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Help me with algebra proof

  • Thread starter kramer733
  • Start date
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1. Homework Statement

Question 1.

http://people.math.carleton.ca/~mezo/A2math1102-11.pdf

2. Homework Equations

r1 = a modn
r2 = b modn
r = (a+b) modn



3. The Attempt at a Solution


I used the division algorithm

So:

a = (q1)n + r1
b = (q2)n + r2
(a+b) = (q3)n + r

i isolated for r1 and r2 to get more equations

r1 = a - (q1)n
r2 = b - (q2)n
r = (a+b) - (q3)n

r1+r2 = (a+b) -n(q1+q2)

I then added a and b together to get the following:

a+b = r1+r2 + n(q1+q2)

After that, i`m completely lost.

This is for question 1 by the way. I've been toying around with this question for 15 hours and i still can't do anything with it. I thought i proved it before but then i realized it wasn't a proof. Please help. I'm dying.



Also another thing that i'm having trouble with is 3. I don't understand why it's worth 10 marks. All i see is me setting equation 1 and 2 equal to each other. I don't understand what's so special about that.
 
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21,992
3,272
1. Homework Statement

Question 1.

http://people.math.carleton.ca/~mezo/A2math1102-11.pdf

2. Homework Equations

r1 = a modn
r2 = b modn
r = (a+b) modn



3. The Attempt at a Solution


I used the division algorithm

So:

a = (q1)n + r1
b = (q2)n + r2
(a+b) = (q3)n + r
Substitute a and b from the first and second equation in the third equation. What do you get??



i isolated for r1 and r2 to get more equations

r1 = a - (q1)n
r2 = b - (q2)n
r = (a+b) - (q3)n

r1+r2 = (a+b) -n(q1+q2)

I then added a and b together to get the following:

a+b = r1+r2 + n(q1+q2)

After that, i`m completely lost.

This is for question 1 by the way. I've been toying around with this question for 15 hours and i still can't do anything with it. I thought i proved it before but then i realized it wasn't a proof. Please help. I'm dying.
 
324
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(q1)n+(q2)n + r1 + r2 = (q3)n+r

Is this what you meant?
 
21,992
3,272
324
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Yes, that is good. Now rearrange a bit and you got the proof!!
I don't think it's that obvious to me. To be honest, i've already tried doing that and i'm not sure but maybe i don't truly know how mods truly work. if i isolated for r1 + r2, then i'd get the following:

r1 + r2 = n(q3 - q2 - q1) + r

r1 = amodn
r2 = bmodn
r = (a+b)modn

I'm not exactly sure how it works i guess. Could you help me? It'd be greatly appreciated.
 
21,992
3,272
Well, a=b (mod n) if there is a q such that qn=a-b. Agreed??

Well, you've shown now that r1 + r2 - r = n(q3 - q2 - q1). Doesn't this prove it?
 
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Well, a=b (mod n) if there is a q such that qn=a-b. Agreed??

Well, you've shown now that r1 + r2 - r = n(q3 - q2 - q1). Doesn't this prove it?
To be honest, i don't really understand why a=b (mod n) if there is a q such that qn = a-b"

Where'd that come from? Could you explain please? I'm sorry but it's really just not that obvious to me.
 
21,992
3,272
To be honest, i don't really understand why a=b (mod n) if there is a q such that qn = a-b"

Where'd that come from? Could you explain please? I'm sorry but it's really just not that obvious to me.
What is the definition of a=b (mod n) ????
 
21,992
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a = q1n + a

You mean that?
Don't you mean [itex]a=q_1 n+b[/itex]??

Well, that's the same thing I posted, no?? Just rearrange, and you get [itex]a-b=qn[/itex]...
 
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Don't you mean [itex]a=q_1 n+b[/itex]??

Well, that's the same thing I posted, no?? Just rearrange, and you get [itex]a-b=qn[/itex]...
Actually i don't know but i think i posted it wrong. shouldn't it be ,

b = qn + a, a = bmodn. That's how our professor wrote it in the form as atleast. Now i'm confused.


I also really don't see how r1 + r2 - r = n(q3 - q2 - q1) proves it. Like I just don't see what this result means in relation to [itex]a-b=qn[/itex]. Like it's telling us to prove (a+b)modn = (amodn+bmodn)modn. I'm just not seeing the connection
 
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