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Divisibility problem. Prove or disprove if 𝑎|𝑏c, then 𝑎|𝑏 or 𝑎|𝑐

Help understanding divisibility proofsDivisibility problemDivisibility question: if $a=be+r$, then $e$ $= ⌊bc⌋$Prove or disprove this implicationBasic question on Number Theory and DivisibilityDeceptively simple divisibility problemProve or disprove f an integer is divisible by 4, then it is divisible by 8Number Theory question about proof of divisibilityProving divisibility with sumProve or disprove: if $10$ divides $n^{4}$, then $10$ divides $n$

1

$begingroup$

I understand the problem very well. I just don't how to go at it.

Prove or Disprove: For all 𝑎, 𝑏, 𝑐 ∈ ℤ+, if 𝑎|𝑏c, then 𝑎|𝑏 or 𝑎|𝑐.

divisibility

share|cite|improve this question

asked Mar 23 '16 at 17:58

user3233845user3233845

61

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $12|3times4=12$, but $12not|3,12not|4$.
  $endgroup$
  – user249332
  Mar 23 '16 at 18:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How to go at it: Write down three integers $a,b,c$. Check whether $a|bc$. If it does, check whether $a|b$ or $a|c$. Repeat with three other integers. Repeat as often as needed until you begin to perceive a pattern. Test the pattern with more repetitions. Try to refine the pattern. Try to find a counterexample. Or try to find a proof.
  $endgroup$
  – Lee Mosher
  Mar 23 '16 at 18:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If a divides b times c does that mean a has to completely divide b or completely divide c? What if part of a divides b and the rest of b divides c but the whole of a divides neither b nor c? Is that possible? What if $k|b$ and $j|c$ and so $a= jk$ and $jk|bc$. Does it follow that $jk|b$ or $jk|c$?
  $endgroup$
  – fleablood
  Mar 23 '16 at 18:08
  

  
  
  
  

add a comment | 

1

$begingroup$

I understand the problem very well. I just don't how to go at it.

Prove or Disprove: For all 𝑎, 𝑏, 𝑐 ∈ ℤ+, if 𝑎|𝑏c, then 𝑎|𝑏 or 𝑎|𝑐.

divisibility

share|cite|improve this question

asked Mar 23 '16 at 17:58

user3233845user3233845

61

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  $12|3times4=12$, but $12not|3,12not|4$.
  $endgroup$
  – user249332
  Mar 23 '16 at 18:00
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  How to go at it: Write down three integers $a,b,c$. Check whether $a|bc$. If it does, check whether $a|b$ or $a|c$. Repeat with three other integers. Repeat as often as needed until you begin to perceive a pattern. Test the pattern with more repetitions. Try to refine the pattern. Try to find a counterexample. Or try to find a proof.
  $endgroup$
  – Lee Mosher
  Mar 23 '16 at 18:02
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If a divides b times c does that mean a has to completely divide b or completely divide c? What if part of a divides b and the rest of b divides c but the whole of a divides neither b nor c? Is that possible? What if $k|b$ and $j|c$ and so $a= jk$ and $jk|bc$. Does it follow that $jk|b$ or $jk|c$?
  $endgroup$
  – fleablood
  Mar 23 '16 at 18:08
  

  
  
  
  

add a comment | 

$begingroup$

I understand the problem very well. I just don't how to go at it.

Prove or Disprove: For all 𝑎, 𝑏, 𝑐 ∈ ℤ+, if 𝑎|𝑏c, then 𝑎|𝑏 or 𝑎|𝑐.

divisibility

share|cite|improve this question

asked Mar 23 '16 at 17:58

user3233845user3233845

61

$endgroup$

share|cite|improve this question

asked Mar 23 '16 at 17:58

user3233845user3233845

61

share|cite|improve this question

asked Mar 23 '16 at 17:58

user3233845user3233845

61

asked Mar 23 '16 at 17:58

user3233845user3233845

61

asked Mar 23 '16 at 17:58

user3233845user3233845

61

2 Answers
2

active

oldest

votes

0

$begingroup$

Counter-example: $$large8|6times 4$$ but $$large8not | , ,6$$ nor $$large8not | , ,4$$

share|cite|improve this answer

answered Mar 23 '16 at 18:02

SchrodingersCatSchrodingersCat

22.4k52862

$endgroup$

add a comment | 

0

$begingroup$

The trick is to realize that maybe $a$ can be factored to $a = jk$. Then it's possible that $j|b$ and $k|c$ but those two statements are "unrelated". It follows that $jk|bc$ but there is no "relation" between $k$ and $b$ or between $j$ and $c$ so there is no reason to believe either $jk|b$ or $jk|c$.

A counter example is easy to make: Pick any two numbers where one divide the other. Say $2|6$. Then take another pair that are relative prime to both of these were one divides the other. Say $5|35$. Then obviously $2*5|6*35$ but equally obviously $2*5not mid 6$ (because $5 not mid 6$) and $2*5 not mid 35$ (because $2 not mid 35$).

(They don't have to be relatively prime. You just need the factors of $a$ "distributed" among $b$ and $c$. If $a = 4 = 2^2$ you can have $2|b$ but $4not mid b$ [say $b= 6$ and $2|c$ but $4not mid b$ [say $c = 10$]. Then you get $4|60$ but $4 not mid 6$ and $4not mid 10$.)

