Real Numbers, Exponent Rules, and Algebraic Simplification

Classification of Real Numbers

  • Definitions and Criteria:

    • A Rational Number is any number that can be expressed as a ratio or fraction ab\frac{a}{b} of two integers, where aa is the numerator, bb is the denominator, and b≠0b \neq 0. Rational numbers include all integers, finite (terminating) decimals, and repeating decimals.
    • An Irrational Number is any real number that cannot be expressed as a simple fraction of two integers. Their decimal representations are non-terminating and non-repeating.

  • Analysis of Numbers:

    • 148148: Rational. It is a whole integer that can be written as 1481\frac{148}{1}.
    • −17.8203-17.8203: Rational. It is a negative terminating decimal, equivalent to −17820310000\frac{-178203}{10000}.
    • 8\sqrt{8}: Irrational. Because 88 is not a perfect square, 8=22≈2.82842717...\sqrt{8} = 2\sqrt{2} \approx 2.82842717..., which is non-terminating and non-repeating.
    • 64\sqrt{64}: Rational. The square root simplifies directly to the integer 88 (since 8×8=648 \times 8 = 64), which can be written as 81\frac{8}{1}.
    • 38\frac{3}{8}: Rational. It is explicitly written as a ratio of two integers, 33 and 88, with a non-zero denominator.

Evaluation of Logical Statements on Number Systems

  • Statement A: "The product of 6⋅6\sqrt{6} \cdot \sqrt{6} is an irrational number."

    • Truth Value: False.
    • Proof / Explanation: By radical rules, 6⋅6=36=6\sqrt{6} \cdot \sqrt{6} = \sqrt{36} = 6. Since 66 is an integer (61\frac{6}{1}), the product is rational, not irrational.

  • Statement B: "402\frac{40}{2} is an example of an integer."

    • Truth Value: True.
    • Proof / Explanation: Simplifying the quotient gives 402=20\frac{40}{2} = 20. The value 2020 is a whole number with no fractional or decimal component, placing it in the set of integers Z\mathbb{Z}.

  • Statement C: "A natural number is also a rational number."

    • Truth Value: True.
    • Proof / Explanation: The natural numbers N={1,2,3,4,… }\mathbb{N} = \{1, 2, 3, 4, \dots\} are a subset of the rational numbers Q\mathbb{Q}. Any natural number nn can be written as n1\frac{n}{1}.

  • Statement D: "An irrational number is any number that can be written as a simple ratio."

    • Truth Value: False.
    • Proof / Explanation: This statement defines a rational number. Irrational numbers are defined as numbers that cannot be written as a simple ratio of two integers.

  • Statement E: "If xx is a rational number, then 3x3x is a rational number."

    • Truth Value: True.
    • Proof / Explanation: The set of rational numbers is closed under multiplication. Since 33 is rational (31\frac{3}{1}) and xx is rational (ab\frac{a}{b}), their product 3x=3ab3x = \frac{3a}{b} is guaranteed to be a rational number.

Algebraic Simplification Using Rules of Exponents

  • Fundamental Laws of Exponents:

    • Product Rule: am⋅an=am+na^m \cdot a^n = a^{m+n}
    • Quotient Rule: aman=am−n\frac{a^m}{a^n} = a^{m-n}
    • Power of a Power Rule: (am)n=am⋅n(a^m)^n = a^{m \cdot n}
    • Power of a Product Rule: (ab)n=anbn(ab)^n = a^n b^n
    • Negative Exponent Rule: a−n=1ana^{-n} = \frac{1}{a^n}

  • Problem 1: Simplify −14x5⋅3x−2-14x^5 \cdot 3x^{-2}

    • Step 1: Multiply numerical coefficients:     −14×3=−42-14 \times 3 = -42
    • Step 2: Apply product rule to variable xx:     x5⋅x−2=x5+(−2)=x3x^5 \cdot x^{-2} = x^{5 + (-2)} = x^3
    • Final Simplified Result: −42x3-42x^3

  • Problem 2: Simplify 42h228h11\frac{42h^2}{28h^{11}}

    • Step 1: Reduce numerical coefficients by finding greatest common factor (14):     4228=3×142×14=32\frac{42}{28} = \frac{3 \times 14}{2 \times 14} = \frac{3}{2}
    • Step 2: Apply quotient rule to variable hh:     h2h11=h2−11=h−9=1h9\frac{h^2}{h^{11}} = h^{2 - 11} = h^{-9} = \frac{1}{h^9}
    • Step 3: Combine coefficient and variable results:     32h9\frac{3}{2h^9}

  • Problem 3: Simplify (−8x2y3)3(-8x^2 y^3)^3

    • Step 1: Distribute outer power to numerical coefficient and each variable term:     (−8)3⋅(x2)3⋅(y3)3(-8)^3 \cdot (x^2)^3 \cdot (y^3)^3
    • Step 2: Evaluate coefficient power:     (−8)3=−8×−8×−8=−512(-8)^3 = -8 \times -8 \times -8 = -512
    • Step 3: Apply power of a power rule to exponent powers:     (x2)3=x2×3=x6(x^2)^3 = x^{2 \times 3} = x^6(y3)3=y3×3=y9(y^3)^3 = y^{3 \times 3} = y^9
    • Final Simplified Result: −512x6y9-512x^6 y^9

Advanced Parameter Equivalence in Exponent Equations

  • Problem Statement: Determine the exact values of parameters AA, BB, and CC to satisfy the equation:   (3xBy−6)2⋅Axy3=−36x5yC(3x^B y^{-6})^2 \cdot A x y^3 = \frac{-36 x^5}{y^C}

  • Step-by-Step Algebraic Solution:

    • Step 1: Expand LHS expression (3xBy−6)2(3x^B y^{-6})^232⋅(xB)2⋅(y−6)2=9x2By−123^2 \cdot (x^B)^2 \cdot (y^{-6})^2 = 9 x^{2B} y^{-12}
    • Step 2: Multiply by Ax1y3\mathbf{A x^1 y^3}(9⋅A)⋅(x2B⋅x1)⋅(y−12⋅y3)=(9A)x2B+1y−9(9 \cdot A) \cdot (x^{2B} \cdot x^1) \cdot (y^{-12} \cdot y^3) = (9A) x^{2B + 1} y^{-9}
    • Step 3: Express RHS with negative exponent for yy−36x5yC=−36x5y−C\frac{-36 x^5}{y^C} = -36 x^5 y^{-C}
    • Step 4: Set LHS equal to RHS and equate corresponding components:     (9A)x2B+1y−9=−36x5y−C(9A) x^{2B + 1} y^{-9} = -36 x^5 y^{-C}

  • Solving for Parameters:

    • Coefficient comparison:     9A=−36  ⟹  A=−369=−49A = -36 \implies A = \frac{-36}{9} = -4
    • Exponents of xx comparison:     2B+1=5  ⟹  2B=4  ⟹  B=22B + 1 = 5 \implies 2B = 4 \implies B = 2
    • Exponents of yy comparison:     −9=−C  ⟹  C=9-9 = -C \implies C = 9

  • Final Values:

    • A=−4A = -4
    • B=2B = 2
    • C=9C = 9