PROOF

B

bghalmeida

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Prove that for any integer [MATH]n[/MATH], at least one of the integers [MATH]n[/MATH], [MATH]n+2[/MATH], [MATH]n+4[/MATH] is divisible by [MATH]3[/MATH].

Besides the proof, I would like to understand why this is true, because when [MATH]n=0[/MATH], none of the numbers are divisible by [MATH]3[/MATH].
 
Why would expect to see a proof it you found a counter example. You need to first understand what a proof is and what a counterexample is. You can't have both!!

Just for the record, 0 IS divisible by 3.

Hint: n is a multiple of 3, or n is a multiple of 3 + 1, or n is a multiple of 3 +2.
 
Well you can't prove it if you have demonstrated a counter-example.

Of course jomo has demonstrated that your counter example is flawed.
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Just to be sure you get my point, please tell us what 0/3 equals and if there is a remainder?
 
Okay, I totally forgot, sorry about that. But actually, I have a proof I developed on this and I would like to see other people's proof, to see if mine is easy to follow or not.

I'm going to send you:

(If you have any other ideas or if my proof is in any way wrong, please tell me)

PROOF In order to prove the statement, let's look at the division theorem: if [MATH]a, b \in \Z, b > 0[/MATH], then there exist a unique [MATH]q, r \in \Z[/MATH] such that [MATH]a = q\cdot b + r[/MATH], [MATH]0 ≤ r < b[/MATH].


So, according to the theorem, it is possible to write a integer a in the form of [MATH]q\cdot b+r[/MATH].


To start the proof, it is necessary to know that 0, 1 and 2 are the only possible remainders for a division by 3.


Now, let k be any integer number to 3k, which is any multiple of 3.


If n be replaced with 3k, then:


[MATH]n=3k\\n+2=3k+2\\n+4=3k+4[/MATH]

A multiple of 3 plus 2 or 4 is not equal to a multiple of 3, because 2 and 4 are not multiples of 3, but 3k is, so the statement holds for n=3k.


If n be replaced with 3k+1, then:


[MATH]n=3k+1\\n+2=3k+3\\n+4=3k+5[/MATH]

It is possible to know that a multiple of 3 plus 1 or 5 is not equal to an multiple of 3, for the same reasons stated above. However, 3k+3 is equal to an multiple of 3, because 3 is a multiple of 3. So the statement holds for n=3k+1.


If n be replaced with 3k+2, then:


[MATH]n=3k+2\\n+2=3k+4\\n+4=3k+6[/MATH]

Again, 3k+2 and 3k+4 cannot result in a multiple of 3, but 3k+6 can, because 6 is a multiple of 3. So the statement holds for n=3k+2.


So, by replacing n with 3k, 3k+1 and 3k+2, I proved that for any integer n, at least one of the integers n, n+2, n+4 is divisible by 3.
 
I strongly believe that when one is taught their times tables it should start with 0, not 1.

For example, when it comes to the 3 times table it should be taught as 3*0=0, 3*1, = 3, etc. When students learn the times table from 1, some students say things like 0 is not a multiple of 3. Some also do not know how many times 3 goes into 2.
 
Yes, this was my mistake. But besides it, what do you think about my proof?

Jomo said:
I strongly believe that when one is taught their times tables it should start with 0, not 1.

For example, when it comes to the 3 times table it should be taught as 3*0=0, 3*1, = 3, etc. When students learn the times table from 1, some students say things like 0 is not a multiple of 3. Some also do not know how many times 3 goes into 2.
Click to expand...
 
I suspect that you can do a more concise proof. Start as you do with your theorem Define that a divides b means the remainder is zero.

[MATH]\exists \text { integer } k \text { such that}\\ n = 3k + r, \text { where } r \in \mathbb Z \text { and } 0 \le r < 3.[/MATH]Three cases.

[MATH]r = 0 \implies 3 \text { divides } n.[/MATH]
[MATH]r = 1 \implies n + 2 = 3k + 1 + 2 = 3(k + 1) + 0 \implies \\ 3 \text { divides } n + 2.[/MATH][MATH]r = 2 \implies n + 4 = 3k + 2 + 4 = 3(k + 2) + 0 \implies \\ 3 \text { divides } n + 4.[/MATH][MATH]\therefore 3 \text { divides } n, \ n + 2, \text { or } n + 4.[/MATH]
Same idea as yours, just shorter.
 
