ANSWER on Question #66814, Math / Abstract Algebra
Consider the set X = R { − 1 } X = R\{-1\} X = R { − 1 } ∗ * ∗ X X X
x 1 ∗ x 2 = x 1 + x 2 + x 1 × x 2 , ∀ x 1 , x 2 ∈ X x_1 * x_2 = x_1 + x_2 + x_1 \times x_2, \quad \forall x_1, x_2 \in X x 1 ∗ x 2 = x 1 + x 2 + x 1 × x 2 , ∀ x 1 , x 2 ∈ X
1) check whether ( X , ∗ ) (X,*) ( X , ∗ )
2) Prove that x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ n times = ( 1 + x ) n − 1 \underbrace{x * x * x * \cdots * x}_{n \text{ times}} = (1 + x)^n - 1 n times x ∗ x ∗ x ∗ ⋯ ∗ x = ( 1 + x ) n − 1 ∀ n ∈ N \forall n \in \mathbb{N} ∀ n ∈ N ∀ x ∈ X \forall x \in X ∀ x ∈ X
SOLUTION
1) By the definition, the ( X , ∗ ) (X,*) ( X , ∗ )
a) ∀ ( a , b , c ∈ X ) : ( a ∗ b ) ∗ c = a ∗ ( b ∗ c ) \forall (a,b,c \in X): (a * b) * c = a * (b * c) ∀ ( a , b , c ∈ X ) : ( a ∗ b ) ∗ c = a ∗ ( b ∗ c )
b) ∃ e ∈ X ∀ a ∈ X : a ∗ e = e ∗ a = a − identity element \exists e \in X \ \forall a \in X: a * e = e * a = a - \text{identity element} ∃ e ∈ X ∀ a ∈ X : a ∗ e = e ∗ a = a − identity element
c) ∀ a ∈ X ∃ a − 1 ∈ X : a ∗ a − 1 = a − 1 ∗ a = e − inverse element \forall a \in X \ \exists a^{-1} \in X: a * a^{-1} = a^{-1} * a = e - \text{inverse element} ∀ a ∈ X ∃ a − 1 ∈ X : a ∗ a − 1 = a − 1 ∗ a = e − inverse element
We begin to check the conditions. We will do this in order:
a)
( x 1 ∗ x 2 ) ∗ x 3 = x 1 ∗ ( x 2 ∗ x 3 ) ( x 1 ∗ x 2 ) ∗ x 3 = ( x 1 + x 2 + x 1 × x 2 ) ∗ x 3 = = x 1 + x 2 + x 1 × x 2 + x 3 + ( x 1 + x 2 + x 1 × x 2 ) × x 3 = = x 1 + x 2 + x 1 × x 2 + x 3 + x 1 × x 3 + x 2 × x 3 + x 1 × x 2 × x 3 = = x 1 + x 2 + x 3 + x 1 × x 2 + x 1 × x 3 + x 2 × x 3 + x 1 × x 2 × x 3 \begin{array}{l}
(x_1 * x_2) * x_3 = x_1 * (x_2 * x_3) \\
(x_1 * x_2) * x_3 = (x_1 + x_2 + x_1 \times x_2) * x_3 = \\
= x_1 + x_2 + x_1 \times x_2 + x_3 + (x_1 + x_2 + x_1 \times x_2) \times x_3 = \\
= x_1 + x_2 + x_1 \times x_2 + x_3 + x_1 \times x_3 + x_2 \times x_3 + x_1 \times x_2 \times x_3 = \\
= x_1 + x_2 + x_3 + x_1 \times x_2 + x_1 \times x_3 + x_2 \times x_3 + x_1 \times x_2 \times x_3
\end{array} ( x 1 ∗ x 2 ) ∗ x 3 = x 1 ∗ ( x 2 ∗ x 3 ) ( x 1 ∗ x 2 ) ∗ x 3 = ( x 1 + x 2 + x 1 × x 2 ) ∗ x 3 = = x 1 + x 2 + x 1 × x 2 + x 3 + ( x 1 + x 2 + x 1 × x 2 ) × x 3 = = x 1 + x 2 + x 1 × x 2 + x 3 + x 1 × x 3 + x 2 × x 3 + x 1 × x 2 × x 3 = = x 1 + x 2 + x 3 + x 1 × x 2 + x 1 × x 3 + x 2 × x 3 + x 1 × x 2 × x 3
On the other hand
x 1 ∗ ( x 2 ∗ x 3 ) = x 1 ∗ ( x 2 + x 3 + x 2 × x 3 ) = = x 1 + x 2 + x 3 + x 2 × x 3 + x 1 × ( x 2 + x 3 + x 2 × x 3 ) = = x 1 + x 2 + x 3 + x 2 × x 3 + x 1 × x 2 + x 1 × x 3 + x 1 × x 2 × x 3 \begin{array}{l}
x_1 * (x_2 * x_3) = x_1 * (x_2 + x_3 + x_2 \times x_3) = \\
= x_1 + x_2 + x_3 + x_2 \times x_3 + x_1 \times (x_2 + x_3 + x_2 \times x_3) = \\
