text: # Thread: Modulo exercice
  
  1. ## Modulo exercice
  
  Hey again !
  
  I know it is pretty simple but i dont seem to prove it.
  If a=bmodk and c=dmodk then why a-c=(b-d)modk
  
  2. ## Re: Modulo exercice
  
  Originally Posted by grainofsand
  Hey again !
  
  I know it is pretty simple but i dont seem to prove it.
  If a=bmodk and c=dmodk then why a-c=(b-d)modk
  $a=k \cdot n +b \land c = k \cdot m+d \Rightarrow a-c = k \cdot(n-m)+(b-d) \Rightarrow$
  
  $\Rightarrow a-c \equiv b-d \pmod k$
  
  3. ## Re: Modulo exercice
  
  thank you so much !
explanation: The document contains a math question and its corresponding answer. The question asks for a proof of a property in modular arithmetic, and the response provides a step-by-step proof.
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # Thread: Modulo exercice
  
  1. ## Modulo exercice
  
  Hey again !
  
  I know it is pretty simple but i dont seem to prove it.
  If a=bmodk and c=dmodk then why a-c=(b-d)modk
  
  2. ## Re: Modulo exercice
  
  Originally Posted by grainofsand
  Hey again !
  
  I know it is pretty simple but i dont seem to prove it.
  If a=bmodk and c=dmodk then why a-c=(b-d)modk
  $a=k \cdot n +b \land c = k \cdot m+d \Rightarrow a-c = k \cdot(n-m)+(b-d) \Rightarrow$
  
  $\Rightarrow a-c \equiv b-d \pmod k$
  
  3. ## Re: Modulo exercice
  
  thank you so much !
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and its corresponding answer. The question asks for a proof of a property in modular arithmetic, and the response provides a step-by-step proof.
  
  Label: question-and-answer