JeffM said:
I suspect that you can do a more concise proof. Start as you do with your theorem Define that a divides b means the remainder is zero.

[MATH]\exists \text { integer } k \text { such that}\\ n = 3k + r, \text { where } r \in \mathbb Z \text { and } 0 \le r < 3.[/MATH]Three cases.

[MATH]r = 0 \implies 3 \text { divides } n.[/MATH]
[MATH]r = 1 \implies n + 2 = 3k + 1 + 2 = 3(k + 1) + 0 \implies \\ 3 \text { divides } n + 2.[/MATH][MATH]r = 2 \implies n + 4 = 3k + 2 + 4 = 3(k + 2) + 0 \implies \\ 3 \text { divides } n + 4.[/MATH][MATH]\therefore 3 \text { divides } n, \ n + 2, \text { or } n + 4.[/MATH]
Same idea as yours, just shorter.
Click to expand...
Thanks for your help, I will change my proof.
 
bghalmeida said:
Okay, I totally forgot, sorry about that. But actually, I have a proof I developed on this and I would like to see other people's proof, to see if mine is easy to follow or not.

I'm going to send you:

(If you have any other ideas or if my proof is in any way wrong, please tell me)

PROOF In order to prove the statement, let's look at the division theorem: if [MATH]a, b \in \Z, b > 0[/MATH], then there exist a unique [MATH]q, r \in \Z[/MATH] such that [MATH]a = q\cdot b + r[/MATH], [MATH]0 ≤ r < b[/MATH].


So, according to the theorem, it is possible to write a integer a in the form of [MATH]q\cdot b+r[/MATH].


To start the proof, it is necessary to know that 0, 1 and 2 are the only possible remainders for a division by 3.


Now, let k be any integer number to 3k, which is any multiple of 3.


If n be replaced with 3k, then:


[MATH]n=3k\\n+2=3k+2\\n+4=3k+4[/MATH]

A multiple of 3 plus 2 or 4 is not equal to a multiple of 3, because 2 and 4 are not multiples of 3, but 3k is, so the statement holds for n=3k.


If n be replaced with 3k+1, then:


[MATH]n=3k+1\\n+2=3k+3\\n+4=3k+5[/MATH]

It is possible to know that a multiple of 3 plus 1 or 5 is not equal to an multiple of 3, for the same reasons stated above. However, 3k+3 is equal to an multiple of 3, because 3 is a multiple of 3. So the statement holds for n=3k+1.


If n be replaced with 3k+2, then:


[MATH]n=3k+2\\n+2=3k+4\\n+4=3k+6[/MATH]

Again, 3k+2 and 3k+4 cannot result in a multiple of 3, but 3k+6 can, because 6 is a multiple of 3. So the statement holds for n=3k+2.


So, by replacing n with 3k, 3k+1 and 3k+2, I proved that for any integer n, at least one of the integers n, n+2, n+4 is divisible by 3.
Click to expand...
This is far more complicated than is neccesary. Given any integer, n, dividing n by 3 gives an integer quotient, k, with remainder 0, 1, or 2. That means that every integer is either 3k or 3k+ 1 or 3k+ 2 for some integer, k. if n is of the form 3k, then it is divisible by 3. If n is of the form 3k+ 1 then n+2= 3k+ 1+ 2= 3k+ 3= 3(k+1) is divisible by 3. If n is of the 3k+ 2 then n+1= 3k+ 2+ 1= 3k+ 3= 3(k+1) is divisible by 3.
 
HallsofIvy said:
This is far more complicated than is neccesary. Given any integer, n, dividing n by 3 gives an integer quotient, k, with remainder 0, 1, or 2. That means that every integer is either 3k or 3k+ 1 or 3k+ 2 for some integer, k. if n is of the form 3k, then it is divisible by 3. If n is of the form 3k+ 1 then n+2= 3k+ 1+ 2= 3k+ 3= 3(k+1) is divisible by 3. If n is of the 3k+ 2 then n+1= 3k+ 2+ 1= 3k+ 3= 3(k+1) is divisible by 3.
Click to expand...
Yes, I gave this idea in an earlier post. The OP then wrote a dissertation size paper on my comment.
 
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