Show That 6+√2 Is Irrational

Video Explanation

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Question: Show that the number 6+√2 is irrational.

Assume that 6+√2 is rational. Then it can be written as a fraction of two integers in lowest terms:

6+√2 = p/q, where p and q are integers with no common factors and q ≠ 0.

Rearrange the equation to isolate √2:

√2 = (p/q) − 6

Now square both sides:

2 = [(p/q) − 6]²

Expand the right side:

2 = (p²/q²) − (12p/q) + 36

Multiply both sides by q²:

2q² = p² − 12pq + 36q²

0 = p² − 12pq + 36q² − 2q²

0 = p² − 12pq + 34q²

The equation p² − 12pq + 34q² = 0 has no solutions in integers p and q (because the discriminant becomes negative).

Discriminant, D = (−12q)² − 4(34q²) = 144q² − 136q² = 8q²

Since 8q² is not a perfect square (unless q=0 which is not possible), this is a contradiction.

∴ The number 6+√2 cannot be expressed as a ratio of two integers in lowest terms, so 6+√2 is irrational.

By assuming 6+√2 is rational and reaching a contradiction, we conclude that 6+√2 is irrational.

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