# Welcome to Math Video Lessons. Stanley Ocken. Department of Mathematics The City College of New York Fall 2013

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1 Welcome to Math Video Lessons Prof. Department of Mathematics The City College of New York Fall 2013 An important feature of the following Beamer slide presentations is that you, the reader, move step-by-step and at your own pace through these notes. To do so, use the arrow keys or the mouse to move from slide to slide, forwards or backwards. Also use the index dots at the top of this slide (or the index at the left, accessible from the Adobe Acrobat Toolbar) to access the different sections of this document. To prepare for the Chapter 1.2 Quiz (September 16th at the start of class), please Read all the following material carefully, especially the included Examples. Memorize and understand all included Definitions and Procedures. Work out the Exercises section, which explains how to check your answers. Do the Quiz Review and check your answers by referring back to the Examples.

2 Powers of numbers and letters. Powers help us write and understand large numbers. Let s begin with powers of means the product of two 10 s. It s easy to understand 100 = 10 2 = It s much harder to think about the number 100, 000, 000, 000, 000. But if we rewrite it as 10 14, the product of 14 tens, it s easier to handle. You know that 100 = 10 2, 1000 = 10 3, and = Similarly, is the decimal number consisting of 1 followed by 14 zeros. To formulate basic rules for powers, copy the behavior of powers of 10. Product of powers rule: 10 m 10 n = 10 m+n. Example: = because = = Power to a power rule: (10 m ) n = 10 mn. Example: (10 2 ) 3 = (100) 3 = = because (10 2 ) 3 = = Power of a product rule: (8 9) m = 8 m 9 m Example: (8 9) 2 = (8 9) (8 9) = (8 8) (9 9) =

3 We like powers of ten because they are the basis for writing decimal numbers. To talk about other powers, we need to define the parts of a power expression is the 6th power of 2. In the symbol 10 6, 10 is the base and 6 is the exponent. However, some people refer to 6 as the power. Power expressions don t need to involve numbers. For example x m is the m th power of x. If m = 4, then x 4 = xxxx, the product of 4 x s. We say: x 4 is the 4 th power of x. We rewrite the three rules from the previous slide as Basic power identities Product of powers identity: x m x n = x m+n. To rewrite the product of powers of the same base, add exponents. Power of power identity: (x m ) n = x mn. To rewrite a power of a power, multiply exponents Power of product identity: (xy) m = x m y m. A power of a product equals the product of powers.

4 Since these are identities, we can substitute any expressions for x, m and n. For the time being, however, we assume that m and n are whole numbers. For example: substitute 2x + 3 for x, 2x + 2 for y, 7 for m, and 5 for n: Product of powers of same base: (2x + 3) 7 (2x + 3) 5 = (2x + 3) 7+5 = (2x + 3) 12 ; Power of a power:((2x + 3) 7 ) 5 = (2x + 3) 7 5 = (2x + 3) 35 ; Power of a product: ((2x + 3)(2x + 2)) 7 = (2x + 3) 7 (2x + 2) 7. Let s figure out an identity about dividing powers by looking at two examples: x 5 = x xxxx x 2 1 x x = xxx 1 = x 3 = x 5 2 x 2 = x2 x 5 x 5 = 1 x x x xxxx = 1 xxx = 1. x 3 The first example shows the following: Procedure x m x n = x m n To divide powers, subtract exponents: Does this procedure work for the second example? Substituting 2 for m and 5 for n gives x2 x 5 = x 2 5 = x 3.

5 But we already know that x2 x 5 = 1 x 3, and this suggests that x 3 = 1 x 3. Definition of negative powers x n = 1 x n This definition is an identity. To substitute an expression for x, use parentheses. Here are some examples. Substitute (x 2 + 7) for x to get (x 2 + 7) 1 = 1 x Substitute (xy + 3) for x to get(xy + 3) 5 1 = (xy + 3) 5 You probably already know that x 0 = 1 and x 1 = x, as well as identities involving involving fractions that will follow easily once we have introduced fraction notation. All the basic identities are listed on the next slide for your convenience. In those identities, m and n can now be any integers, positive or negative. We will see later that they make sense and remain true if m and n are allowed to be fractions.

