ . The solution is imaginary, so the parabola is entirely above or below the x-axis.

Since the first term is positive, we know this is a parabola facing up, so it’s entirely above the x-axis.

Therefore any $\mathbb{R}$ will work.

$x^2+x+1>2$

$x^2 + x -1 >0$

If this were an equality and we solve for x, we’d get $x = \frac{-1 \pm \sqrt{5}}{2}$ . This is a real solution so the parabola does cross the x-axis and there are two solutions. The parabola faces up so we want the two extreme branches. Therefore $x < \frac{-1-\sqrt{5}}{2}$ or $x > \frac{-1+\sqrt{5}}{2}$ .

$x^2-x+10 >16$

$x^2 -x-6 > 0$

$(x-3)(x+2) > 0$

The critical points are $x = 3$ and $x = -2$ .

If $x = -3$ , we would have $- * - = + > 0$ .

If $x = 0$ , we would have $- * + = - < 0$ .

If $x = 4$ , we would have $+ * + = + > 0$ .

Therefore, we want $x < -2$ or $x > 3$ .

$x^2+x+1>0$

If this were $x^2 -x + 1 = 0$ , the solution would be $\frac{-2 \pm \sqrt{-3}}{2}$ . The solution is imaginary, so the parabola is entirely above or below the x-axis.

Since the first term is positive, we know this is a parabola facing up, so it’s entirely above the x-axis.

Therefore any $\mathbb{R}$ will work.

$(x- \pi) (x+5)(x-3) >0$

The critical points are $x = \pi$ , $x = -5$ , and $x = 3$ .

If $x = -6$ , we would have $- * - * - = - < 0$ .

If $x = 0$ , we would have $- * + * - = + > 0$ .

If $x = 3.1$ , we would have $- * + * + = - < 0$ .

If $x = 4$ , we would have $+ * + * + = + > 0$ .

Therefore, the solution is $-5 < x < 3$ or $x > \pi$ .

$(x- \sqrt[3]{2} ) (x- \sqrt{2} ) > 0$

The critical points are $x = \sqrt[3]{2}$ and $x = \sqrt{2}$ .

If $x = 0$ , we would have $- * - = + > 0$ .

If $x = 1.3$ , we would have $+ * - = - < 0$ .

If $x = 2$ , we would have $+ * + = + > 0$ .

The solution is $< \sqrt[3]{2}$ or $x > \sqrt{2}$ .

$\sqrt{x}{2} < 8$

2 is positive. $2^3 = 8$ . Any exponent greater than 3 should be greater than 8. So $x > 3$ .

$x + 3^x < 4$

By visual inspection, it’s clear that if $x = 1$ , $1 + 3^1 = 4$.

So any x less than 1 should be less than 4.

$x<3$

$\frac{1}{x} + \frac{1}{1-x} > 0$

The critical points are $x = 0$ and $x = 1$ .

If x is negative, then $\frac{1}{x} < 0$ and $\frac{1}{1-x} > 0$ , but the absolute value of $\frac{1}{x}$ will ways be bigger.

x	$\frac{1}{x}$	$\frac{1}{1-x}$
-1	-1	$\frac{1}{2}$
-2	$-\frac{1}{2}$	$\frac{1}{3}$
-3	$-\frac{1}{3}$	$\frac{1}{4}$
-4	$-\frac{1}{4}$	$\frac{1}{5}$

Therefore, when $x < 0$ , $\frac{1}{x} + \frac{1}{1-x} < 0$ .

If $x > 1$ , then $\frac{1}{x} > 0$ and $\frac{1}{1-x} < 0$ , but the absolute value of $\frac{1}{1-x}$ will ways be bigger.

x	$\frac{1}{x}$	$\frac{1}{1-x}$
2	$\frac{1}{2}$	-1
3	$-\frac{1}{3}$	$-\frac{1}{2}$
4	$-\frac{1}{4}$	$-\frac{1}{3}$
5	$-\frac{1}{5}$	$-\frac{1}{4}$

Therefore, when $x > 1$ , $\frac{1}{x} + \frac{1}{1-x} < 0$ .

So let’s consider $0 < x < 1$ . Consider $x = \frac{1}{2}$ .

$\frac{1}{\frac{1}{2}} + \frac{1}{(1 - \frac{1}{2})}$

Notice that for $0 < x < 1$ both terms must be positive. Therefore, the only solution is $0 < x < 1$ .

Here I have a disagreement which the answer key. It claims that the solution is $x > 1$ or $0 < x < 1$ . However, I don’t see how $x> 1$ could be a solution. Below is a plot of $\frac{1}{x} + \frac{1}{1-x}$ . It is clearly negative where $x > 1$ .

$\frac{x-1}{x+1} > 0$

The critical points here are $x = 1$ and $x = -1$ .

If $x = -2$ , we would have $\frac{-}{-} = + > 0$ .

If $x = 0$ , we would have $\frac{-}{+} = - < 0$ .

If $x = 2$ , we would have $\frac{+}{+} = + > 0$ .

