Proving: n^2 % 3 = 0 Implies n % 3 = 0

  • Context: Undergrad 
  • Thread starter Thread starter Dr-NiKoN
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around proving the statement that if \( n^2 \mod 3 = 0 \), then \( n \mod 3 = 0 \). Participants explore various approaches to this problem, including algebraic manipulations, prime factorization, and logical reasoning. The scope includes mathematical reasoning and proof techniques.

Discussion Character

  • Mathematical reasoning
  • Exploratory
  • Debate/contested

Main Points Raised

  • One participant suggests starting with the assumption that \( n \) is a whole number and applying the fundamental theorem of algebra to explore the prime factorization of \( n \) and \( n^2 \).
  • Another participant proposes working from the assumption that \( n \) is not a multiple of 3, leading to the forms \( n = 3m + 1 \) or \( n = 3m + 2 \), and shows that squaring these forms results in expressions that are not multiples of 3.
  • A participant mentions the use of logical implications and the contrapositive, questioning how these relate to the proof.
  • There is a discussion about the validity of using prime factorization, with one participant stating it cannot be used as it is covered in a later chapter.
  • Another participant clarifies that the negation of "n is a multiple of 3" is not equivalent to "n is even," emphasizing the importance of understanding logical implications in the context of the proof.
  • One participant suggests exhausting the possibilities for \( n \mod 3 \) as a potential method to approach the proof.

Areas of Agreement / Disagreement

Participants express differing views on the best approach to prove the statement, with some favoring algebraic methods while others focus on logical reasoning. There is no consensus on a single method or solution, and the discussion remains unresolved.

Contextual Notes

Some participants mention limitations in their approaches, such as the inability to use prime factorization and the need to clarify logical implications. The discussion also reflects varying levels of familiarity with proof techniques.

Dr-NiKoN
Messages
93
Reaction score
0
I'm supposed to prove that if:
n^2 % 3 = 0
Then
n % 3 = 0

I'm not sure how to proceed here?

Does it got something to do with setting n^2 to x?
x % 3 = 0
and
\frac{x}{n} % 3 = 0

Not sure how that helps..

The actual problem is to prove that if n^2 / 3 is a whole number, then n/3 is also a whole number. But, I can't figure out any other way to present it.
 
Last edited:
Physics news on Phys.org
The actual problem is to prove that if n^2 / 3 is a whole number, then n/3 is also a whole number. But, I can't figure out any other way to present it.
Seems like that was good enough...

Assuming that [itex]n[/itex] is a whole number, we can apply the fundamental theorem of algebra, so [itex]n[/itex] has a unique prime factorization:
[tex]n=p_1^{a_1}\times p_2^{a_2}...[/tex]
Where all the [itex]p_i[/itex] are primes, and all the [itex]a_i[/itex] are whole numbers. Then [itex]n^2[/itex] has the following prime factorization:
[tex]n^2=p_1^{2a_1} \times p_2^{2a_2}...[/tex]

Should be pretty easy from there...
 
Work from the other side. If n is not a multiple of 3 (i.e. if n/3 is not a whole number), then n must be of the form n= 3m+1 or n= 3m+2 for some whole number m.

Now square them: n2= (3m+1)2= 9m2+ 6m+ 1=
3(3m2+ 2m)+ 1. In other words, it is a multiple of 3 plus 1, not a multiple of 3.

if n= 3m+2, then n2= (3m+2)2= 9m2+ 12m+ 4=
3(3m2+ 4m+ 1)+ 1. Again, it is not a multiple of 3.

Okay, if n is not a multiple of 3, then n2 cannot be. What does that tell you about the case if n2 is a multiple of 3?
 
Or use the alternative characterizations of the primes:

p divides ab iff p divides either a or b.

Let =3, a=b=n
 
I can't use prime factorization, since that is in a later chapter.

Work from the other side. If n is not a multiple of 3 (i.e. if n/3 is not a whole number), then n must be of the form n= 3m+1 or n= 3m+2 for some whole number m.
Doesn't this give me:
p = n^2 is even
q = n is even
(in logic, don't know the latex for it)
if NOT q then NOT p

How does this prove:
if q then p

I tried working with the implication law, but can't seem to figure out how you get there. Or is this just some common law of logic?
 
Dr-NiKoN said:
Doesn't this give me:
p = n^2 is even
q = n is even
(in logic, don't know the latex for it)
if NOT q then NOT p
HallsOfIvy's method gives you:
p = n^2 is divisible by 3
q = n is divisible by 3
if NOT q then NOT p

Dr-NiKoN said:
How does this prove:
if q then p

It doesn't, but it proves if p then q... Which is what you originally wanted, no?
 
You could exhaust over the possibilities for n % 3.
 
"Doesn't this give me:
p = n^2 is even
q = n is even"

Well, no- the "negation" of "n is a multiple of 3" is not "n is even"!

It is a "law of logic" that "if p then q" is equivalent to "if not q then not p". That's called the contrapositive and is a special case of "proof by contradiction".
 
Thanks guys,

this actually helped me to solve a couple of other problems(proving the contrapositive that is). :)
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
Replies
9
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 9 ·
Replies
9
Views
2K