= x_1 + x_2 + x_3 + x_2 \times x_3 + x_1 \times x_2 + x_1 \times x_3 + x_1 \times x_2 \times x_3
\end{array} x 1 ∗ ( x 2 ∗ x 3 ) = x 1 ∗ ( x 2 + x 3 + x 2 × x 3 ) = = x 1 + x 2 + x 3 + x 2 × x 3 + x 1 × ( x 2 + x 3 + x 2 × x 3 ) = = x 1 + x 2 + x 3 + x 2 × x 3 + x 1 × x 2 + x 1 × x 3 + x 1 × x 2 × x 3
As we can see
( x 1 ∗ x 2 ) ∗ x 3 = x 1 + x 2 + x 3 + x 1 × x 2 + x 1 × x 3 + x 2 × x 3 + x 1 × x 2 × x 3 = = x 1 ∗ ( x 2 ∗ x 3 ) \begin{array}{l}
(x_1 * x_2) * x_3 = x_1 + x_2 + x_3 + x_1 \times x_2 + x_1 \times x_3 + x_2 \times x_3 + x_1 \times x_2 \times x_3 = \\
= x_1 * (x_2 * x_3)
\end{array} ( x 1 ∗ x 2 ) ∗ x 3 = x 1 + x 2 + x 3 + x 1 × x 2 + x 1 × x 3 + x 2 × x 3 + x 1 × x 2 × x 3 = = x 1 ∗ ( x 2 ∗ x 3 )
Conclusion, the first condition is satisfied.
b) ∃ e ∈ X ∀ a ∈ X : a ∗ e = e ∗ a = a \exists e \in X \ \forall a \in X: a * e = e * a = a ∃ e ∈ X ∀ a ∈ X : a ∗ e = e ∗ a = a
As an identity element, we choose e = 0 e = 0 e = 0
We check that the condition is satisfied with such a choice of an identity element
a ∗ 0 = a + 0 + a × 0 = a + 0 + 0 = a a * 0 = a + 0 + a \times 0 = a + 0 + 0 = a a ∗ 0 = a + 0 + a × 0 = a + 0 + 0 = a 0 ∗ a = 0 + a + 0 × a = 0 + a + 0 = a 0 * a = 0 + a + 0 \times a = 0 + a + 0 = a 0 ∗ a = 0 + a + 0 × a = 0 + a + 0 = a
As we can see,
a ∗ 0 = 0 ∗ a = a a * 0 = 0 * a = a a ∗ 0 = 0 ∗ a = a
What corresponds to the definition of an identity element.
Conclusion, the second condition is satisfied
c) ∀ a ∈ X ∃ a − 1 ∈ X : a ∗ a − 1 = a − 1 ∗ a = e \forall a \in X \ \exists a^{-1} \in X: a * a^{-1} = a^{-1} * a = e ∀ a ∈ X ∃ a − 1 ∈ X : a ∗ a − 1 = a − 1 ∗ a = e
Our task is to find the inverse element. For this we use the definition
a ∗ a − 1 = e a * a^{-1} = e a ∗ a − 1 = e
As was shown in b), the neutral element in this "group" is zero
Then,
a ∗ a − 1 = a + a − 1 + a × a − 1 = 0 ↔ a − 1 ( 1 + a ) = − a ↔ a − 1 = − a a + 1 a * a^{-1} = a + a^{-1} + a \times a^{-1} = 0 \leftrightarrow a^{-1}(1 + a) = -a \leftrightarrow a^{-1} = -\frac{a}{a + 1} a ∗ a − 1 = a + a − 1 + a × a − 1 = 0 ↔ a − 1 ( 1 + a ) = − a ↔ a − 1 = − a + 1 a a − 1 = − a a + 1 ∈ X a^{-1} = -\frac{a}{a + 1} \in X a − 1 = − a + 1 a ∈ X
It remains to verify the execution of its properties
a ∗ a − 1 = a ∗ ( − a a + 1 ) = a + ( − a a + 1 ) + a × ( − a a + 1 ) = a * a^{-1} = a * \left(-\frac{a}{a + 1}\right) = a + \left(-\frac{a}{a + 1}\right) + a \times \left(-\frac{a}{a + 1}\right) = a ∗ a − 1 = a ∗ ( − a + 1 a ) = a + ( − a + 1 a ) + a × ( − a + 1 a ) = = a − a a + 1 − a 2 a + 1 = a ( a + 1 ) a + 1 − a a + 1 − a 2 a + 1 = a 2 + a − a − a 2 a + 1 = 0 = a - \frac{a}{a + 1} - \frac{a^2}{a + 1} = \frac{a(a + 1)}{a + 1} - \frac{a}{a + 1} - \frac{a^2}{a + 1} = \frac{a^2 + a - a - a^2}{a + 1} = 0 = a − a + 1 a − a + 1 a 2 = a + 1 a ( a + 1 ) − a + 1 a − a + 1 a 2 = a + 1 a 2 + a − a − a 2 = 0 a − 1 ∗ a = ( − a a + 1 ) ∗ a = − a a + 1 + a + ( − a a + 1 ) × a = a^{-1} * a = \left(-\frac{a}{a + 1}\right) * a = -\frac{a}{a + 1} + a + \left(-\frac{a}{a + 1}\right) \times a = a − 1 ∗ a = ( − a + 1 a ) ∗ a = − a + 1 a + a + ( − a + 1 a ) × a = = − a a + 1 + a ( a + 1 ) a + 1 − a 2 a + 1 = − a + a 2 + a − a 2 a + 1 = 0 = -\frac{a}{a + 1} + \frac{a(a + 1)}{a + 1} - \frac{a^2}{a + 1} = \frac{-a + a^2 + a - a^2}{a + 1} = 0 = − a + 1 a + a + 1 a ( a + 1 ) − a + 1 a 2 = a + 1 − a + a 2 + a − a 2 = 0
Conclusion, the third condition is satisfied
ANSWER
( X , ∗ ) (X,*) ( X , ∗ )
2) Prove that x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ n times = ( 1 + x ) n − 1 \underbrace{x*x*x*\cdots*x}_{n \text{ times}} = (1 + x)^n - 1 n times x ∗ x ∗ x ∗ ⋯ ∗ x = ( 1 + x ) n − 1 ∀ n ∈ N \forall n \in \mathbb{N} ∀ n ∈ N ∀ x ∈ X \forall x \in X ∀ x ∈ X
This formula will be proved by the method of mathematical induction
1 step: Basis of induction
n = 1 n = 1 n = 1
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ n times = ( 1 + x ) n − 1 ↔ { ( 1 + x ) 1 − 1 = 1 + x − 1 = x x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ n times = x \underbrace{x * x * x * \cdots * x}_{n \text{ times}} = (1 + x)^n - 1 \leftrightarrow \left\{ \begin{array}{c} (1 + x)^1 - 1 = 1 + x - 1 = x \\ \underbrace{x * x * x * \cdots * x}_{n \text{ times}} = x \end{array} \right. n times x ∗ x ∗ x ∗ ⋯ ∗ x = ( 1 + x ) n − 1 ↔ ⎩ ⎨ ⎧ ( 1 + x ) 1 − 1 = 1 + x − 1 = x n times x ∗ x ∗ x ∗ ⋯ ∗ x = x
As we can see
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ 1 times = x = ( 1 + x ) 1 − 1 \underbrace{x * x * x * \cdots * x}_{1 \text{ times}} = x = (1 + x)^1 - 1 1 times x ∗ x ∗ x ∗ ⋯ ∗ x = x = ( 1 + x ) 1 − 1
Conclusion, the formula is true for n = 1 n = 1 n = 1
n = 2 n = 2 n = 2
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ n times = ( 1 + x ) n − 1 ↔ { ( 1 + x ) 2 − 1 x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ n times = x ∗ x \underbrace{x * x * x * \cdots * x}_{n \text{ times}} = (1 + x)^n - 1 \leftrightarrow \left\{ \begin{array}{c} (1 + x)^2 - 1 \\ \underbrace{x * x * x * \cdots * x}_{n \text{ times}} = x * x \end{array} \right. n times x ∗ x ∗ x ∗ ⋯ ∗ x = ( 1 + x ) n − 1 ↔ ⎩ ⎨ ⎧ ( 1 + x ) 2 − 1 n times x ∗ x ∗ x ∗ ⋯ ∗ x = x ∗ x ( 1 + x ) 2 − 1 = 1 + 2 x + x 2 − 1 = 2 x + x 2 (1 + x)^2 - 1 = 1 + 2x + x^2 - 1 = 2x + x^2 ( 1 + x ) 2 − 1 = 1 + 2 x + x 2 − 1 = 2 x + x 2 x ∗ x ⏟ b y the definition = x + x + x × x = 2 x + x 2 \underbrace{x * x}_{by \text{ the definition}} = x + x + x \times x = 2x + x^2 b y the definition x ∗ x = x + x + x × x = 2 x + x 2
