Math / Numbers & Operations

Question4. Assuming that $\operatorname{gcd}(a, b)=1$ , prove the following: (b) $\operatorname{gcd}(2 a+b, a+2 b)=1$ or 3 .

Studdy Solution

STEP 1

1. We are given that $\operatorname{gcd}(a, b) = 1$ .
2. We need to prove that $\operatorname{gcd}(2a + b, a + 2b)$ is either $1$ or $3$ .

STEP 2

1. Use the properties of gcd.
2. Apply the Euclidean algorithm.
3. Conclude the proof by considering possible gcd values.

STEP 3

Since $\operatorname{gcd}(a, b) = 1$ , $a$ and $b$ are coprime. This implies that any linear combination of $a$ and $b$ will also have gcd that divides both coefficients of the linear combination.

STEP 4

Apply the Euclidean algorithm to find $\operatorname{gcd}(2a + b, a + 2b)$ .
First, express $2a + b$ and $a + 2b$ in terms of each other:
$\operatorname{gcd}(2a + b, a + 2b) = \operatorname{gcd}(a + 2b, (2a + b) - 2(a + 2b))$
Simplify the expression:
$= \operatorname{gcd}(a + 2b, 2a + b - 2a - 4b)$ $= \operatorname{gcd}(a + 2b, b - 4b)$ $= \operatorname{gcd}(a + 2b, -3b)$

STEP 5

Since $\operatorname{gcd}(a, b) = 1$ , and $-3b$ is a multiple of $b$ , we have:
$\operatorname{gcd}(a + 2b, -3b) = \operatorname{gcd}(a + 2b, 3)$

STEP 6

Conclude the proof by considering the possible gcd values. Since $\operatorname{gcd}(a + 2b, 3)$ must divide 3, the possible values are 1 or 3.
Thus, $\operatorname{gcd}(2a + b, a + 2b) = 1$ or $3$ .
The proof is complete. The gcd is either $1$ or $3$ .

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Question4. Assuming that $\operatorname{gcd}(a, b)=1$ , prove the following: (b) $\operatorname{gcd}(2 a+b, a+2 b)=1$ or 3 .

Studdy Solution

STEP 1

1. We are given that $\operatorname{gcd}(a, b) = 1$ .
2. We need to prove that $\operatorname{gcd}(2a + b, a + 2b)$ is either $1$ or $3$ .

STEP 2

1. Use the properties of gcd.
2. Apply the Euclidean algorithm.
3. Conclude the proof by considering possible gcd values.

STEP 3

Since $\operatorname{gcd}(a, b) = 1$ , $a$ and $b$ are coprime. This implies that any linear combination of $a$ and $b$ will also have gcd that divides both coefficients of the linear combination.

STEP 4

STEP 5

Since $\operatorname{gcd}(a, b) = 1$ , and $-3b$ is a multiple of $b$ , we have:
$\operatorname{gcd}(a + 2b, -3b) = \operatorname{gcd}(a + 2b, 3)$

STEP 6

Conclude the proof by considering the possible gcd values. Since $\operatorname{gcd}(a + 2b, 3)$ must divide 3, the possible values are 1 or 3.
Thus, $\operatorname{gcd}(2a + b, a + 2b) = 1$ or $3$ .
The proof is complete. The gcd is either $1$ or $3$ .

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Question4. Assuming that gcd⁡(a,b)=1\operatorname{gcd}(a, b)=1gcd(a,b)=1, prove the following: (b) gcd⁡(2a+b,a+2b)=1\operatorname{gcd}(2 a+b, a+2 b)=1gcd(2a+b,a+2b)=1 or 3 .

Studdy Solution

STEP 1

1. We are given that gcd⁡(a,b)=1\operatorname{gcd}(a, b) = 1gcd(a,b)=1.2. We need to prove that gcd⁡(2a+b,a+2b)\operatorname{gcd}(2a + b, a + 2b)gcd(2a+b,a+2b) is either 111 or 333.

STEP 2

1. Use the properties of gcd.2. Apply the Euclidean algorithm.3. Conclude the proof by considering possible gcd values.

STEP 3

Since gcd⁡(a,b)=1\operatorname{gcd}(a, b) = 1gcd(a,b)=1, aaa and bbb are coprime. This implies that any linear combination of aaa and bbb will also have gcd that divides both coefficients of the linear combination.

STEP 4

STEP 5

Since gcd⁡(a,b)=1\operatorname{gcd}(a, b) = 1gcd(a,b)=1, and −3b-3b−3b is a multiple of bbb, we have:gcd⁡(a+2b,−3b)=gcd⁡(a+2b,3)\operatorname{gcd}(a + 2b, -3b) = \operatorname{gcd}(a + 2b, 3)gcd(a+2b,−3b)=gcd(a+2b,3)

STEP 6

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Question4. Assuming that $\operatorname{gcd}(a, b)=1$ , prove the following: (b) $\operatorname{gcd}(2 a+b, a+2 b)=1$ or 3 .

1. We are given that $\operatorname{gcd}(a, b) = 1$ .
2. We need to prove that $\operatorname{gcd}(2a + b, a + 2b)$ is either $1$ or $3$ .

1. Use the properties of gcd.
2. Apply the Euclidean algorithm.
3. Conclude the proof by considering possible gcd values.

Since $\operatorname{gcd}(a, b) = 1$ , $a$ and $b$ are coprime. This implies that any linear combination of $a$ and $b$ will also have gcd that divides both coefficients of the linear combination.

Since $\operatorname{gcd}(a, b) = 1$ , and $-3b$ is a multiple of $b$ , we have:
$\operatorname{gcd}(a + 2b, -3b) = \operatorname{gcd}(a + 2b, 3)$