- #1

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Demonstrate that [itex]\sqrt{2}+\sqrt{3}[/itex] is irrational.

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- #1

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Demonstrate that [itex]\sqrt{2}+\sqrt{3}[/itex] is irrational.

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- #2

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This feels like homework. However, I will give a proof just to make sure I still can:Demonstrate that [itex]\sqrt{2}+\sqrt{3}[/itex] is irrational.

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For purposes of contradiction, assume [itex]\sqrt{2}+\sqrt{3}[/itex] is

[itex]\sqrt{2}+\sqrt{3}=\frac{n}{m}[/itex]

[itex]m(\sqrt{2}+\sqrt{3})=n[/itex]

[itex]m(\sqrt{2}+\sqrt{3})(\sqrt{2}-\sqrt{3})=n(\sqrt{2}-\sqrt{3})[/itex]

[itex]m(4+9)=n(\sqrt{2}-\sqrt{3})[/itex]

[itex]13m=n(\sqrt{2}-\sqrt{3})[/itex]

Since 13 is prime and m does not divide n, 13|n and [itex]m|(\sqrt{2}-\sqrt{3})[/itex]. This implies, then, that [itex]\sqrt{2}-\sqrt{3}[/itex] is in fact an integer. Disregarding the fact that this is not the case, we still have, then, that [itex]l=\sqrt{2}-\sqrt{3}[/itex] for some

To use the latter contradiction, you would need to further show that [itex]\sqrt{12}[/itex] is irrational, so I would stick with the former contradiction (that [itex]\sqrt{2}-\sqrt{3}[/itex] is an integer)

- #3

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- #4

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If Eval had done the first part correctly, he/she would have arrived at the otherwise immediate

√3 - √2 = m/n. Adding this equation to √3 + √2 = n/m, you obtain that √3 is rational, which is a contradiction.

Another way is just squaring both sides of √3 + √2 = n/m, giving √6 rational, and again a contradiction.

- #5

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Moved to homework. Please do not give out full solutions.

- #6

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It is not to homwork, is to pass an exam.

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- #7

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It is not to homwork, is to pass an exam.

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It is a homework-type question. It still belongs in homework and the homework rules apply.

- #8

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- #9

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If Eval had done the first part correctly, he/she would have arrived at the otherwise immediate

√3 - √2 = m/n. Adding this equation to √3 + √2 = n/m, you obtain that √3 is rational, which is a contradiction.

Another way is just squaring both sides of √3 + √2 = n/m, giving √6 rational, and again a contradiction.

To avoid misinforming the person asking the question, I would like to note that a conjugate is not an inverse. A conjugate of (a+b) is (a-b) whereas the inverse of (a+b) is (a-b)/(a

If √3 + √2 = n/m, then 1/(√3 + √2) = m/n. Then, multiplying the top and bottom of the lefthand side by its conjugate, we get (√3 - √2)/(9

Also, squaring both sides of √3 + √2 = n/m would imply that √6 is rational after several more steps and is probably the easiest method. However, in all the methods given by Norwegian, I would like to point out that you would further have to show that √3 and √6 are also irrational, respectively, so you will have some more work to do.

The easiest way to prove this would be to assume that they are rational (so the ration of two integers).

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- #10

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If √3 + √2 = n/m, then 1/(√3 + √2) = m/n. Then, multiplying the top and bottom of the lefthand side by its conjugate, we get (√3 - √2)/(9^{2}+2^{2}) = (√3 - √2)/13 = m/n,not√3 - √2 = m/n.

Hi Eval,

I invite you to reconsider the above. In particular, I want you to observe the elementary identity (√3 + √2)(√3 - √2)=1 (which by the way shows that conjugation equals inverse in this case).

- #11

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Thanks for the catch, I'll edit my previous post so that I don't mislead people. Blame it on the lack of sleep I have had, my brain is failing me :/

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