prove that √5−3√2 is irrational
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If possible, let us consider (√5 - 3√2) is rational.
Then, √5 - 3√2 = a/b, where a and b are integers with non-zero b.
⇒ 3√2 = √5 - a/b
⇒ (3√2)² = (√5 - a/b)²,
squaring both sides
⇒ 18 = 5 - 2√5 a/b + a²/b²,
using the identity (a - b)² = a² - 2ab + b²
⇒ 2√5 a/b = a²/b² + 5 - 18
⇒ 2√5 a/b = a²/b² - 13
⇒ √5 = (a²/b² - 13) × (b/2a)
⇒ √5 = (a/2b - 13b/2a) ...(i)
Since, a/b is rational as considered, b/a is also rational and thus the R.H.S. of (i) is also rational which leads to a contradiction to the fact that √5 is irrational.
∴ (√5 - 3√2) is irrational. [Proved]
#MarkAsBrainliest
If possible, let us consider (√5 - 3√2) is rational.
Then, √5 - 3√2 = a/b, where a and b are integers with non-zero b.
⇒ 3√2 = √5 - a/b
⇒ (3√2)² = (√5 - a/b)²,
squaring both sides
⇒ 18 = 5 - 2√5 a/b + a²/b²,
using the identity (a - b)² = a² - 2ab + b²
⇒ 2√5 a/b = a²/b² + 5 - 18
⇒ 2√5 a/b = a²/b² - 13
⇒ √5 = (a²/b² - 13) × (b/2a)
⇒ √5 = (a/2b - 13b/2a) ...(i)
Since, a/b is rational as considered, b/a is also rational and thus the R.H.S. of (i) is also rational which leads to a contradiction to the fact that √5 is irrational.
∴ (√5 - 3√2) is irrational. [Proved]
#MarkAsBrainliest
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