Why is \sqrt{2}+\sqrt{3} irrational?

In summary, to prove that √2 + √3 is irrational, assume the opposite and use a contradiction proof. Show that if it is rational, then √2 - √3 and √6 must also be rational, leading to a contradiction. This can be done by manipulating equations and using the fact that rational numbers are closed under addition, subtraction, and multiplication.
  • #1
inverse
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Demonstrate that [itex]\sqrt{2}+\sqrt{3}[/itex] is irrational.

Thanks
 
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  • #2
inverse said:
Demonstrate that [itex]\sqrt{2}+\sqrt{3}[/itex] is irrational.

Thanks
This feels like homework. However, I will give a proof just to make sure I still can:

For purposes of contradiction, assume [itex]\sqrt{2}+\sqrt{3}[/itex] is rational. That is, there are some integers n and m (m≠0) such that [itex]\sqrt{2}+\sqrt{3}=\frac{n}{m}[/itex] and gcd(n,m)=1 (in other words, n and m are relatively prime). We have, then, that:
[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 l, but [itex]\sqrt{2}+\sqrt{3}=l+2\sqrt{3}[/itex]. Since [itex]\sqrt{12}[/itex] is irrational and l is rational, their sum is irrational, further contradicting the original assumption.

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
I fail to understand how I show from the expression [itex]\sqrt{2}-\sqrt{3}[/itex]that must comply something for m and n is irrational.
 
  • #4
Yes, probably homework, but we should not give misleading advice, so:

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
Moved to homework. Please do not give out full solutions.
 
  • #6
It is not to homwork, is to pass an exam.

Thanks
 
  • #7
inverse said:
It is not to homwork, is to pass an exam.

Thanks

It is a homework-type question. It still belongs in homework and the homework rules apply.
 
  • #8
I have proposed a model for solving methodology see, because if I have a reference (Worked) can not practice more similar exercises.
 
  • #9
This post contains incorrect info (after the quote, of course)
Norwegian said:
Yes, probably homework, but we should not give misleading advice, so:

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)/(a2-b2). Applying this:

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)/(92+22) = (√3 - √2)/13 = m/n, not √3 - √2 = m/n. Then you can add this to (√3 + √2)/13 = n/(13m) to get that (√3)/13 = m/n+n/13m = (13m2+n2)/(13nm), so √3 = (13m2+n2)/(nm) implying that √3 is the ratio of two integers, which is a contradiction. This is no more immediate than the proof I gave :P

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).
 
Last edited:
  • #10
Eval said:
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)/(92+22) = (√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
Ah, right, sorry. It is not exactly an identity, but yes, you are right. It is because I was still thinking (a+b) as opposed to (sqrt(a)+sqrt(b)). The identitiy there is a-b, which in thios case happens to be 1 (which is extremely useful).

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 :/
 

What is an irrational demonstration?

An irrational demonstration is a scientific experiment or study that does not follow a logical or rational thought process. It may involve flawed reasoning, faulty data, or other factors that make it unreliable or invalid.

How do you identify an irrational demonstration?

An irrational demonstration can be identified by looking for inconsistencies, logical fallacies, or gaps in the evidence. It may also be flagged if it contradicts well-established scientific principles or if it lacks peer review or replicability.

Why is it important to avoid irrational demonstrations?

It is important to avoid irrational demonstrations because they can lead to false conclusions and potentially harm the scientific community's credibility. They can also waste time, resources, and funding that could be better used for valid and reliable research.

Can an irrational demonstration be useful?

In some rare cases, an irrational demonstration may lead to unexpected discoveries or insights. However, these should be approached with caution and thoroughly tested and validated before being accepted as reliable evidence.

What can be done to prevent irrational demonstrations?

To prevent irrational demonstrations, scientists should follow strict research protocols, conduct thorough peer review, and aim for replicable results. It is also essential to remain open-minded and critical of one's own work to avoid biased or flawed thinking.

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