Question

Prove that 6-√2 is an irrational number​

Answer 1

Answer:

Answer:

Given 6 + √2

To prove: 6 + √2 is an irrational number.

Proof:

Let us assume that 6 + √2 is a rational number.

So it can be written in the form a/b

6 + √2 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving

6 + √2 = a/b

we get,

=> √2 = a/b – 6

=> √2 = (a-6b)/b

=> √2 = (a-6b)/b

This shows (a-6b)/b is a rational number.

But we know that √2 is an irrational number, it is contradictsour to our assumption.

Our assumption 6 + √2 is a rational number is incorrect.

6 + √2 is an irrational number

Hence, proved.

Answer 2

Step-by-step explanation:

Answer:

Given 6 + √2

To prove: 6 + √2 is an irrational number.

Proof:

Let us assume that 6 + √2 is a rational number.

So it can be written in the form a/b

6 + √2 = a/b

Here a and b are coprime numbers and b ≠ 0

Solving

6 + √2 = a/b

we get,

=> √2 = a/b – 6

=> √2 = (a-6b)/b

=> √2 = (a-6b)/b

This shows (a-6b)/b is a rational number.

But we know that √2 is an irrational number, it is contradictsour to our assumption.

Our assumption 6 + √2 is a rational number is incorrect.

6 + √2 is an irrational number

Hence, proved.

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