Question

given that ✓5 is irrational prove that 2 under root 5 minus 3 is an irrational number​

Answer 1

Answer:

Step-by-step explanation:

$Give \:that \sqrt5\: is\: irrational.\\Prove that 2\sqrt5-3\: is\: an \:irrational\: number$

$Let \:us\: assume\: that\: 2\sqrt5-3 is\: a\: rational\: number\\i.e. 2\sqrt5-3 = p/q (where \:p \:and\: q \:are\: integers )\\Now,\\2\sqrt5-3 = p/q2\sqrt5 = p/q +32\sqrt5 = (p+3q)/q\sqrt5 = (p+3q)/2q$

We can see that LHS is an irrational number which can never be equal to RHS which is a rational number. Therefore, our assumption that 2√5-3 is a rational number is wrong. 2√5-3 is not a rational number, it is an irrational number.

Hence proved.

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Answer 2

Answer:

Since p and q are integers ∴p+3q2q is a rational number. ∴√5 is a rational number which is a contradiction as √5 is an irrational number . Hence our assumption is wrong and hence 2√5-3 is an irrational number.

Step-by-step explanation:

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