# lec0229 - Some Algebra Rules . Laws of exponent and...

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Some Algebra Rules . .. Laws of exponent and Logarithm: If a > 0, a x a y = a x + y ( a x ) y = a xy a x / a y = a x y a x = 1 / ( a x ) If b > 0 and b 1, log b ( xy ) = log b x + log b y log b ( x / y ) = log b x log b y log b ( x p ) = p log b x . Rules of inequalities: a > b a + c > b + c if c > 0, then a > b a c > b c if a > b and b > c a > c ; if a > b and c > d a + c > b + d . Useful algebra rules: ab = 0 a = 0 or b = 0 if bd 0, then a / b = c / d ad = bc ; ( a + b ) 2 = a 2 + 2 ab + b 2 ; ( a + b )( a b ) = a 2 b 2 .

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A couple of summation rules = n i i f 1 ) ( = = m i i f 1 ) ( + + = n m i i f 1 ) ( , if 1 m < n. = b a i i f ) ( = = b i i f 1 ) ( - - = 1 1 ) ( a i i f , if b a > 1. = n i i f 1 ) ( = = - + n i i n f 1 ) 1 (
Prove by induction that n ! > 2 n for all n 4. (Note: n ! = n ( n 1)···2·1, for n 1; 0! = 1 by convention.) Proof: We use induction on n 4. (Basis Step) Consider n = 4. In this case, n ! = 4! = 4·3·2·1 = 24, and 2 n = 2 4 = 16 < 24. So the Basis Step is proved. (Induction Hypothesis) Consider the statement for an arbitrary value of n. Suppose n ! > 2 n for some n 4. (Induction Step) Consider the statement for n+1. We need to prove ( n + 1)! > 2 n +1 ( n + 1)! = ( n + 1) · n !, > ( n + 1) ·2 n , by the Induction Hypothesis > 2 ·2 n , because n + 1 5 > 2 = 2 n +1 . By induction, we have proved the inequality n ! > 2 n for all n 4.

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Inductively (Recursively) Defined Objects The induction principles can be used to define structures in terms of themselves, usually “smaller” in size. We first consider recursively defined functions. For example, to define the Factorial function n !, where n is a natural number, 0! = 1 n ! = n · ( n – 1)!, for n > 0. So, we can compute the following using the recursive definition:
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lec0229 - Some Algebra Rules . Laws of exponent and...

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