Definition Associativity allows us to write an expression of...

Definition Associativity allows us to write an expression of the form $x_1 \cdot x_2 \cdots x_n$ (where $x_1, \dots, x_n \in G$) without ambiguity. Let $x \in G$. For any positive integer $n$, define $x^n = \underbrace{x \cdot x \cdots x}_{n \text{ times}}$. Then define $x^{-n} = (x^{-1})^n$ and $x^0 = e$. Exercise Let $x \in G$, and let $m$ and $n$ be integers. Prove the following. 1. $x^{-n} = (x^n)^{-1}$ 2. $x^m \cdot x^n = x^{m+n}$ 3. $(x^m)^n = x^{mn}$ Note: In each case, $m$ and $n$ may (independently) be positive, negative, or zero. Since the definition of $x^n$ is different when $n$ is positive, negative, and zero, the proof must deal with all these cases separately.

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00:02 In this question we need to prove that group g is a billion.

00:11 So let g be a group in which a b over 3 equal to a over 3 multiply b over 3 and a b over 5 equal to a over 5 multiplied b over 5 multiple b over 5 multiple b b over 5 multiple b 5 for all ab is element of group g to prove that g is a billion we can write that ab equal to b so as ab over 5 equal to a b equal to a over 5 multiply b over 5 so we can write this equation like ab multiply ab multiply ab in five different groups so that is equal to ab over 5 that is a f over 5 multiply b over 5 now rearrange the equation we can get a multiply b a multiply b a multiply b a multiply b a multiply b a multiply b equal to a over 5 multiply b over 5 left multiplicity by a over minus 1 and right multiplicity by b over minus 1 we get a over minus 1 multiply b a xiply b a multiply b a multiply b multiply b over minus 1 that is equal to a over minus 1 multiply a over 5 multiply b over 5 multiply b over minus 1 so we can write b a multiply b a multiply b a multiply b a multiply b a equal to a over 4 triple b over 4 so b a over 4 equal to a over 4 now we can reduce it further we can write a multiply b multiply b multiply b b equal to a over 3 fly b over 3 next we can write a over minus 1 multiply a multiply b a reply b a multiply b a multiply b multiply b over minus 1 equal to a over minus 1 multiply a over 3 multiply v over 3 multiply b over minus 1 b a over 2 equal to a over 2 b a over 2 b a over 4 equal to a over 4 multiply b over 4 b a over 4 equal to b a over 2 multiply v a over 2 that is equal to a over 2 multiply b a over multi so we can write a over 4 multiply b over 4 equal to a over 2 multiple b over 2 reply a over 2 multiply b over 2 we can write a over minus 2 multiplied a over 4 b over 4 multiply b over minus 2 equal to a over minus 2 multiplied a over 2 b over 2 0 multiply v over 2 multiply v over 2 t over minus 2 so we can write a over 2 multiply b over 2 equal to b over 2 multiply a over 2 equal to ab over 2 a over 2 multiply b over 2 equal to ab multiply ab a b multiply a b over minus 1 multiply a a over 2, b over 2, multiply b over minus 1 equal to a over minus 1, multiply ab multiply b over minus 1...

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