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Math Help - exponents: proofs

  1. #1
    Senior Member tukeywilliams's Avatar
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    exponents: proofs

    Prove (i)  x^{n}y^{n} = (xy)^{n} for nonzero real numbers  x,y and integers  n

    Prove (ii)  x^{m+n} = x^{m}x^{n} for nonzero real numbers  x and integers  m,n .

    Use the definition:  x^{-m} = \frac{1}{x^{m}} for  m > 0 .





    23. (i) Proof: We use induction on  n . Case #1:  n > 0. Base case: For  n = 0,  x^{0}y^{0} = 1 = (xy)^{0}. Inductive step: Suppose now as inductive hypothesis that  x^{k}y^{k} = (xy)^{k} for any real numbers  x, y and for a non-negative integers  n . Then  xy(x^{k}y^{k}) = x^{k+1}y^{k+1} = xy(xy)^{k} = (xy)^{k+1}. Case #2:  n < 0 . Base case: For  n = 0 ,  \frac{1}{x^{0}}\frac{1}{y^{0}} = \frac{1}{(xy)^{0}} = 1 for  n > 0 . Suppose now as induction hypothesis that  \frac{1}{x^{k}}\frac{1}{y^{k}} = \frac{1}{(xy)^{k}} for some positive integer  k . Then  \frac{1}{xy}\frac{1}{x^{k}}\frac{1}{y^{k}} = \frac{1}{x^{k+1}}\frac{1}{y^{k+1}} = \frac{1}{xy}\frac{1}{(xy)^{k}} = \frac{1}{(xy)^{k+1}}. Conclusion: Hence, by induction,  x^{n}y^{n} = (xy)^{n} for any real numbers  x, y and for integers  n .  \square


    (ii) Proof: We use induction on  n . Case #1:  m,n > 0 . Base case: For  n = 0 ,  x^{m+0} = x^{m} = x^{m}x^{0}. Inductive step: Suppose now as inductive hypothesis that  x^{m+k} = x^{m}x^{k} for any real numbers  x,y and for non-negative integers  m,n. Then  x \cdot x^{m+k} = x^{m+k+1} = x\cdot x^{m}x^{k} = x^{m}x^{k+1}. Case #2:  m,n < 0. Base case: For  n = 0,  \frac{1}{x^{n}} = \frac{1}{x^{n}}. Inductive step: Suppose now as inductive hypothesis that  \frac{1}{x^{m+k}} = \frac{1}{x^{m}x^{k}} for some positive integer  k. Then  \frac{1}{x}\frac{1}{x^{m+k}} = \frac{1}{x^{m+k+1}} = \frac{1}{x}\frac{1}{x^{m}x^{k}} = \frac{1}{x^{m}x^{k+1}}. Case #3:  m >0, n< 0. Base case: For  n = 0,  x^m = x^m. (a)  x^{m+n} > 0 or (b)  x^{m+n} < 0. Inductive step: If (a), then suppose as inductive hypothesis that   x^{m+k} = \frac{x^{m}}{x^{k}}. Then  x \cdot x^{m+k} = x^{m+k+1} = x \cdot \frac{x^{m}}{x^{k}} = \frac{x^{m+1}}{x^{k}} = \frac{x^{m}}{x^{k+1}}. Conclusion: Hence, by induction,  x^{m+n} = x^{m}x^{n} for any real numbers  x and for  m,n.  \square






    For (ii) am I considering the cases correctly? Am I doing this incorrectly?



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  2. #2
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