6 Power identities x 0 = 1 x 1 = x x m x n = x m+n x m x n = xm n (x m ) n = x mn (xy) m = x m y m x n = 1 ( ) x n x m = xm y y n ( ) x 1 = y y x ( ) x m ( y m y = = y x) m x m

7 Exercise: Rewrite and verify all the above identities with m = 2; n = 3; x = 4, and y = 2. Some incorrect identities involving whole number powers Make sure you don t confuse the following two examples: x 3 x 2 = x 3+2 = x 5. However ( x 3) 2 = x 3 2 = x 6. Here are some additional gentle warnings: x m + x n does not simplify (unless m = n. In that case, x m + x m = 2x m ). (x + y) m is not x m + y m. (x + y) m can be rewritten as a simplified sum if m = 2, 3, 4,..., but you need to multiply out to figure out the answer. For example (x + y) 3 = (x + y)(x + y) 2 = (x + y)(x 2 + 2xy + y 2 ) = x 3 + 3x 2 y + 3xy 2 + y 3 In particular, (x + y) 3 is not x 3 + y 3. Please don t invent identities!

8 Rewriting fractions that involve powers of variables The basic fraction identity A B = A 1 B = A B 1. Numerical example: 7 6 = = Every identity involving fractions is derived from the above statement. For example, substituting x 4 for A and y 5 for B gives (x 4 ) (y 5 ) = 1 (x4 ) (y 5 ) = (x4 ) (y 5 ) 1 = (x 4 ) (y 5 ) = x 4 y 5 The last step applied the power to a power rule and omitted useless parentheses. This is a pattern that you need to know in your bones. For example, = Two methods for rewriting products of powers: An expression that is a product of powers can be written in two ways: as a fraction with only positive powers, or without a fraction, but possibly including negative powers.

9 Example 1: Rewrite a5 b 5 c 6 without a fraction line. Solution: a5 b 5 c 6 = a 5 b 5 1 c 6 = a5 b 5 c 6. 1 Example 2: Rewrite c 6 without a fraction line. d7 1 Solution: Start out with c 6 d 7 Definition of negative powers: Power of product identity Power to a power identity = (c 6 d 7 ) 1 = (c 6 ) 1 (d 7 ) 1 = c 6 d 7. This is the answer. Example 3: Write a5 b 5 c 6 without a fraction line. d7 Solution: a5 b 5 c 6 d 7 = (a5 b 5 )(c 6 d 7 ) 1 = a 5 b 5 c 6 d 7 The critical move was to apply the power of a product identity by substituting c 6 for A, d 7 for B, and 1 for m = 1 in (AB) m = A m B m to see that (c 6 d 7 ) 1 = c 6 d 7.

10 You need to perform this move in your head. Some more examples: (a 5 b 6 ) 2 = a 5 2 b 6 2 = a 10 b 12 (a 5 b 6 c 8 ) 2 = a 10 b 12 c 16 Procedure Moving powers between numerator and denominator: If a power is a factor of a fraction s numerator or denominator, you may rewrite that fraction by using the following identities: x m E F = E x m F A proof of the first identity: x m E F Example 4: x 12 y 3 = y3 x and x12 and x m E F = E x m F = x m E F = 1 x m E F = 5 y 3 x 4 = y3 y3 x 5 = x 4 x E x m F Example 5: x 3 y 5 z 3 = y5 x 3 (z 3 ) = y5 x 3 z 3 Warning: In the above examples, x m or x m is a factor of the numerator or denominator. If that s not true, the power can t be moved across the fraction line. Here are two examples:

11 x 2 +xy y 3 Example 6: cannot be rewritten as numerator x 2 + xy. xy x 2 y 3 because x 2 is not a factor of the a+6 Example 7: In c 2 (d 8 ), you can rewrite c 2 in the denominator as c 2 in the numerator. However, you must rewrite this fraction as c2 (a+6), not as c2 a+6. d 8 d 8 To see why, use parentheses to substitute c for x, 2 for m, a + 6 for E, and d 8 for F in to get. Parentheses are necessary! E x m F = xm E F Example 8: Rewrite (a+6) c 2 (d 8 ) = c2 (a+6) d 8 ( 2a 2 b 3 c 4 ) 2 without parentheses. ( 2a 2 b 3 c 4 ) 2 Solution (long version): The original problem is Use ( ) A m B = A m B with A = 2a 2 and B = b 3 c 4 to get = (2a2 ) 2 m (b 3 c 4 ) 2 Use (AB) 2 = A 2 B 2 to rewrite the numerator = 22 (a 2 ) 2 (b 3 c 4 ) 2 Use the same identity to rewrite the denominator = 22 (a 2 ) 2 Use (A m ) n = (A) mn to get the answer: (b 3 ) 2 (c 4 ) 2 = 4a4 b 6 c 8

12 ( ) 3 Example 9: Rewrite 2a 1 b a 2 b as a reduced fraction without negative exponents. 3 You can t rush into this problem. The most important principle is that you are trying to get rid of negative powers. This can be done is several ways. Strategy 1: Rewrite negative power factors across the fraction line as positive powers. Here is the original problem Rewrite a 1 and b 3 across the fraction line = Apply power times power rule = Apply (A/B) 3 = (B/A) 3 = ( ) 3 2a 1 b a 2 b 3 ( 2bb 3 a 2 a 1 ) 3 ( 2b 4 a 3 ) 3 ( a 3 2b 4 ) 3 Apply (A/B) 3 = A 3 /B 3 = (a3 ) 3 (2b 4 ) 3 Apply (A m ) n = A mn = a9 2 3 (b 4 ) 3 = a9 8b 12

13 After a while, you should be able to skip the explanatory remarks and write just: ( ) 3 ( ) 3 ( ) 2a 1 b a 2 b = 2b 3 b 3 aa = 2b 4 3 ( ) 2 a = a 3 3 (a 3 2b = 3 ) 3 = a9 = a9 4 (2b 4 ) (b 4 ) 3 8b 12 Strategy 2: First flip the fraction to eliminate the outside negative power. ( ) 3 ( ) 3 2a 1 b a 2 b = a 2 b 3 3 2a 1 b = (a 2 ) 3 (b 3 ) 3 = a6 b 9 = a6 a 3 = a9 2 3 (a 1 ) 3 b 3 8a 3 b 3 8b 3 b 9 8b 12 Strategy 3: Combine the last two methods. First eliminate the outside negative power. Next eliminate all inside negative powers. ( ) 3 ( ) 3 ( ) 3 ( ) 2a 1 b a 2 b = a 2 b 3 3 2a 1 b = a 2 a 2bb = a b = (a 3 ) 3 = a9 4 (2b 4 ) 3 8b 12

14 Algebra and Geometry: Powers and measurement If a square has side length s, its area is s s = s 2. Thus A = s 2. For example, a square with side length 5 centimeters (cm) has area 5 2 = 25 square centimeters (cm 3 ). If we know s, it s easy to find A. These statements are illustrated by the square at the right, with side length 5 and area 5 2 = 25. As stated in the introduction, we will usually not mention specific units of measurement. Now let s go in reverse. Suppose we know the area A and would like to figure out the side length s. In other words, we need to solve the equation s 2 = A for the side length s, which must be a positive real number. Then s = A, the square root of A. For example, the side length of a square with area 9 is 9 = 3. In general, the side length will be irrational. For example, the side length of a square with area 2 is while the side length of a square with area 3 is A similar discussion applies to a cube with side length s. Its volume is V = s 3. If we know V and want to find s, then s = 3 V is called the cube root of V.

