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Chapter 17: Problem 2

Prove In any ring, if \(a b=-b a\), then \((a+b)^{2}=(a-b)^{2}=a^{2}+b^{2}\).

Short Answer

Expert verified
\((a+b)^2 = (a-b)^2 = a^2 + b^2\) when \(ab = -ba\) in a ring.

Step by step solution

01

Understanding the given condition

We are given that in a ring, for elements \(a\) and \(b\), the equation \(ab = -ba\) holds. This means \(ab + ba = 0\). We need to prove two equations: \((a+b)^2 = a^2 + b^2\) and \((a-b)^2 = a^2 + b^2\).
02

Expanding \((a+b)^2\)

Start by expanding \((a+b)^2\) using the distributive property:\((a+b)^2 = (a+b)(a+b) = a^2 + ab + ba + b^2.\)Given the condition \(ab = -ba\), we substitute \(ba = -ab\):\(a^2 + ab + (-ab) + b^2 = a^2 + 0 + b^2 = a^2 + b^2.\)
03

Expanding \((a-b)^2\)

Now expand \((a-b)^2\) using the distributive property:\((a-b)^2 = (a-b)(a-b) = a^2 - ab - ba + b^2.\)Substitute \(ba = -ab\) once more:\(a^2 - ab - (-ab) + b^2 = a^2 - ab + ab + b^2 = a^2 + 0 + b^2 = a^2 + b^2.\)
04

Conclusion of the proof

For both expansions, \((a+b)^2\) and \((a-b)^2\), we reached the conclusion that they both simplify to \(a^2 + b^2\) using the fact that \(ab + ba = 0\). Thus, the statement is proven.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Distributive Property
The distributive property is a fundamental rule in algebra and ring theory. It states that for any elements in a ring, the product of an element with the sum or difference of two other elements can be distributed:
  • Distributing multiplication over addition: \( a(b + c) = ab + ac \)
  • Distributing multiplication over subtraction: \( a(b - c) = ab - ac \)
In our exercise, when we expanded \((a+b)^2\) and \((a-b)^2\), we used the distributive property to rewrite the expressions:
  • \((a+b)(a+b) = a^2 + ab + ba + b^2\)
  • \((a-b)(a-b) = a^2 - ab - ba + b^2\)
This property shows how multiplication interacts with addition or subtraction, allowing us to break down complex expressions into simpler parts. The simplification largely relies on understanding these interactions.
Expansion of Squares
The expansion of squares is a technique used to simplify expressions where a binomial is squared. For both \((a+b)^2\) and \((a-b)^2\),expanding the squares involves applying the distributive property to multiply each term in the binomial by itself or its components:
  • For \((a+b)^2\), you multiply: \((a+b)(a+b) = a^2 + ab + ba + b^2\)
  • For \((a-b)^2\), you calculate: \((a-b)(a-b) = a^2 - ab - ba + b^2\)
By substituting \(ba = -ab\) (due to the given condition), the middle terms cancel each other out!Thus, both expansions simplify to \(a^2 + b^2\), proving the desired equality for the exercise.
Commutative Ring
A commutative ring is a type of algebraic structure where addition and multiplication operations behave according to specific rules.
  • Addition is commutative: \(a + b = b + a\)
  • Multiplication is commutative: \(a b = b a\) (This property doesn't hold in our condition, where \(ab = -ba\))
  • Multiplication distributes over addition: \(a(b+c) = ab + ac\)
Although our exercise is based in the context of a ring (possibly a non-commutative one), knowing the rules of a commutative ring helps understand why \(ab + ba = 0\)leads to particular simplifications, allowing us to equate different forms of \((a + b)^2\) and \((a - b)^2\).
Algebraic Proof
An algebraic proof is a logical argument, using algebra rules, to demonstrate the truth of a statement. In this exercise, we needed to prove two identities involving the elements of a ring. The steps involved were:
  • Recognizing the given condition: \(ab = -ba\) This implies that \(ab + ba = 0\).
  • Using this condition, we showed both expansions \((a+b)^2\) and \((a-b)^2\) lead to \(a^2 + b^2\).
Each step was checked for logical consistency, using properties of rings like the distributive law, until the desired results were clearly derived. An algebraic proof like this hinges on simplifying expressions through known rules, ensuring that the mathematical relationships hold at every transformation step.

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Most popular questions from this chapter

By part 5, the set of all the nonzero elements in a field is a multiplicative group. Now use Lagrange's theorem to prove that in a finite field with \(m\) elements, \(x^{m-1}=1\) for every \(x \neq 0\)

The set \(S\) of all the invertible elements in a ring is a multiplicative group.

In each of the following, a set \(A\) with operations of addition and multiplication is given. Prove that \(A\) satisfies all the axioms to be a commutative ring with unity. Indicate the zero element, the unity, and the negative of an arbitrary \(a\). \(A=\\{x+y \sqrt{2}: x, y \in \mathbb{Z}\\}\) with conventional addition and multiplication.

Prove that each of the following is true in a nontrivial ring with unity. If \(a \neq \pm 1\) and \(a^{2}=1\), then \(a+1\) and \(a-1\) are divisors of zero.

Prove that each of the following is true in a nontrivial ring with unity. Suppose \(a\) is neither 0 nor a divisor of zero. If \(a b=a c\), then \(b=c\).
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