{"id":187,"date":"2024-07-01T14:11:35","date_gmt":"2024-07-01T14:11:35","guid":{"rendered":"https:\/\/routledgelearning.com\/teachingsecondarymathematics\/?post_type=content&p=187"},"modified":"2024-08-07T10:38:23","modified_gmt":"2024-08-07T10:38:23","slug":"chapter-9-proof","status":"publish","type":"content","link":"https:\/\/routledgelearning.com\/teachingsecondarymathematics\/content\/resources\/chapter-9-proof\/","title":{"rendered":"Chapter 9 – Proof"},"content":{"rendered":"\n
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\n\tHome<\/a>\n<\/span>\/<\/span>\n\tResources<\/a>\n<\/span>\/<\/span>\n\tChapter 9 \u2013 Proof\n<\/span><\/div><\/div><\/div>\n\n\n\n
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Chapter 9 – Proof<\/h1>\n\n\n

Beginning with the Standards in 1989, the National Council of Teachers of Mathematics suggested a change in the role of proof in our mathematics curriculum. The call is for a decrease in attention given to Euclidean geometry as an axiomatic system and two-column proofs. At the same time, it is recommended that short sequences of theorems be developed and that deductive arguments be expressed orally and in paragraph or sentence form. Suppose the desire is to prove the base angles of an isosceles triangle are congruent using the typical two-column proof as well as paragraph form. In one figure, you are given triangle DEF with DE \u2245 DF and asked to prove \u2220DEF \u2245\u2220DFE. The power of proof can foster an increase in mathematical communication and in-depth knowledge. Simply knowing the numerical solution does not imply an understanding of the mathematics and problem-solving behind the solution. Updated versions of the Standards include the importance of mathematical reasoning and proof as key elements in the curriculum.DOWNLOAD<\/p>\n<\/div>\n\n\n\n

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Exercises<\/h2>\n\n\n\n
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Exercise 9.1<\/h3>\n\n\n\n
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  1. 36 in = 1 yd; 9 in = 0.25 yd (divide both sides by 4); 9<\/mn><\/msqrt><\/math>  in = 0.25<\/mn><\/msqrt><\/math>  yd; 3 in = 0.5 yd (positive square root of both sides). Is it true that 3 inches equals half a yard? What is wrong with this \u201cproof\u201d?<\/li>\n<\/ol>\n<\/div>\n\n\n\n
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    Exercise 9.2<\/h3>\n\n\n\n
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    1. Use a dynamic geometry program to create a setting like the medians of a triangle discussed in the chapter. Move the figure to confirm, visually, that an invariant has been created.<\/li>\n\n\n\n
    2. Graphing circles centered at (2, -5) and (5, -2), each with a radius of 2, and the line y = -x yields a figure in which the line appears tangent to the circles. Zooming shows that is not the case. Create a similar environment using symbolic manipulating, function-plotting software.<\/li>\n<\/ol>\n<\/div>\n\n\n\n
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      Exercise 9.3<\/h3>\n\n\n\n
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      1. Create a lesson for an algebra class that includes a historical proof.<\/li>\n\n\n\n
      2. Create a lesson for a geometry class that includes a historical proof.<\/li>\n\n\n\n
      3. Create a lesson for a pre-calculus class that includes a historical proof.<\/li>\n\n\n\n
      4. Create a lesson for a calculus class that includes a historical proof.<\/li>\n\n\n\n
      5. Create a lesson for a pre-algebra class that includes a historical proof.<\/li>\n<\/ol>\n<\/div>\n\n\n\n
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        Exercise 9.4<\/h3>\n\n\n\n
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        1. Show that sin2<\/sup>x + cos2<\/sup>x = 1 graphically. Describe how this development could be used in the secondary curriculum.<\/li>\n\n\n\n
        2. Use software\/application or a graphing calculator to show two different identities graphically, and describe the results for each.<\/li>\n<\/ol>\n<\/div>\n\n\n\n
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          Exercise 9.5<\/h3>\n\n\n\n
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          1. Prove that 2 + 4 + 6 + . . . + 2n<\/em> = n<\/em>(n<\/em> + 1)<\/li>\n\n\n\n
          2. Prove that<\/li>\n<\/ol>\n\n\n\n\n \n \n 1<\/mn>\n \n 1<\/mn>\n \u00d7<\/mo>\n 3<\/mn>\n <\/mrow>\n <\/mfrac>\n +<\/mo>\n \n 1<\/mn>\n \n 3<\/mn>\n \u00d7<\/mo>\n 5<\/mn>\n <\/mrow>\n <\/mfrac>\n +<\/mo>\n \n 1<\/mn>\n \n 5<\/mn>\n \u00d7<\/mo>\n 7<\/mn>\n <\/mrow>\n <\/mfrac>\n +<\/mo>\n \u2026<\/mo>\n \n 1<\/mn>\n \n (<\/mo>\n 2<\/mn>\n n<\/mi>\n \u2212<\/mo>\n 1<\/mn>\n )<\/mo>\n \u00d7<\/mo>\n (<\/mo>\n 2<\/mn>\n n<\/mi>\n +<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n <\/mfrac>\n =<\/mo>\n \n 1<\/mn>\n \n 2<\/mn>\n n<\/mi>\n +<\/mo>\n 1<\/mn>\n <\/mrow>\n <\/mfrac>\n <\/mrow>\n<\/math>\n<\/div>\n\n\n\n
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            Exercise 9.6<\/h3>\n\n\n\n
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            1. Do a proof that shows how the divisibility rule for 9 would work with a four-digit number WXYZ<\/em>.<\/li>\n\n\n\n
            2. Should students be required to prove a divisibility rule? Why or why not?<\/li>\n\n\n\n
            3. Why does the 6 rule break into an even 3 rule? Explain why a similar rule could or could not be devised for divisibility by 15.<\/li>\n\n\n\n
            4. Describe a divisibility rule for some number other than those discussed here in the text and prove why it works.<\/li>\n\n\n\n
            5. Are divisibility rules limited to integers?<\/li>\n<\/ol>\n<\/div>\n\n\n\n
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              Exercise 9.7<\/h3>\n\n\n\n

