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 =
, where <em>b</em> ≠ 0. Which by isolation of π gives π =
.
The right side of the equation is <u>irrational</u>, because the left side of the
equation is <u>irrational</u>, this is a contradiction.
Therefore, the assumption is wrong and the number 4·π is<u> irrational</u>.
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}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%204%20%5Ccdot%20%5Cpi%20%3D%20%5Cmathbf%7B%5Cfrac%7Ba%7D%7Bb%7D%7D)
Where;
b ≠ 0
Making π the subject of the above formula, by isolating, it, we have;
![\displaystyle \pi = \mathbf{\frac{a}{4 \cdot b}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%20%5Cpi%20%3D%20%5Cmathbf%7B%5Cfrac%7Ba%7D%7B4%20%5Ccdot%20b%7D%7D)
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