Question

Prove the following statements. These exercises are cumulative, covering all techniques addressed in Chapters \(4-7\). Suppose \(x, y \in \mathbb{R} .\) Then \(x^{3}+x^{2} y=y^{2}+x y\) if and only if \(y=x^{2}\) or \(y=-x\).

Short answer

The equation \(x^3+x^2y=y^2+xy\) is true if and only if \(y=x^2\) or \(y=-x\).

Step-by-step solution

Part 1: If \(y=x^2\) or \(y=-x\), then \(x^3+x^2y=y^2+xy\)

Assume \(y = x^2\). Substitute \(y\) into the equation and find that \(x^3 + x^2(x^2) = (x^2)^2 + x(x^2) => x^3 + x^4 = x^4 + x^3\), thus the equation holds for \(y = x^2\). Now, assume \(y = -x\). Substitute \(y\) into the equation and find that \(x^3 + x^2(-x) = (-x)^2 + x(-x) => x^3 - x^2 = x^2 - x^2\), thus the equation also holds for \(y = -x\). Therefore, if \(y = x^2\) or \(y = -x\), then \(x^3+x^2y=y^2+xy\) holds true.

Part 2: If \(x^3+x^2y=y^2+xy\) then \(y=x^2\) or \(y=-x\)

Assume \(x^3+x^2y=y^2+xy\). Rearrange to obtain \(x^3 - xy + x^2y - y^2 = 0\), which can be written as \(x^2(x - y) + y(x - y) = 0\), factor out \(x-y\) to get \((x - y)(x^2 + y) = 0\), according to the zero product property, one of the factors must equal zero, therefore \(x = y\) or \(x^2 = -y\). Since \(x, y\) are real numbers, \(x^2\) is non-negative for any real number \(x\), thus it can't be equal to \(-y\) unless both \(x\) and \(y\) are zero. Therefore, the solution is \(y = x\), because there is only one left that is equivalent to \(y = x^2 = -x\). So, if \(x^3+x^2y=y^2+xy\) is true, then \(y=x^2\) or \(y=-x\).

Conclusion

The equation \(x^3+x^2y=y^2+xy\) is true if and only if \(y=x^2\) or \(y=-x\). Therefore, whenever we see the equation \(x^3 + x^2y = y^2 + xy\), one of \(y=x^2\) or \(y=-x\) can be concluded, and vice versa.

Key concepts

Proof by Substitution

When we talk about proof by substitution, we are referring to a method used in mathematics to show that two expressions are equal by substituting one variable with another. It’s like swapping out a piece of a puzzle to see if the picture still looks the same. This technique comes in handy when dealing with equalities and functions.

In the original exercise, we see this technique at play when assuming values for the variable \(y\) based on the given condition. For \(y = x^2\) and \(y = -x\), the value of \(y\) is substituted into the original equation to show that it holds true. Just like confirming a hunch, if the equation balances out, our assumption is correct!

Factorization in Algebra

Now, lets delve into the concept of factorization in algebra. This method involves breaking down a complex expression into simpler elements that, when multiplied together, give back the original expression. Think of it as deconstructing a building into individual blocks that are easier to handle.

Factorization helps in solving equations by simplifying expressions and exposing potential solutions which are the roots of the equation. In our problem, the equation is rearranged and factored into \((x - y)(x^2 + y) = 0\) to reveal the possible values of \(y\) that make the equation true, utilizing the zero product property—our next star concept!

Zero Product Property

The zero product property is a fundamental principle to understand when working with equations. It states that if the product of two numbers is zero, then at least one of the numbers must be zero. It's a lot like saying, if there's no cake at a birthday party, then either the cake was never made or it disappeared.

In particular, our exercise used this property to conclude that either \(x - y\) or \(x^2 + y\) must be zero after the equation is factored. Using this property makes it easier to pinpoint the roots of the equation and enables us to conclude that \(y = x^2\) or \(y=-x\), wrapping up our proof tightly.

Mathematical Logic

At its heart, mathematical logic is the backbone of all proof techniques in mathematics. It's the structure of reasoning that allows us to navigate from assumptions to conclusions in a formal and valid way. Like a GPS for math problems, it guides us to the correct destination.

Our exercise is underpinned by logical reasoning every step of the way. From assuming a condition to be true, like \(y = x^2\) or \(y = -x\), to showing that it satisfies the given equation, and reversing the process to show necessity. The logical flow makes sure the 'if and only if' condition is verified for all possible scenarios, establishing the proof on a solid ground of mathematical logic.