15 Radical notation and fractional exponents In the last slide, the powers and roots involved were positive real numbers, used for measurement. But it s tricky to extend the discussion to negative numbers. The equation x 2 = 4 has two solutions x = 2 and x = 2, written x = ±2. The square root of 4 is the positive solution. Thus 4 = 2, not ±2. The equation x 2 = 4 has no solutions, since the square of any real number is positive. Thus 4 is undefined. Squares and square roots For all real numbers u > 0, the equation x 2 = u has two solutions: x = u (positive) and x = u (negative). For all real numbers u < 0, x 2 = u has no solutions and so u is undefined. The situation is completely different when we deal with third powers and roots. Cubes and cube roots For all real numbers u, the equation x 3 = u has one and only one solution x = 3 u, which has the same sign as u.

16 Example 10: 4 = 2 The solution of x 2 = 4 is x = ± 4 = ±2. However, x 2 = 4 has no solutions. x 3 = 8 has one solution x = 3 8 = 2. x 3 = 8 has one solution x = 3 8 = 2. Similar statements hold for solutions of x n = u depending on whether n is even or odd. n For n > 3, we call u the n th root of u. Even powers and roots Suppose n is even. For all real numbers u > 0, the equation x n = u has two solutions: x = n u (positive) and x = n u (negative). For all real numbers u < 0, x n = u has no solutions and so n u is undefined. Odd powers and roots Assume n is odd. For all real numbers u, the equation x n = u has one and only one solution x = n u, which has the same sign as u.

17 Example 11: The pattern for odd roots: 3 8 = 2 because 2 3 = 8. The solution of x 3 = 8 is x = 3 8 = = 2 because ( 2) 3 = 8. The solution of x 3 = 8 is x = 3 8 = 2. The pattern for even roots: 4 81 = 3 because 3 4 = 81 and 3 is positive. The equation x 4 = 81 has two solutions x = ± 4 81 = ± is undefined because x 4 = 81 has no solution. That s because x 4 = x 2 x 2, the product of two positive numbers, is positive. Special advice involving numerical expressions with radicals: Expressions such as 1, 4, and 9 should be rewritten immediately as 1, 2, and 3, respectively. I am emphasizing this point, since you may have been told otherwise in your high school courses.

18 Until now, x n has been defined only when n is an integer, positive or negative. We now show that the roots we have been discussing can be represented as fractional powers. In the power to a power rule (u m ) n = u mn, it was assumed that m and n are whole numbers. If, however, we write m = 1/n, then the power to a power rule says (u 1/n ) n = u 1/n n = u 1 = u. In other words: x = u 1/n is a solution of x n = u. However, we have seen that x = n u is a solution of x n = u. Conclusion: u 1/n = n u The above conclusion is not entirely correct. After all, we have usually understood that variables represent real numbers, although we have not stated this explicitly. If n is even, then u 1/n = u is defined if u < 0. However, if variables represent positive number, than we can be comfortable writing identities with fractional powers as on the next slide.

19 Identities involving radicals are obtained by rewriting power identities. For instance, if we use m = 1/2 to rewrite (xy) m = x m y m, we get (xy) 1/2 = x 1/2 y 1/2 and so xy = x y. Square root identities xy = x y if x 0 and y 0. x y = x y if x 0 and y > 0. Warning: x + y = x + y is NOT an identity. Example 12: 12 = 4 3 = Example 13: 12 = = 3 2. We usually prefer to write a fraction without a 3 radical sign in the denominator. In this case, multiply by 3 3 to obtain a complete solution as follows: 3 12 = 3 2 = = (3) = 3 2.