              Note that in today\u2019s technological world, you can find each of these solved somewhere on the Internet. The challenge is to avoid doing that, and YOU do the proof. THAT is how you grow your own abilities.<\/p>\n\n\n\n

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              1. Prove that<\/li>\n<\/ol>\n\n\n\n\n \n \n 1<\/mn>\n 2<\/mn>\n <\/msup>\n +<\/mo>\n \n 2<\/mn>\n 2<\/mn>\n <\/msup>\n +<\/mo>\n \n 3<\/mn>\n 2<\/mn>\n <\/msup>\n +<\/mo>\n \u2026<\/mo>\n +<\/mo>\n \n n<\/mi>\n 2<\/mn>\n <\/msup>\n =<\/mo>\n \n \n n<\/mi>\n (<\/mo>\n n<\/mi>\n +<\/mo>\n 1<\/mn>\n )<\/mo>\n (<\/mo>\n 2<\/mn>\n n<\/mi>\n +<\/mo>\n 1<\/mn>\n )<\/mo>\n <\/mrow>\n 6<\/mn>\n <\/mfrac>\n <\/mrow>\n<\/math>\n\n\n\n
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                1. Prove that rm<\/sup>rn<\/sup> = rm + n<\/sup> for r, a real number, and m and n as counting numbers.<\/li>\n\n\n\n
                2. Prove that if n is a natural number, a is a real number and a > \u2013 1, then
                  (1 + a)n<\/sup>  >  1 +na.<\/li>\n\n\n\n
                3. Find and prove a unique statement using PMI.<\/li>\n<\/ol>\n<\/div>\n<\/div>\n\n\n\n

                  Problem Solving Challenges<\/h2>\n\n\n\n
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                  1. Suppose that the surface of the Earth is smooth and spherical and that the distance around the equator is 25,000 miles. A steel band is made to fit tightly around the Earth at the equator, then the band is cut and a piece of band 18 feet long is inserted. Assuming the equator is a circle and the band is a concentric circle, to the nearest foot, what will be the gap, all the way around, between the band and the Earth\u2019s surface? (Use 3.14 as an approximate value of Pi.)  <\/li>\n<\/ol>\n\n\n\n

                    Hint: Try a smaller problem with a smaller Earth.<\/p>\n\n\n\n

                    Answer<\/summary>\n

                    Answer\/solution: 3 feet<\/p>\n\n\n\n

                    Circumference = Pi x diameter<\/p>\n\n\n\n

                    25,000 miles = 132,000,000 ft<\/p>\n\n\n\n

                    132,000,000 = 3.14 x D<\/p>\n\n\n\n

                    D = 42,038,216.56 ft.<\/p>\n\n\n\n

                    If 18 feet is added to the circumference, then<\/p>\n\n\n\n

                    132,000,018 = 3.14 x D<\/p>\n\n\n\n

                    D = 42,038,222.29 ft<\/p>\n\n\n\n

                    The new diameter \u2013 old diameter = 42038222.29 \u2013 42038216.56 = 5.73<\/p>\n\n\n\n

                    5.73 is the added length for the new diameter. Half of this would be the gap on each side of the Earth between the Earth and the new band, which is 2.865 and that rounds to 3 feet.<\/p>\n<\/details>\n<\/div>\n\n\n\n

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                    1. Two = One                                               <\/li>\n<\/ol>\n\n\n\n

                      Observe the following algebraic proof.<\/p>\n\n\n\n

                      Given:<\/p>\n\n\n\n

                           A and B are real numbers;<\/p>\n\n\n\n

                           A = B<\/p>\n\n\n\n

                      A = B<\/p>\n\n\n\n

                                                        A2<\/sup> = AB<\/p>\n\n\n\n

                                                  A2<\/sup> \u2013 B2<\/sup> = AB \u2013 B2<\/sup><\/p>\n\n\n\n

                                       (A + B)(A \u2013 B) = B(A \u2013 B)<\/p>\n\n\n\n

                               Dividing both sides by (A \u2013 B) yields<\/p>\n\n\n\n

                                                 (A + B) = B<\/p>\n\n\n\n

                                Since A = B, substituting B for A yields<\/p>\n\n\n\n

                                                   B + B = B<\/p>\n\n\n\n

                                                       2B = B<\/p>\n\n\n\n

                                      Dividing both sides by B yields<\/p>\n\n\n\n

                                                         2 = 1<\/p>\n\n\n\n

                      Since we know that 2 does not equal 1, clearly state the mistake made in the above algebraic proof. <\/p>\n\n\n\n

                      Hint:  What happens when you divide by zero?<\/p>\n\n\n\n

                      Answer<\/summary>\n

                      Answer\/solution: The error occurs when division is done with (A \u2013 B).  Since A = B, A \u2013 B = 0 and the step involves division by zero which is undefined in the set of real numbers.<\/p>\n<\/details>\n<\/div>\n<\/div>\n\n\n\n

                      Additional Learning Activities<\/h2>\n\n\n\n
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                      There is no Additional Learning Activities for this chapter.<\/p>\n<\/div>\n<\/div>\n\n\n\n

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