# Exponentiation: Theorems, Proofs, Problems Pre/Calculus 11, Veritas Prep.

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1 Exponentition: Theorems, Proofs, Problems Pre/Clculus, Verits Prep. Our Exponentition Theorems Theorem A: n+m = n m Theorem B: ( n ) m = nm Theorem C: (b) n = n b n ( ) n n Theorem D: = b b n Theorem E: n m = n m Theorem F: 0 = Theorem G: n = n Theorem H: n/m = m n Like ll theorems, these do not come out of nowhere. They come from definition nd logicl deduction. Let us strt, then, with definition. Wht is exponentition, nywy? Wht do we men when we write number with superscript? I contend tht we men something like repeted multipliction. When we write, for exmple, 5 4, we men this simply s more convenient wy of writing (or, even more simply, 625). Formlly, let s define exponentition in the following wy 2 : n (The triple-equls-sign here is used to show tht this is definition tht this eqution is not the consequence of some erlier theorem or xiom, but tht with it we re defining wht we men by n.) Let us prove these theorems. Theorem A: n+m = n m Proof: We ll strt with the left side of the eqution, pply the definition of exponentition, do some lgebr, nd eventully end up with the right side. n+m = n+ (by the definition of exponentition) = (} {{ } )( } {{ } ) (becuse multipliction is ssocitive) = n m (pplying the definition of exponentition gin) (done! this stylized A is my end-of-proof symbol.) Theorem B: ( n ) m = nm Proof: To prove this, we ll need to pply the definition of exponentition twice. (Well, ctully three times, but the lst time doesn t count.) For the most prt. If you re mthemticin, we could hve lengthy converstion bout everything I sy on this sheet, but for our purposes this definition is sufficient. If you disgree (nd this is in n imginry converstion between myself nd mthemticin), I will sy but this: do you believe we should tech students Dedekind cuts before we let them utter the words, rel number? 2 Note tht we hven t put ny specifictions on wht nd n cn nd/or should be must they be integers? rel numbers? complex numbers? The wy our definition works, wht with the business, it must only hold for n being positive integer, but these theorems hold for n being ny rel number. Wht we ll see in our derivtions is tht, even though we strt by considering n only s positive integer, we cn extend our ide of exponentition in very logicl wy such tht it holds for ny rel number. But this is in footnote for reson.

2 ( n ) m = (} {{ } ) m (by definition) = (} {{ } ) (} {{ } ) ( } {{ } ) (by definition, gin) } {{ } = n* (multipliction) = nm (finish by re-pplying the definition in reverse) Theorem C: (b) n = n b n Proof: See pttern? We ll pply the definition of exponentition, do some lgebr, nd eventully get wht we wnt. (b) n = } b b {{ b } (by definition) = (} {{ } ) (b} b {{ } b) (multipliction is commuttive we cn rerrnge) = n b n (definition) ( ) n n Theorem D: = b b n Proof: How do you think this should go? Theorem E: n m = n m. Proof: This is where things strt to get interesting. The cler first step is to write out the entire quntity: n m = But where we go from here will depend on the reltive size of n nd m. n could be bigger thn m, it could be smller, or the two could be equl. Let s strt with the first cse tht of n > m. In tht cse, both the top nd the bottom will hve t lest m of the s, nd the top will hve more it ll hve n m more (becuse m + (n m) = n). Put 2

