Question

prove that 3 root 7 is an irrational

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Answer 1

Answer:

We prove that 3√7 is not rational no. . i.e., it is an irrational no. ⇒ 3√7 is not rational no. . i.e., 3√7 is an irrational no.

Step-by-step explanation:

Answer 2

Step-by-step explanation:

let's say 3✓7 is a rational number

therefore we can write 3✓7= a/b where b≠0

squaring both sides

we have 9×7 = a²/b²

63b² = a² eq(1)

which means 63 is divisible by a²

so let's consider a = 63 p , for some integer p

Now adding the value of a in eq (1) we get

63b² = (63p)²

63b² = 63p×63p

b² = 63p² ( after cancelling 63)

therefore we come to a conclusion that b² is also divisible by 63 and vice versa.

This proves that a and b has at least one common factor i.e 63.

but we are unable to verify if a and b are co prime by contradiction. Therefore our assumption that 3✓7 is a rational is wrong.

Thus, 3✓7 is irrational