As we can see
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ 2 times = 2 x + x 2 = ( 1 + x ) 2 − 1 \underbrace{x * x * x * \cdots * x}_{2 \text{ times}} = 2x + x^2 = (1 + x)^2 - 1 2 times x ∗ x ∗ x ∗ ⋯ ∗ x = 2 x + x 2 = ( 1 + x ) 2 − 1
Conclusion, the formula is true for n = 2 n = 2 n = 2
2 step: Induction hypothesis
Suppose that the formula is true for n = k n = k n = k ∀ k ∈ N \forall k \in \mathbb{N} ∀ k ∈ N
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ k times = ( 1 + x ) k − 1 \underbrace{x * x * x * \cdots * x}_{k \text{ times}} = (1 + x)^k - 1 k times x ∗ x ∗ x ∗ ⋯ ∗ x = ( 1 + x ) k − 1 3 step: Inductive transition
It is necessary to prove that the formula is true for n = k + 1 n = k + 1 n = k + 1 ∀ k ∈ N \forall k \in \mathbb{N} ∀ k ∈ N
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ k + 1 times = ( 1 + x ) k + 1 − 1 \underbrace{x * x * x * \cdots * x}_{k + 1 \text{ times}} = (1 + x)^{k + 1} - 1 k + 1 times x ∗ x ∗ x ∗ ⋯ ∗ x = ( 1 + x ) k + 1 − 1
using the inductive hypothesis
In our case,
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ k + 1 times = ( x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ k times ) ∗ x = ( ( 1 + x ) k − 1 ) ∗ x = = ( x 1 = ( 1 + x ) k − 1 x 2 = x ) = x 1 + x 2 + x 1 × x 2 = = ( 1 + x ) k − 1 + x + ( ( 1 + x ) k − 1 ) × x = ( 1 + x ) k − 1 + x + x × ( 1 + x ) k − x = = 1 × ( 1 + x ) k + x × ( 1 + x ) k − 1 + x − x = ( 1 + x ) k × ( 1 + x ) − 1 = = ( 1 + x ) k + 1 − 1 \begin{aligned}
\underbrace{x * x * x * \cdots * x}_{k + 1 \text{ times}} &= \left(\underbrace{x * x * x * \cdots * x}_{k \text{ times}}\right) * x = ((1 + x)^k - 1) * x = \\
&= \left(x_1 = (1 + x)^k - 1 \atop x_2 = x\right) = x_1 + x_2 + x_1 \times x_2 = \\
&= (1 + x)^k - 1 + x + ((1 + x)^k - 1) \times x = (1 + x)^k - 1 + x + x \times (1 + x)^k - x = \\
&= 1 \times (1 + x)^k + x \times (1 + x)^k - 1 + x - x = (1 + x)^k \times (1 + x) - 1 = \\
&= (1 + x)^{k + 1} - 1
\end{aligned} k + 1 times x ∗ x ∗ x ∗ ⋯ ∗ x = ( k times x ∗ x ∗ x ∗ ⋯ ∗ x ) ∗ x = (( 1 + x ) k − 1 ) ∗ x = = ( x 2 = x x 1 = ( 1 + x ) k − 1 ) = x 1 + x 2 + x 1 × x 2 = = ( 1 + x ) k − 1 + x + (( 1 + x ) k − 1 ) × x = ( 1 + x ) k − 1 + x + x × ( 1 + x ) k − x = = 1 × ( 1 + x ) k + x × ( 1 + x ) k − 1 + x − x = ( 1 + x ) k × ( 1 + x ) − 1 = = ( 1 + x ) k + 1 − 1
Conclusion,
x ∗ x ∗ x ∗ ⋯ ∗ x ⏟ k + 1 times = ( 1 + x ) k + 1 − 1 \underbrace{x * x * x * \cdots * x}_{k + 1 \text{ times}} = (1 + x)^{k + 1} - 1 k + 1 times x ∗ x ∗ x ∗ ⋯ ∗ x = ( 1 + x ) k + 1 − 1
Q.E.D
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#339914 on Dec 2023
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