3. Prove that
![\sqrt[3]{6} \: \: is \: irrational.](https://tex.z-dn.net/?f=+%5Csqrt%5B3%5D%7B6%7D++%5C%3A++%5C%3A+is+%5C%3A+irrational. "\sqrt[3]{6} \: \: is \: irrational.")
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Hey !
Solution :
Assumption : Let ∛ 6 be a rational number.
Fact : √ 6 is an irrational number
Proof : Rational number can be expressed in the form of p / q where p,q are co - prime and have a HCF of 1.
=> ∛ 6 = p / q
=> √ 6 = p / 3 × q
=> √ 6 = p / 3q ---( Equation 1 )
We know that ( p / 3q ) is a rational number.
But Equation 1 is contradicting the fact that √ 6 is irrational.
Hence our assumption was wrong.
Hence ∛ 6 is irrational.
Hope it helped :-)
Solution :
Assumption : Let ∛ 6 be a rational number.
Fact : √ 6 is an irrational number
Proof : Rational number can be expressed in the form of p / q where p,q are co - prime and have a HCF of 1.
=> ∛ 6 = p / q
=> √ 6 = p / 3 × q
=> √ 6 = p / 3q ---( Equation 1 )
We know that ( p / 3q ) is a rational number.
But Equation 1 is contradicting the fact that √ 6 is irrational.
Hence our assumption was wrong.
Hence ∛ 6 is irrational.
Hope it helped :-)
Steph0303:
:-)
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