In mathematics, particularly in number theory and algebra, the concept of the conjugate product often applies to expressions involving square roots or irrational numbers. To explain why the conjugate product of terms that include a square always results in a rational number, we can look at the properties of conjugates and the way they interact with squares.
Conjugate Products
Let's consider two numbers \( a \) and \( b \), where \( a \) includes a square root. For example, let:
\[ a = \sqrt{x} \quad \text{and} \quad b = \sqrt{y} \]
The conjugate of a term involving a square root often involves changing the sign in front of one of the terms. For example, the conjugate of \( a + b \) is \( a - b \). When we multiply \( a + b \) by its conjugate \( a - b \), we use the difference of squares:
\[ (a + b)(a - b) = a^2 - b^2 \]
Result in Rational Numbers
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Squares of the Terms: The terms in the product \( a^2 \) and \( b^2 \) become: \[ a^2 = (\sqrt{x})^2 = x \quad \text{and} \quad b^2 = (\sqrt{y})^2 = y \] Here, \( x \) and \( y \) are presumably rational numbers (though they can also be integers, etc.).
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Difference of Squares: The product leads us to: \[ a^2 - b^2 = x - y \] Since both \( x \) and \( y \) are rational (assuming they are positive rational numbers), their difference \( x - y \) is also rational.
Conclusion
Thus, when we take the conjugate product of terms that include square roots, the resulting expression involves squares of those roots, which yields rational results under normal assumptions about \( x \) and \( y \). Consequently, the conjugate product results in a rational number:
\[ (a+b)(a-b) = a^2 - b^2 \text{ is rational if } a^2, b^2 \text{ are rational.} \]
This is why the conjugate product of terms that include a square always results in a rational number.
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