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Prove that root 2 + 3/ root 2 is irrational

User Mxk
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2 Answers

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Answer: Let (√2 + 3)/√2 be a rational number which is equal to x.

Step-by-step explanation: Thus √2 is in p/q form and hence a rational number. This leads to contradiction for a fact that √2 is actually an irational number. That's why (√2 + 3)/√2 is an irractional.

User Omabena
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Let's assume that √2 + 3/√2 is a rational number. That means it can be expressed as a ratio of two integers a and b, where b is not equal to zero, and a and b have no common factors other than 1.

√2 + 3/√2 = a/b

Multiplying both sides by b√2 gives:

(√2 + 3/√2) * b√2 = a

Simplifying the left side of the equation:

b√2 * √2 + 3b = a

2b + 3b√2 = a

Rearranging the equation:

3b√2 = a - 2b

Squaring both sides:

18b^2 = a^2 - 4ab + 4b^2

a^2 - 4ab + 2b^2 = 18b^2

a^2 - 4ab - 16b^2 = 0

The left side of the equation is a quadratic expression in terms of a, and it can be solved using the quadratic formula:

a = [4b ± √(16b^2 + 64b^2)]/2

a = [4b ± 8b√2]/2

a = 2b ± 4b√2

We know that a and b have no common factors other than 1. However, if a is even, then 2 is a common factor of a and 2b, which contradicts our assumption. Therefore, a must be odd.

If we substitute a = 2n + 1, where n is an integer, into the equation a^2 - 4ab - 16b^2 = 0, we get:

(2n + 1)^2 - 4(2n + 1)b - 16b^2 = 0

4n^2 + 4n + 1 - 8nb - 4b - 16b^2 = 0

4n^2 + (4 - 8b)n + (1 - 4b - 16b^2) = 0

The discriminant of this quadratic equation is:

(4 - 8b)^2 - 4(4)(1 - 4b - 16b^2) = 32b^
User Niro
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