Real number systems: For natural n, if sqrt[n] is not natural, then also not rational

[pisrationalhahaha]

Hi , I have a problem dealing with this exercise

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\(\displaystyle \mbox{Given that }\, n\, \in\, \mathbb{N},\)

\(\displaystyle \mbox{prove the following:}\)

. . .\(\displaystyle \sqrt{\strut n\,}\, \notin\, \mathbb{N}\, \implies\, \sqrt{\strut n\,}\, \notin\, \mathbb{Q}\)

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When I was trying to solve it , I tried to use the method of contradiction but then I did not know what to do. It is so complicated. Its the first time for me to deal with such exercises.

 

[tkhunny]

Please display your efforts.

 

[ksdhart2]

Hi , I have a problem dealing with this exercise

---

\(\displaystyle \mbox{Given that }\, n\, \in\, \mathbb{N},\)

\(\displaystyle \mbox{prove the following:}\)

. . .\(\displaystyle \sqrt{\strut n\,}\, \notin\, \mathbb{N}\, \implies\, \sqrt{\strut n\,}\, \notin\, \mathbb{Q}\)

---

When I was trying to solve it , I tried to use the method of contradiction but then I did not know what to do. It is so complicated. Its the first time for me to deal with such exercises.

I'm assuming that the N and Q in your image are meant to be \(\displaystyle \mathbb{N}\) (the natural numbers) and \(\displaystyle \mathbb{Q}\) (the rational numbers), respectively. If this is not correct, please include an explanation of what the sets N and Q stand for.

With that assumption in mind, I think a proof by contradiction could work here. The given statement can be reformulated as:

\(\displaystyle \forall (n \in \mathbb{N}) \left( \sqrt{n} \notin \mathbb{N} \rightarrow \sqrt{n} \notin \mathbb{Q} \right)\)

The above statement must be true for every n. If the negation were true, you'd need to show that it holds for at least one n. Hence, your proof by contradiction should start by assuming:

\(\displaystyle \exists(n \in \mathbb{N}) \left( \sqrt{n} \notin \mathbb{N} \not \rightarrow \sqrt{n} \notin \mathbb{Q} \right)\)

Where do you think you'd go from here?

 

[pisrationalhahaha]

I'm assuming that the N and Q in your image are meant to be \(\displaystyle \mathbb{N}\) (the natural numbers) and \(\displaystyle \mathbb{Q}\) (the rational numbers), respectively. If this is not correct, please include an explanation of what the sets N and Q stand for.

With that assumption in mind, I think a proof by contradiction could work here. The given statement can be reformulated as:

\(\displaystyle \forall (n \in \mathbb{N}) \left( \sqrt{n} \notin \mathbb{N} \rightarrow \sqrt{n} \notin \mathbb{Q} \right)\)

The above statement must be true for every n. If the negation were true, you'd need to show that it holds for at least one n. Hence, your proof by contradiction should start by assuming:

\(\displaystyle \exists(n \in \mathbb{N}) \left( \sqrt{n} \notin \mathbb{N} \not \rightarrow \sqrt{n} \notin \mathbb{Q} \right)\)

Where do you think you'd go from here?

I dont know how to continue. Its the first time for me with propositions and contradiction method.

 

[ksdhart2]

That's alright. At least now you know what areas to study and brush up on. If you're finding your class notes, textbook, and/or any handouts provided to be insufficient, you might try using Google to look up additional resources. I Googled "beginner's guide to set theory" and found this PDF from UMass. The first section thereof should cover many things that will prove useful for this problem. Unfortunately, I'm a bit short on time at the moment, but you might also want to Google for logic resources. Perhaps "beginner's guide to mathematical logic" would help you find something?

In particular, I think you should look for information about the contrapositive of a logical statement. What does that term mean? If given a logical statement, what do you always automatically know about its contrapositive? I'd also look for information about subsets. Every natural number is also an integer, which is also a rational number. Hence, \(\displaystyle \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q}\). What do these two things tell you about any natural number? So, if we suppose square root of n is a rational number, what also must be true?