Select from the drop-down menus to correctly complete the proof.

To prove that 2√⋅7 is irrational, assume the product is rational and set it equal to a/b, where b is not equal to 0. Isolating the radical gives 2√ = a/7b. The right side of the equation is (irrational, rational). Because the left side of the equation is (irrational, rational), this is a contradiction. Therefore, the assumption is wrong, and the product is (rational, irrational).

Question

Mathematics

Sheldon Pennington

2 October, 16:24

Select from the drop-down menus to correctly complete the proof.

To prove that 2√⋅7 is irrational, assume the product is rational and set it equal to a/b, where b is not equal to 0. Isolating the radical gives 2√ = a/7b. The right side of the equation is (irrational, rational). Because the left side of the equation is (irrational, rational), this is a contradiction. Therefore, the assumption is wrong, and the product is (rational, irrational).

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Answers (1)

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Keira Hickman · Answer 1

We need to prove 2*√7 is an irrational number.

Steps:

1) To prove that 2√7 is an irrational, assume the product is rational and set it equal to a/b, where b is not equal to 0.

2) Isolating the radical gives 2 = a/√7b.

3) The right side of the equation is irrational.

4) Because the left side of the equation is rational, this is a contradiction.

5) Therefore, the assumption is wrong, and the product is irrational.