# Prove that ## gcd(a, n)=gcd(b, n) ##.

• Math100
In summary, the greatest common divisor (gcd) of two integers, a and n, is the largest positive integer that divides both a and n without a remainder. This can be proven by repeatedly subtracting the smaller number from the larger number until they are equal, or by using prime factorization. Proving this statement is important as it helps us understand the relationship between two numbers and their gcd, and has many applications in various fields of mathematics.
Math100
Homework Statement
If ## a\equiv b \mod n ##, prove that ## gcd(a, n)=gcd(b, n) ##.
Relevant Equations
None.
Proof:

Suppose ## a\equiv b \mod n ##.
Then ## n\mid (a-b)\implies kn=a-b ## for some ## k\in\mathbb{Z} ##.
Let ## gcd(a, n)=d ## and ## gcd(b, n)=h ##.
Note that ## d\mid a \land d\mid n\implies d\mid (a-kn)\implies d\mid b ## and ## h\mid b \land h\mid n\implies h\mid (b+kn)\implies h\mid a ##.
Thus ## d\mid n \land d\mid b\implies d\mid gcd(b, n)\implies d\mid h ## and ## h\mid n \land h\mid a\implies h\mid gcd(a, n)\implies h\mid d ##.
Therefore, if ## a\equiv b \mod n ##, then ## gcd(a, n)=gcd(b, n) ##.

Math100 said:
Homework Statement:: If ## a\equiv b \mod n ##, prove that ## gcd(a, n)=gcd(b, n) ##.
Relevant Equations:: None.

Proof:

Suppose ## a\equiv b \mod n ##.
Then ## n\mid (a-b)\implies kn=a-b ## for some ## k\in\mathbb{Z} ##.
Let ## gcd(a, n)=d ## and ## gcd(b, n)=h ##.
Note that ## d\mid a \land d\mid n\implies d\mid (a-kn)\implies d\mid b ## and ## h\mid b \land h\mid n\implies h\mid (b+kn)\implies h\mid a ##.
Thus ## d\mid n \land d\mid b\implies d\mid gcd(b, n)\implies d\mid h ## and ## h\mid n \land h\mid a\implies h\mid gcd(a, n)\implies h\mid d ##.
Therefore, if ## a\equiv b \mod n ##, then ## gcd(a, n)=gcd(b, n) ##.
I'm not sure whether this is the most elegant way to write it, but it is correct. Maybe a few too many ##\Longrightarrow ##.A common technique is to prove only ##d\,|\,h## and state that ##h\,|\,d## for symmetry reasons.

Also, there is an error in all of it. Any statement about divisors is always only true up to units. Units are invertible numbers, integers in this case. The only units in the ring of integers are ##\pm 1.## Hence we cannot conclude equality unless we add the requirement that the greatest common divisor has to be positive, which might be the case, I don't know. But from ##d\,|\,h ## and ##h\,|\,d## alone, we only get ##d=\pm h.##

Suppose ## a\equiv b \mod n ##.
Then ## n\mid (a-b)\implies kn=a-b ## for some ## k\in\mathbb{Z} ##.
Let ## gcd(a, n)=d ## and ## gcd(b, n)=h ## where ## d ## is a positive integer.
Note that ## d\mid a\land d\mid n\implies d\mid (a-kn)\implies d\mid b ##.
Thus ## d\mid n\land d\mid b\implies d\mid gcd(b, n)\implies d\mid h ##.
Conversely, ## h\mid d ## by symmetry.
Since ## d\mid h ## and ## h\mid d ##,
it follows that ## d=h\implies gcd(a, n)=gcd(b, n) ##.
Therefore, if ## a\equiv b \mod n ##, then ## gcd(a, n)=gcd(b, n) ##.

fresh_42

## 1. What does "gcd" stand for in this context?

"gcd" stands for "greatest common divisor". It is a mathematical term used to refer to the largest number that divides both of the given numbers without leaving a remainder.

## 2. How do you prove that gcd(a, n)=gcd(b, n)?

To prove that gcd(a, n)=gcd(b, n), you can use the Euclidean algorithm. This algorithm involves repeatedly dividing the larger number by the smaller number until the remainder is 0. The last non-zero remainder will be the gcd of the two numbers.

## 3. Can you give an example of proving gcd(a, n)=gcd(b, n)?

Sure, let's say we want to prove that gcd(12, 18)=gcd(6, 18). Using the Euclidean algorithm, we have:
18 = 12 * 1 + 6
12 = 6 * 2 + 0
Therefore, the gcd of 12 and 18 is 6. Similarly, we can show that the gcd of 6 and 18 is also 6. Thus, gcd(12, 18)=gcd(6, 18).

## 4. What is the significance of proving gcd(a, n)=gcd(b, n)?

Proving that gcd(a, n)=gcd(b, n) can be useful in various mathematical proofs and applications. It allows us to make simplifications and solve problems involving fractions, divisibility, and modular arithmetic.

## 5. Are there any other methods for proving gcd(a, n)=gcd(b, n)?

Yes, there are other methods such as the prime factorization method and the extended Euclidean algorithm. However, the basic Euclidean algorithm is the most commonly used method for proving gcd(a, n)=gcd(b, n).

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