Sure, here is a step-by-step proof:

Step 1: Let's assume that x is a nonzero real number such that both x^5 and 20x + 19/x are rational numbers.

Step 2: We know that the sum, difference, product, and quotient of rational numbers are always rational. Therefore, if x^5 is rational and 20x + 19/x is rational, then their difference must also be rational.

Step 3: Let's calculate this difference: x^5 - (20x + 19/x) = x^5 - 20x - 19/x.

Step 4: We can rewrite this as: x^5 - 20x^2 + 19/x = x^5 - 20x^2 + 19x^-1.

Step 5: Now, we can see that this is a polynomial in x. The coefficients of the polynomial are rational numbers (1, -20, and 19), and the exponents are integers (5, 2, and -1).

Step 6: We know that a polynomial with rational coefficients evaluated at a rational number is always rational. Therefore, if x^5 - 20x^2 + 19x^-1 is rational, then x must be rational.

Step 7: Therefore, we have proved that if x is a nonzero real number such that both x^5 and 20x + 19/x are rational numbers, then x is a rational number.

Question

Sure, here is a step-by-step proof:

Step 1: Let's assume that x is a nonzero real number such that both x^5 and 20x + 19/x are rational numbers.

Step 2: We know that the sum, difference, product, and quotient of rational numbers are always rational. Therefore, if x^5 is rational and 20x + 19/x is rational, then their difference must also be rational.

Step 3: Let's calculate this difference: x^5 - (20x + 19/x) = x^5 - 20x - 19/x.

Step 4: We can rewrite this as: x^5 - 20x^2 + 19/x = x^5 - 20x^2 + 19x^-1.

Step 5: Now, we can see that this is a polynomial in x. The coefficients of the polynomial are rational numbers (1, -20, and 19), and the exponents are integers (5, 2, and -1).

Step 6: We know that a polynomial with rational coefficients evaluated at a rational number is always rational. Therefore, if x^5 - 20x^2 + 19x^-1 is rational, then x must be rational.

Step 7: Therefore, we have proved that if x is a nonzero real number such that both x^5 and 20x + 19/x are rational numbers, then x is a rational number.

Knowee AI · Accepted Answer

Sure, here is a step-by-step proof:

Step 1: Let's assume that x is a nonzero real number such that both x^5 and 20x + 19/x are rational numbers.

Step 2: We know that the sum, difference, product, and quotient of rational numbers are always rational. Therefore, if x^5 is rational and 20x + 19/x is rational, then their difference must also be rational.

Step 3: Let's calculate this difference: x^5 - (20x + 19/x) = x^5 - 20x - 19/x.

Step 4: We can rewrite this as: x^5 - 20x^2 + 19/x = x^5 - 20x^2 + 19x^-1.

Step 5: Now, we can see that this is a polynomial in x. The coefficients of the polynomial are rational numbers (1, -20, and 19), and the exponents are integers (5, 2, and -1).

Step 6: We know that a polynomial with rational coefficients evaluated at a rational number is always rational. Therefore, if x^5 - 20x^2 + 19x^-1 is rational, then x must be rational.

Step 7: Therefore, we have proved that if x is a nonzero real number such that both x^5 and 20x + 19/x are rational numbers, then x is a rational number.

1. Suppose x is a nonzero real number such that both x^5 and 20x +19/x are rational numbers. Prove that x is a rational number.

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