So the solution is $x < -1$ or $x > 1$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 3

September 27, 2014 etoix Leave a comment

Problem

Prove the following:

$\frac{a}{b} = \frac{ac}{bc}$ , if $b, c \neq 0$ .
$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$ , if $b, d \neq 0$ .
$(ab)^{-1} = a^{-1} b^{-1}$ , if $a, b \neq 0$ . (To do this you must remember the defining property of $(ab)^{-1}$ .)
$\frac{a}{b} \cdot \frac{c}{d} = \frac{ac}{db}$ , if $b, d \neq 0$ .
$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$ , if $b, c, d \neq 0$ .
If $b, d \neq 0$ , then $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$ . Also determine when $\frac{a}{b} = \frac{b}{a}$ .

Solution

$\frac{a}{b} = \frac{ac}{bc}$ , if $b, c \neq 0$

We’ll work on the right side.

$\frac{a}{b} = \frac{ac}{bc}$

Rewrite.

$\frac{a}{b} = a c b^{-1} c^{-1}$

Commutative property.

$\frac{a}{b} = a c c^{-1} b^{-1}$

Multiplicative inverse.

$\frac{a}{b} = a b^{-1}$

Rewrite.

$\frac{a}{b} = \frac{a}{b}$

$\frac{a}{b} + \frac{c}{d} = \frac{ad + bc}{bd}$

Rewrite.

$\frac{a}{b} + \frac{c}{d} = (ad + bc) (bd)^{-1}$

Distributive.

$\frac{a}{b} + \frac{c}{d} = ad (bd)^{-1} + bc (bd)^{-1}$

In the next part we’ll prove that $(bd)^{-1} = b^{-1} d^{-1}$ . For now, let’s assume.

$\frac{a}{b} + \frac{c}{d} = ad b^{-1} d^{-1} + bc b^{-1} d^{-1}$

Commutative.

$\frac{a}{b} + \frac{c}{d} = ad d^{-1} b^{-1} + b b^{-1} c d^{-1}$

Multiplicative inverse.

$\frac{a}{b} + \frac{c}{d} = a b^{-1} + c d^{-1}$

Rewrite.

$\frac{a}{b} + \frac{c}{d} = \frac{a}{b} + \frac{c}{d}$

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

Multiply by ab.

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

Associative.

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

Commutative.

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

Multiplicative inverse three times.

$1 = 1$

We have an identity, therefore, $(ab)^{-1} = a^{-1} b^{-1}$ , if $a, b \neq 0$ .

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

Rewrite.

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

Apply what we approved in part 3.

$a b^{-1} c d^{-1} = ac d^{-1} b^{-1}$

Commutative property twice.

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

We have an identity.

$\frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc}$ , if $b, c, d \neq 0$

Rewrite.

$\frac{a}{b} (\frac{c}{d})^{-1} = \frac{ad}{bc}$

Distribute the exponent.

$\frac{a}{b} c^{-1} (\frac{1}{d})^{-1} = \frac{ad}{bc}$

Rewrite.

$a b^{-1} c^{-1} (d^{-1})^{-1} = \frac{ad}{bc}$

The -1 exponent is simply a flip of the fraction. $(d^{-1})^{-1}$ can be rewritten as $(\frac{1}{d})^{-1}$ , which could be rewritten again as simply d. So $(d^{-1})^{-1} = d$ .

Substitute.

$a b^{-1} c^{-1} d = \frac{ad}{bc}$

Rewrite.

$\frac{ad}{bc} = \frac{ad}{bc}$

We have an identity.

$\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$

Assume $\frac{a}{b} = \frac{c}{d}$ . This is the same as $a b^{-1} = c d^{-1}$ . If we multiply both sides by bd we get $a b^{-1} b d = c d^{-1} b d$ which is $ad = bc$ .

Conversely, assume $ad = bc$ . Multiply both sides by $b^{-1} d^{-1}$ and we get $ad b^{-1} d^{-1} = bc b^{-1} d^{-1}$ . Simplify and we get $\frac{a}{b} = \frac{c}{d}$ .

Therefore, $\frac{a}{b} = \frac{c}{d}$ if and only if $ad = bc$ .

Also determine when $\frac{a}{b} = \frac{b}{a}$ .

Assume $\frac{a}{b} = \frac{b}{a}$ , then $a^2 = b^2$ , so $a = \pm b$ .

Conversely, let’s start with $a = b$ . $ab^{-1} = 1$ . Likewise, $b a^{-1} = 1$ . So $\frac{a}{b} = \frac{b}{a} = 1$ .

For the case where $a = -b$ , $ab^{-1} = -1$ . Likewise, $b a^{-1} = -1$ . So $\frac{a}{b} = \frac{b}{a} = -1$ .

Therefore, $\frac{a}{b} = \frac{b}{a}$ if and only if $a = b$ .

Calculus by Spivak

Calculus by Spivak, Chapter 1, Problem 2

September 26, 2014 etoix Leave a comment

Problem

What is wrong with the following “proof”? Let $x = y$ . Then

$x^2 = xy$

$x^2 - y^2 = xy - y^2$

$(x+y)(x-y) = y(x-y)$

$x+y = y$

$2y =y$

$2 = 1$

Solution

Take a closer a look at the transition from line 3 to line 4.

$(x+y)(x-y) = y(x-y) \to x+y = y$

Here we’re dividing both sides by $x-y$ .

Since $x = y$ , then $x -y = 0$ . So in this step in the “proof”, we’re actually dividing by 0, which is undefined and leads to illogical result, $1 = 2$ .

e^(ix)

Category Archives: Calculus by Spivak

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