Question

Proof that 3 - \sqrt{8} is irrational.

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Answer 1

Step-by-step explanation:

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It's definitely irrational.

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It's the same as 6sqrt2, and if that's a rational number (call it q ) then q/6 = sqrt2 , which we already know to be irrational.

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• Should there be any doubt about this; if q is rational, it can be expressed as a fraction q=m/n with m,n both integers and n≠0 .
• Then q/6=m/6n , where m,6n are integers, and 6n≠0 .
• This is using the early-school notion in which “fractions” are taken for granted, with rules for doing arithmetic on them in this form. If you've seen the construction of the rational numbers as equivalence classes of ordered pairs of integers, it should be easy to see how to translate this into the appropriate terms, and what you need to prove.

• How is √ 2 known to be irrational?

• Almost all proofs of this can be adapted very easily to direct proofs that √38=√72 . The only catch is that you have to be just a little more careful about finding how many times 2 divides into things.

• It's very slightly easier to show that 3√2=√18 is irrational, and then argue as above to show that this number is irrational if and only if 3√8=2.3√2 is irrational.

The hard bits are things you may be taking for granted; the fundamental theorem of arithmetic, which shows that prime factorisations are unique (up to order and units), and the reasons why irrational square roots ought to exist at all. I expect that you won't be expected to deal with these questions, though.

Answer 2

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see the following attachment..

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