Question

prove that cube root 7 is an irrational number​

Answer 1

Answer:

Proof: Suppose that the cube root of 7 is rational. Then there are integers p and q such that p^3/q^3 = 7. ...

Now p and q have a common factor (7),

which contradicts that p and q have no common factors.

Therefore, the cube root of 7 is irrational.

Answer 2

∛7 is an irrational number​

Step-by-step explanation:

To prove: Prove that cube root of 7 is an IRRATIONAL number

proof:

Assume that cube rt 7 is rational.

Then (7)^(1/3) = a/b where a and b are integers and a/b is reduced to lowest terms.

Then $a=b[7^\frac{1}{3} ]$

Since a is a multiple of b and a is an integer, b divides a.

Since b divides a, a = nb and n is an integer.

Therefore $7^\frac{1}{3} = \frac{a}{b}$ $=\frac{nb}{b}$, so a/b is not reduced to lowest terms.

This led to contradiction

The assumption that $7^\frac{1}{3}$ was rational.

The assumption must be wrong.

Therefore ∛7 if irrational.

#Learn more:

Prove that root 7 is an irrational number and hense show that 3-2root 7 is irrational​

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