20 We defined above x 1/n = n x. Since m n = m 1 n = 1 n m we can substitute 1/n for n in the identity (x m ) n = x mn to obtain x m n = x m (1/n) = (x m ) 1/n = n x m. The following summarizes this statement and related ideas. Identities with fractional powers and radicals Assume m and n are integers. n x n = x if n is odd or x 0. n x n = x if n is even. (x) m n = (x m ) 1 n = n x m if n is odd or x 0. ) (x) m n = (x 1 m n = ( n x) m if n is odd or x 0. n xy = n x n y if n x and n y are both defined. n x y = n x n y if both n th roots are defined and y is not 0.. Exercise: Please check the last two identities by letting n = 2 and x = 4 or 4.

21 Example 14: = = (2 1 4 ) 3 = ( 4 2) 3. Another possibility: = = ( 2 3) 1 4 = 4 8. This answer looks nicer. ( ) x 3 4 = x = ( 4 x) 3 or 4 x 3 ( ) = = 3 2 = = Definition = ( ) 3 = 1 = In radical expressions such as 12 or a + b or 5 12 or 5 a + b, the radicand is the number or expression (12 or a + b in these examples) under the radical sign. Warning: n x + y IS NOT equal to n x + n y. In particular x + y IS NOT equal to x + y.

22 Simplifying expressions with radicals. The rule xy = x y (for x and y positive) allows us to simplify square root expressions. Example 15: 12 = 4 3 = 4 3 = = = = Procedure To simplify a square root, factor out from the radicand the largest possible number that is a perfect square. You can do this by trial and error or by using the prime power factorization of the radicand. In all cases the strategy is to factor our squares from the radicand. Example 16: Simplify each of the following. 9 = 3 2 = 3 18 = = = 2 3 = = 9 5 = 9 5 = = 4 27 = = = 6 3

23 The general strategy is to factor out the highest possible even power from each power in the radicand. Example 17: = = = = Example 18: 288 = 32 9 = = = = 12 2 Example 19: Simplify Solution: First work out the square roots: 216 = = = = = = = 6 6 Or, a bit faster: 216 = 36 6 = 36 6 = 6 6 Similarly, 96 = 16 6 = 16 6 = 4 6 Then complete the original problem: = = = 38 6

24 Here are examples with variables. The strategy is still to factor out even powers, since x 2n = (x 2n ) 1/2 = (x) 2n 1/2 = x n. To avoid technical difficulties arising from the fact that x 2 = x, not x, assume all variables represent positive numbers. Example 20: x 21 = x 20 x = x 20 x = x 10 x Example 21: x 11 y 13 = x 10 y 12 xy = x 10 y 12 xy = x 5 y 6 xy Example 22: Simplify x 3 y 3 x 11 y x 2 y x 13 y 15. Again we simplify the radicals separately: x 11 y 11 = x 10 y 10 xy = x 5 y 5 xy and x 13 y 15 = x 12 y 14 xy = x 6 y 7 xy. Therefore x 3 y 3 x 11 y x 2 y x 13 y 15 = x 3 y 3 x 5 y 5 xy + 7x 2 yx 6 y 7 xy = 1x 8 y 8 xy + 7x 8 y 8 xy = 8x 8 y 8 xy A similar strategy works for cube roots: a power x n has a nice cube root if n is a multiple of 3. 3 Example 23: x 30 = (x 30 ) 1/3 = (x) 30 1/3 = x 10

25 A power of a variable is a perfect cube provided that power is a multiple of 3. In the following we factor out the largest power of x that is a multiple of 3 Example 24: 3 x 31 = 3 x 30 x = 3 x 30 3 x = x 10 3 x 3 x 32 = 3 x 30 x 2 = 3 x 30 3 x 2 = x 10 3 x 2 3 x 33 = 3 (x 11 ) 3 = x 11 Finally, we do some problems involving rewriting powers with fractional exponents. This doesn t involve anything new. Again, assume variables represent positive numbers. Example 25: Simplify (s 2 ) 1/7 s 1/2 Solution: First find (s 2 ) 1/7 = (s) 2 (1/7) = s 2/7. Then (s 2 ) 1/7 s 1/2 = s 2/7 s 1/2 = (s) 2/7+1/2 = s 11/14 since = = =