3 differently, we cn think of our frction s being something like this: = n m = n- If you re confused, count up the totl number of s on top there should be n of them, just like wht we strted with. And then wit second! There re m s on top, nd m s on the bottom! We cn cncel them out! And we just end up with: n- = Which, by definition, is just n m. But wht bout these other two cses? First of ll, wht hppens in the cse tht m = n? Well, if m = n, then n will just equl m n (or n m either wy is the sme). But tht clerly is just equl to. m And I hope you will gree tht if n nd m re equl, then n m = 0, nd tht then n m = 0. So clerly, then, since we wnt n = n m, then 0 =. (Which should mke sense., fter ll, is the multiplictive m identity, so if we multiply something no times, we should end up not with zero, but with one. Think bout how if you hve frction in which everything cncels out, you hve, not 0. Sme sort of del.) Finlly. Wht if n < m? Then (by the sme rgument s the first cse) we hve something like: = n m = But wit! We cn cncel things here, too! We get: = } {{ } m- = } {{ } m n m- But this isn t quite wht we wnt. We wnt n m. So let s define wy, our eqution here will become: = m n = (m n) = m+n = n m stuff s being equl to -stuff. Tht Excellent. This hs been long discussion, so let s review wht we ve done. We ve extended our ide of n exponent from just the positive integers to ll integers (i.e., we cn now exponentite by 0 nd negtive numbers!) But in order to do this, we hd to mke slight extension of our definition. Our originl definition only considered the cse of n exponent being positive integer; in the course of this proof, we discovered nturl wy to extend tht definition to 0 nd the negtives. Thus this hs been somewht more thn just proof it hs been prtly proof (for n > m), nd prtly redefinition. Theorem F: 0 = Proof: We lredy discussed this, in the proof of E Theorem G: n = n Proof: Agin, we lredy discussed this in the proof of E Note how my writing style is bit different in this proof thn in the previous ones I m mking n rgument in prose rther thn in nice, clen, two-column formt. Tht s OK. Mny mth textbooks do the sme thing, nd it s the job of the reder (nd wht job it is!) to trnslte tht into more cler, more symbolic formt if necessry.

4 Theorem H: n/m = m n Proof: So fr, we lredy know how to del with exponents when the exponents re positive integers (thnks to our definition), zero (thnks to F), or negtive integers (thnks to G). But wht if we wnt to move beyond the integers nd hve exponents tht re rtionl numbers (i.e., frctions)? This theorem will tell us how to think bout tht. First of ll, I should point out tht we don t relly hve solid definition of wht we men by stuff. (From technicl, mthemticl stndpoint, wht we re doing here is defining wht we men by such nottion.) But intuitively, I hope you gree tht the bsic ide of rdicl/root is tht if I hve n nth root of something, nd I multiply n of those nth roots, I end up with my originl thing. Mening the thing under the rdicl sign. Right? Good. So imgine we hve /m. (Cn we do this? Let s ssume we cn.) Now imgine we hve m copies of it, ll multiplied together: /m /m /m /m By the definition of exponentition, this must be ( /m ) m. But by B, we cn simplify this: /m m = m/m = In other words, whtever this /m is, if we tke m copies of it, we get. So this is just root! (The mth root, to be specific). Formlly: /m = m Wht bout the rest (the n/m prt?) Imgine we hve n/m. Becuse of Theorem B gin, we cn write this s ( n ) /m. Which, becuse of wht we just proved, is m n. So now we know how to exponentite by ny rtionl number. Hoory! Evlute ech of the following expressions. Problems ( / /2 ) 6. 6 / / 8. ( 5 0) 2/ 9. ( 7 /) /4 (. 27 ( ) / ) / ( ( /4 ) /2 ) 2/ 6. ( 2 0) / ( 0 + ) / /2 2. ( 26 ) /2 8 / ( 8 24, ) ( ) /2 27. ( + 2) / 9 /2 27 / / /2 2 0 ( 2 ) /2 (9 4 ) 27

5 Simplify ech of the following expressions so tht they re written only with positive exponents nd no rdicls y x 8 6. x x y 2 y 8. (x 2 y)(4x 5 y 4 ) 9. (2x y) ( b) 4. (6x 2 ) /2 42. (8y 6 x ) / 4. (4x )(2x) x 2 x 2 y ( ) 2 2x y 8x 2 9xy 47. (27x y 9 ) / r 8 s b x 4 y (4x + 2y) b + ( + b) b 5. (7) 2 (5b) /2 (5) /2 (7b) 4 (6) /2 b b /2 ( ) 5/9 r 2/ 55. s /5 56. (c 2/5 d 2/ )(c 6 d ) 4/ 57. (2) /2 (b) 2 (4) /5 (4) /2 (b) 2 (2) /5 58. ( x2 ) /x (b x ) x b x c (c 5/6 ) 42 (c 5 ) 2/ Prove tht ech of the following equtions re true. (How do you do this? Think of it s being like two-column proof in geometry. Strt with one side of the eqution nd step-by-step pply exponentition lws until you end up with the other side of the eqution.) 6. b 2 b /2 = b 62. c 2 d 6 4c d 4 = d5 4c 6. b 7 ( b) 4 = 4 5

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