My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Thanks Tree1Thanks
  • 1 Post By Maschke
Reply
 
LinkBack Thread Tools Display Modes
January 31st, 2018, 05:59 PM   #1
Newbie
 
Joined: Jan 2018
From: Toronto

Posts: 9
Thanks: 0

Mod Opertaor

I don't understand this question - Please help me understand why it is true or false.

If a ≡ x1 (mod m) and b ≡ y1 (mod m), then we will have ab ≡ x1y1 (mod m). True or False and why?

Thank you.
Tricia is offline  
 
January 31st, 2018, 06:05 PM   #2
Senior Member
 
Joined: Feb 2016
From: Australia

Posts: 1,512
Thanks: 506

Math Focus: Yet to find out.
Did you try play around with some examples?

Suppose $m = 2\pi$ x1, y1 are angles for example. What happens when you consider different values for x1 and y1 for the ab expression?
Joppy is offline  
January 31st, 2018, 06:12 PM   #3
Newbie
 
Joined: Jan 2018
From: Toronto

Posts: 9
Thanks: 0

Quote:
Originally Posted by Joppy View Post
Did you try play around with some examples?

Suppose $m = 2\pi$ x1, y1 are angles for example. What happens when you consider different values for x1 and y1 for the ab expression?
I have not tried diffrent values. Not understanding what a or x1, y1 would represent. In addition to what numeric m (mod m) would represent.

I understand when written out like this 33 mod 8, how to get the results, but the above question is really not clear.
Tricia is offline  
January 31st, 2018, 06:34 PM   #4
Senior Member
 
Joined: Aug 2012

Posts: 1,708
Thanks: 455

Quote:
Originally Posted by Tricia View Post

If a ≡ x1 (mod m) and b ≡ y1 (mod m), then we will have ab ≡ x1y1 (mod m). True or False and why?
Here's a complete solution, although playing around with Joppy's hint would be valuable too so you can see more examples of how this works.

The trick to this type of problem is to break it down to what it's really saying. To do that, we have to get picky with the definitions.

If $a = x_1 \pmod m$ then $m | a - x_1$ (The vertical bar means "divides")

Then (by the definition of divides) there is some integer $d_a$ (d for divisor) such that $m d_a = a - x_1$, or

(1) $a = m d_a + x_1$

And likewise there is some integer $d_b$ such that

(2) $b = m d_b + x_2$

Now using (1) and (2) we have

$a b = (m d_a + x_1)(m d_b + x_2)$

$= m^2 d_a d_b + m d_a x_2 + m d_b x_1 + x_1 x_2$

$= m(m d_a d_b + d_a x_2 + d_b x_1) + x_1 x_2$

Now you can see that $a b - x_1 x_2$ is a multiple of $m$ and we're done.

This is how all of these kinds of problems should be approached. Replace each technical term or symbol you are learning about with its precise definition. Whenever you get stuck, just go symbol-by-symbol and ask yourself: Exactly what does this mean? Write down the textbook definition. After that the problem often solves itself. This is one of the standard patterns for doing problems. Always make sure you understand exactly what the definitions say.
Thanks from Tricia

Last edited by Maschke; January 31st, 2018 at 06:54 PM.
Maschke is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
mod, opertaor



Thread Tools
Display Modes






Copyright © 2018 My Math Forum. All rights reserved.