26 Example 26: Simplify Solution 1: ( x 8 y 4 16y 4/3 ) 1/4 = (x8 y 4 ) 1/4 (16y 4/3 ) 1/4 = x 2 y /4 y 1/3 = 161/4 y 1 y 1/3 x 2 = 2y4/3 x 2 ( x 8 y 4 16y 4/3 ) 1/4 Solution 2: ( x 8 y 4 = = = ) 1/4 16y 4/3 ( ) 16y 4/3 1/4 x 8 y 4 ( ) 16y 4/3 ( 4) 1/4 x 8 ( 16y 16/3 x 8 ) 1/4 = (16y16/3 ) 1/4 (x 8 ) 1/4 = 161/4 (y 16/3 ) 1/4 (x 2 = 2y4/3 x 2

27 Exercises for M19500 Chapter 1.2: Exponents and radicals Click on Wolfram Calculator to find an answer checker. Click on Wolfram Algebra Examples to see how to check various types of algebra problems. 1. Do the WebAssign homework. 2. Solve each of the following equations a) x 2 = 7 b) x 2 = 9 c) x 3 = 8 d) x 4 = 16 e) x 2 = 16 f) x 3 = Rewrite each of the following with neither radicals nor fraction exponents: a) 9 b) 7 c) 3 27 d) 3 27 e) 4 16 f) g) 27 2/3 h) ( 27) 2/3 i) ( 27) Simplify a) b) Rewrite without negative exponents: ( ) 3 ( ) ( ) 2 ( a) a 1/6 b 3 x 2 b 1 b) x 2 b 1 a 1/6 b 3 x 1 y a 3/2 y 1/3 a 3/2 y 1/3 x 1 y ) 3

28 6. Rewrite without negative exponents or radical signs: 5 x 3 y 2 10 x 4 y Rewrite as a single term with lowest possible powers in the radicand: a) 4a 3 b 2 c 12a 5 b 7 c 9 3a 4 b 4 c 4 27a 3 b 3 c 3 b) 4a 3 x 2 c 12a 9 x 1 1c 9 3a 4 x 4 c 4 27a 7 x 7 c 3

29 Quiz Review Example 1: Rewrite a5 b 5 c 6 without a fraction line. 1 Example 2: Rewrite c 6 without a fraction line. d7 Example 3: Write a5 b 5 c 6 without a fraction line. d7 Example 4: Rewrite x 12 y 3 and x 5 y 3 without negative exponents. x 4 Example 5: Rewrite x 3 y 5 Example 6: z 3 Does x 2 +xy y 3 without negative exponents. equal xy x 2 y 3? Why or why not? a+6 Example 7: Rewrite without negative exponents. c 2 (d 8 ) ( ) Example 8: Rewrite 2a 2 2 b 3 c without parentheses. 4

30 ( 2a 1 b Example 9: Rewrite ) 3 a 2 b as a reduced fraction without negative exponents. 3 Example 10: 4 =? The solution of x 2 = 4 is x =? Solve x 2 = 4. Solve x 3 = 8. Example 11: Find 3 8. Solve x 3 = 8. Find 3 8. Solve x 3 = 8. Find Solve x 4 = 81. Find Example 12: Rewrite 12 by factoring squares out of the radicand. Example 13: Rewrite 3 12 without a radical in the denominator.

31 Example 14: Rewrite each expression without fractional powers x Example 15: Simplify the following square roots: Example 16: Simplify each of the following Example 17: Simplify

32 Example 18: Simplify 288. Example 19: Simplify Example 20: Simplify x 21. Example 21: Simplify x 11 y 13. Example 22: Simplify x 3 y 3 x 11 y x 2 y x 13 y 15. Example 23: Simplify 3 x 30. Example 24: Simplify each of the following: 3 x x x 33. Example 25: Simplify (s 2 ) 1/7 s 1/2. Example 26: Simplify ( x 8 y 4 16y 4/3 ) 1/4.

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