Can a Natural Number Indivisible by 3 Have a Square Divisible by 3?

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The discussion revolves around the properties of natural numbers and their squares in relation to divisibility by 3. The original poster attempts to prove that if a natural number is not divisible by 3, then its square is also not divisible by 3, and explores implications for rational numbers whose squares equal 3.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the validity of the original poster's proof and the relationship between the statements regarding divisibility. They explore the concept of contrapositives in mathematical proofs.

Discussion Status

The discussion is active, with participants providing feedback on the proof structure and clarifying the logical connections between the statements. There is an exploration of the contrapositive relationship, but no explicit consensus has been reached regarding the overall proof's acceptance.

Contextual Notes

Participants question whether the proof's assumptions and conclusions are consistent, particularly regarding the implications of divisibility and the definitions involved.

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Let q be a natural number, show that if q is not divisible by 3, then neither is q^2

proof:

if q is not divisible by 3 then q = 3k + 2 for some integer k
q^2 = 4 + 3(3k^2 + 4k) = 4 + 3m for some integer m, hence q is not divisible by 3

another case, if q = 3k + 1 for some integer k, then q^2 = 1 + 3(3k^2+2k) = 1 + 3n for some integer n, hence q^2 is not divisible by 3

second part:

Assuming the statement that was to be proved above, deduce that there is no rational number x satisfying x^2 = 3

proof:

assume there is some ration number x satisfying x^2 = 3

we can then express x = p/q where p,q are coprime

x^2 = p^2/q^2 = 3
hence p^2 = 3q^2 hence p = 3m for some integer m, i.e. p is divisible by 3 ***(we proved this in the first part)***

therefore 9m^2 = 3q^2 => q^2 = 3m^2 hence q = 3n for some integer n, hence q is also divisible by 3

this is a condradiction as we assumed p and q were coprime, but have shown they have a factor of 3, hence there does not exist a rational number x such that x^2 = 3

ok, is this proof OK? the part I have labelled *** ("***(we proved this in the first part)***") is it true that we proved this in the first part, as we proved that if a natural number is not divisible by 3, then neither is it's square, but in this proof, I am saying if it's square is divisible by 3, then the number is divisible by 3 - is this the same thing (I think it is)
 
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The proof looks fine... as a formatting thing instead of adding at the end of a line (we proved this in the first part), I would put at the start of the line "by part one,". For example

"hence p^2 = 3q^2 hence p = 3m for some integer m" becomes
"hence p^2 = 3q^2, so by part one p=3m for some integer m"
 
thanks, so proving that q is not divisible by 3 then neither is q^2 is the same thing as proving if q^2 = 3m (i.e. divisible by 3) then so is q?
 
Yes, that's called the contrapositive and is often useful. If you have two statements , A and B, then proving if A then B is the same as proving if not B, then not A.

In your case you have A is "q is not divisible by 3", and B is "q2 is not divisible by 3". You proved if A, then B, so you get for free if not B, then not A.

not B is (after cancelling a double negative)"q2 is divisible by 3" and not A is "q is divisible by 3"
 

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