Question

prove that
$\sqrt{2} + 6 \: is \: irrational$
​

Answer 1

Answer:

Suppose there exist coprime a,b∈Z s.t.

2–√+3–√=ab⟹6–√=a22b2−52=a2−5b22b2

If you already know 6–√ is irrational then you're already done, otherwise prove it as with 2–√ , say:

6–√=pq,(p,q)=1⟹6q2=p2⟹2∣p

and thus we can write

6–√=2p′q⟹2∣qalso , and this is a contradiction

Step-by-step explanation:

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Answer 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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