Number Theory Theorems: Understanding Divisibility Rules

In summary, the theorems being discussed state that if a number divides both b and c, then it also divides their sum when multiplied by any integers x and y. These theorems are similar, but care must be taken when choosing letters to avoid confusion and ensure a clear proof. However, the theorems have been successfully proven by showing that the sum of a number multiplied by integers x and y can be expressed as a multiple of the original number.
  • #1
chimath35
110
0
So could someone please clarify these:

a|b and a|c then a|bx+cy for any x,y integers?

a|b and b|c then a|bx+cy for any x,y integers?

seems the two are very similar, but are those both theorems?
 
Physics news on Phys.org
  • #2
chimath35 said:
So could someone please clarify these:

a|b and a|c then a|bx+cy for any x,y integers?

a|b and b|c then a|bx+cy for any x,y integers?

seems the two are very similar, but are those both theorems?

Yes, they are both theorems. You should try to prove them using the definition of 'divides'.
 
  • Like
Likes 1 person
  • #3
al=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
ae=alx+ady
ae=a(lx+dy)
 
Last edited:
  • #4
al=b
bd=c
ae=bx+cy
ae=alx+bdy
ae=alx+aldy
ae=a(lx+ldy)
 
Last edited:
  • #5
so both of these big E's are integers from the laws of addition and multiplication seeing that all letters here are representing integers
 
  • #6
chimath35 said:
ac=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
a=acx+ady
ae=a(cx+dy)

Sort of. You've got the right idea. But if you are trying to prove a|b and a|c then a|bx+cy for any x,y integers, you are overusing some letters. a|b means pa=b for some integer p and a|c means qa=c for some integer q. No reason to think p=c. Try and present the proofs again without that flaw. It's the same proof really.
 
  • Like
Likes 1 person
  • #7
Dick said:
Sort of. You've got the right idea. But if you are trying to prove a|b and a|c then a|bx+cy for any x,y integers, you are overusing some letters. a|b means pa=b for some integer p and a|c means qa=c for some integer q. No reason to think p=c. Try and present the proofs again without that flaw. It's the same proof really.

I never said p=c.
 
  • #8
chimath35 said:
I never said p=c.

Not in so many words, no. But you said given a|b and a|c means ac=b. a|b means a*(something)=b. That something doesn't have to be c. c already has a meaning in the problem. It's just sloppy.
 
  • Like
Likes 1 person
  • #9
On the second proof I had to use many letters so I could factor out a. It would not have worked otherwise. Show me if there is a simpler way please. I think I had to keep c the same as it is important.
 
  • #10
I mean this proof is correct and complete right?
 
  • #11
chimath35 said:
On the second proof I had to use many letters so I could factor out a. It would not have worked otherwise. Show me if there is a simpler way please. I think I had to keep c the same as it is important.

There are 26 letters you can use. When you want to say a|b means (something)*a=b just use a letter for the (something) that's not already used in the problem. The spirit of your proofs is correct. They could be marked incorrect if you don't do that.
 
  • Like
Likes 1 person
  • #12
Okay thanks please point out where I did that as I tried to be very careful not to use the same letter twice for anything.
 
  • #13
Sorry never mind, I see what you mean. Don't use that c I get it.
 
  • #14
That is suppose to be a common error when learning proofs I see now.
 
  • #15
chimath35 said:
Okay thanks please point out where I did that as I tried to be very careful not to use the same letter twice for anything.

Ok, if you want to prove a|b and a|c then a|bx+cy for any x,y integers, and you want to start by saying that a|b means b=a*(something) don't use a, b, c, x or y for the something. If you do, it's going to confuse someone reading it. I promise.
 
  • Like
Likes 1 person
  • #16
al=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
ae=alx+ady
ae=a(lx+dy)
al=b
bd=c
ae=bx+cy
ae=alx+bdy
ae=alx+aldy
ae=a(lx+ldy)
 
  • #17
Well thanks, I now learned even more not to use a letter that will be used at all in the proof.
 
  • #18
chimath35 said:
al=b
ad=c
ae=bx+cy
we can call bx+cy big E
and here it shows big E equal to a times an integer
ae=alx+ady
ae=a(lx+dy)



al=b
bd=c
ae=bx+cy
ae=alx+bdy
ae=alx+aldy
ae=a(lx+ldy)

I would skip the big E thing because I don't know what it means but yes, you've shown bx+cy=a*(lx+dy) in the first case and bx+cy=a*(lx+ldy) in the second. So in both cases a|(cx+dy). You've proved the theorems. Well done.
 
  • Like
Likes 1 person

FAQ: Number Theory Theorems: Understanding Divisibility Rules

What is number theory and why is it important?

Number theory is a branch of mathematics that studies the properties and relationships of integers. It is important because it has many real-world applications, such as in cryptography, coding theory, and computer science.

What are divisibility rules and why are they useful?

Divisibility rules are shortcuts or tricks that help determine if a number is divisible by another number without actually performing long division. They are useful because they can save time and mental effort when working with large numbers.

How do you use divisibility rules to determine if a number is divisible by another number?

To use a divisibility rule, you need to know the rule for the specific divisor (the number you are dividing by). Then, you apply the rule to the digits of the number you are trying to divide. If the result of applying the rule is true, then the number is divisible by the divisor.

What are some common divisibility rules?

Some common divisibility rules include: a number is divisible by 2 if its last digit is even, a number is divisible by 3 if the sum of its digits is divisible by 3, a number is divisible by 4 if the last two digits form a number divisible by 4, and a number is divisible by 9 if the sum of its digits is divisible by 9.

How can understanding divisibility rules help in problem-solving?

Understanding divisibility rules can help in problem-solving by making it easier to determine if a number is divisible by another number. This can be especially helpful in more complex mathematical problems, such as finding common multiples or factors of numbers.

Similar threads

Solving these Simultaneous Equations
Replies
3
Views
2K
Geodesic on a sphere and on a plane in 2D
Replies
13
Views
765
Proving the Division Property of Prime Numbers in Positive Integers
Replies
3
Views
2K
Mean Value Theorem: Proof & Claim
Replies
11
Views
2K
Taylor's Theorem for finding error for Taylor expansion
Replies
10
Views
919
Calculate this Integral around the Circular Path using Green's Theorem
Replies
3
Views
1K
Simple geometry problem -- Find the perimeter of the triangle ABC
Replies
7
Views
2K
Finding Integer Solutions to Polynomial Equations: Can it be Done Easily?
Replies
9
Views
2K
Prove ##(a+b)\cdot c=a\cdot c+b\cdot c## using Peano postulates
Replies
3
Views
1K
Quadratics Having a Common Tangent
Replies
1
Views
639

Hot Threads

Recent Insights

Back
Top