Question

Select from the drop-down menus to correctly complete the proof. To prove that 4π is irrational, assume the product is rational and set it equal to ​ab​, where b is not equal to 0: 4π=ab. Isolating π gives π=a4b. The right side of the equation is Choose.(rational or Irrational).. . Because the left side of the equation is Choose.(rational or irrational).. , this is a contradiction. Therefore, the assumption is wrong, and the number 4π is Choose(rational or Irrational)... .

Answer 1

The product of a rational number and an irrational number is an irrational

number.

The correctly completed proof is presented as follows;

To prove that 4·π is irrational, with the assumption that, 4·π = $\displaystyle \frac{a}{b}$, where b ≠ 0. Which by isolation of π gives π = $\displaystyle \mathbf{ \frac{a}{4 \cdot b}}$.

The right side of the equation is irrational, because the left side of the

equation is irrational, this is a contradiction.

Therefore, the assumption is wrong and the number 4·π is irrational.

Reasons:

The proof is presented as follows;

With the assumption that 4·π is rational, we have;

$\displaystyle 4 \cdot \pi = \mathbf{\frac{a}{b}}$

Where;

b ≠ 0

Making π the subject of the above formula, by isolating, it, we have;

$\displaystyle \pi = \mathbf{\frac{a}{4 \cdot b}}$

The right side of the equation is rational, however, the left side of the

equation π is irrational, which results in a contradiction, therefore, the

product 4·π is an irrational number.

Learn more about irrational numbers here:

brainly.com/question/1625874

Answer 2

Answer:

1. irrational
2. irrational
3. irrational