But what if $a$ is prime? Then this is true. $a$ can not factor so if $bc$ is a multiple of $a$ either $b$ is, or $c$ is, or both are.

share|cite|improve this answer

answered Mar 23 '16 at 18:26

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

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0

$begingroup$

Counter-example: $$large8|6times 4$$ but $$large8not | , ,6$$ nor $$large8not | , ,4$$

share|cite|improve this answer

answered Mar 23 '16 at 18:02

SchrodingersCatSchrodingersCat

22.4k52862

$endgroup$

add a comment | 

0

$begingroup$

Counter-example: $$large8|6times 4$$ but $$large8not | , ,6$$ nor $$large8not | , ,4$$

share|cite|improve this answer

answered Mar 23 '16 at 18:02

SchrodingersCatSchrodingersCat

22.4k52862

$endgroup$

add a comment | 

$begingroup$

Counter-example: $$large8|6times 4$$ but $$large8not | , ,6$$ nor $$large8not | , ,4$$

share|cite|improve this answer

answered Mar 23 '16 at 18:02

SchrodingersCatSchrodingersCat

22.4k52862

$endgroup$

share|cite|improve this answer

answered Mar 23 '16 at 18:02

SchrodingersCatSchrodingersCat

22.4k52862

answered Mar 23 '16 at 18:02

SchrodingersCatSchrodingersCat

22.4k52862

answered Mar 23 '16 at 18:02

SchrodingersCatSchrodingersCat

22.4k52862

0

$begingroup$

The trick is to realize that maybe $a$ can be factored to $a = jk$. Then it's possible that $j|b$ and $k|c$ but those two statements are "unrelated". It follows that $jk|bc$ but there is no "relation" between $k$ and $b$ or between $j$ and $c$ so there is no reason to believe either $jk|b$ or $jk|c$.

A counter example is easy to make: Pick any two numbers where one divide the other. Say $2|6$. Then take another pair that are relative prime to both of these were one divides the other. Say $5|35$. Then obviously $2*5|6*35$ but equally obviously $2*5not mid 6$ (because $5 not mid 6$) and $2*5 not mid 35$ (because $2 not mid 35$).

(They don't have to be relatively prime. You just need the factors of $a$ "distributed" among $b$ and $c$. If $a = 4 = 2^2$ you can have $2|b$ but $4not mid b$ [say $b= 6$ and $2|c$ but $4not mid b$ [say $c = 10$]. Then you get $4|60$ but $4 not mid 6$ and $4not mid 10$.)

But what if $a$ is prime? Then this is true. $a$ can not factor so if $bc$ is a multiple of $a$ either $b$ is, or $c$ is, or both are.

share|cite|improve this answer

answered Mar 23 '16 at 18:26

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

0

$begingroup$

The trick is to realize that maybe $a$ can be factored to $a = jk$. Then it's possible that $j|b$ and $k|c$ but those two statements are "unrelated". It follows that $jk|bc$ but there is no "relation" between $k$ and $b$ or between $j$ and $c$ so there is no reason to believe either $jk|b$ or $jk|c$.

A counter example is easy to make: Pick any two numbers where one divide the other. Say $2|6$. Then take another pair that are relative prime to both of these were one divides the other. Say $5|35$. Then obviously $2*5|6*35$ but equally obviously $2*5not mid 6$ (because $5 not mid 6$) and $2*5 not mid 35$ (because $2 not mid 35$).

(They don't have to be relatively prime. You just need the factors of $a$ "distributed" among $b$ and $c$. If $a = 4 = 2^2$ you can have $2|b$ but $4not mid b$ [say $b= 6$ and $2|c$ but $4not mid b$ [say $c = 10$]. Then you get $4|60$ but $4 not mid 6$ and $4not mid 10$.)

But what if $a$ is prime? Then this is true. $a$ can not factor so if $bc$ is a multiple of $a$ either $b$ is, or $c$ is, or both are.

share|cite|improve this answer

answered Mar 23 '16 at 18:26

fleabloodfleablood

73.8k22891

$endgroup$

add a comment | 

$begingroup$

The trick is to realize that maybe $a$ can be factored to $a = jk$. Then it's possible that $j|b$ and $k|c$ but those two statements are "unrelated". It follows that $jk|bc$ but there is no "relation" between $k$ and $b$ or between $j$ and $c$ so there is no reason to believe either $jk|b$ or $jk|c$.

A counter example is easy to make: Pick any two numbers where one divide the other. Say $2|6$. Then take another pair that are relative prime to both of these were one divides the other. Say $5|35$. Then obviously $2*5|6*35$ but equally obviously $2*5not mid 6$ (because $5 not mid 6$) and $2*5 not mid 35$ (because $2 not mid 35$).

(They don't have to be relatively prime. You just need the factors of $a$ "distributed" among $b$ and $c$. If $a = 4 = 2^2$ you can have $2|b$ but $4not mid b$ [say $b= 6$ and $2|c$ but $4not mid b$ [say $c = 10$]. Then you get $4|60$ but $4 not mid 6$ and $4not mid 10$.)

But what if $a$ is prime? Then this is true. $a$ can not factor so if $bc$ is a multiple of $a$ either $b$ is, or $c$ is, or both are.

share|cite|improve this answer

answered Mar 23 '16 at 18:26

fleabloodfleablood

73.8k22891

$endgroup$

share|cite|improve this answer

answered Mar 23 '16 at 18:26

fleabloodfleablood

73.8k22891

answered Mar 23 '16 at 18:26

fleabloodfleablood

73.8k22891

answered Mar 23 '16 at 18:26

fleabloodfleablood

73.8k22891