Let m and n be positive integers. Show that mZ+nZ= (m, n)Z.
Advanced Engineering Mathematics
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ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
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Could you explain how to show 9.4 in detail? I included list of theorems and definitions from the textbook.
![**Exercise 9.4**
Let \( m \) and \( n \) be positive integers. Show that:
\[ m\mathbb{Z} + n\mathbb{Z} = (m, n)\mathbb{Z} \]
In this problem, you are asked to demonstrate that the sum of the integer multiples of \( m \) and \( n \) is equivalent to the multiples of their greatest common divisor, denoted as \( (m, n)\mathbb{Z} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2Fc1632137-27f6-4e5f-9221-e2cfbf8a1324%2Fthb3j5f_processed.png&w=3840&q=75)
Transcribed Image Text:**Exercise 9.4**
Let \( m \) and \( n \) be positive integers. Show that:
\[ m\mathbb{Z} + n\mathbb{Z} = (m, n)\mathbb{Z} \]
In this problem, you are asked to demonstrate that the sum of the integer multiples of \( m \) and \( n \) is equivalent to the multiples of their greatest common divisor, denoted as \( (m, n)\mathbb{Z} \).
![**Definition 9.1.** Let \( R \) be a ring. Then a subring \( I \) of \( R \) is said to be an ideal if \( ir, ri \in I \) for all \( i \in I \) and \( r \in R \). We call this the absorption property.
**Theorem 9.1.** Let \( R \) be a ring and \( I \) a subset of \( R \). Then \( I \) is an ideal if and only if
1. \( 0 \in I \);
2. \( i - j \in I \) for all \( i, j \in I \); and
3. \( ir, ri \in I \) for all \( i \in I, r \in R \).
**Example 9.1.** Let \( n \) be any integer. Then \( n\mathbb{Z} \) is an ideal of \( \mathbb{Z} \). Indeed, we already know that it is a subring. But also, if \( nk \in n\mathbb{Z} \), then for any integer \( r, r(nk) = n(rk) \in n\mathbb{Z} \).
**Example 9.2.** Let \( I \) be the set of all polynomials \( f(x) \in \mathbb{R}[x] \) such that \( f(0) = 0 \). We claim that \( I \) is an ideal in \( \mathbb{R}[x] \). Certainly \( I \) contains the zero polynomial. Also, if \( f(0) = g(0) = 0 \), then \( (f-g)(0) = f(0) - g(0) = 0 \), hence \( f(x) - g(x) \in I \). Also, if \( f(0) = 0 \) and \( h(x) \in \mathbb{R}[x] \), then \( h(0)f(0) = h(0)0 = 0 \). Hence, \( h(x)f(x) \in I \).
**Example 9.3.** Let \( I \) be the set of all polynomials in \( \mathbb{Z}[x] \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe15ed467-90ec-4e60-afef-3d3f6119f74d%2Fc1632137-27f6-4e5f-9221-e2cfbf8a1324%2F2z302i_processed.png&w=3840&q=75)
Transcribed Image Text:**Definition 9.1.** Let \( R \) be a ring. Then a subring \( I \) of \( R \) is said to be an ideal if \( ir, ri \in I \) for all \( i \in I \) and \( r \in R \). We call this the absorption property.
**Theorem 9.1.** Let \( R \) be a ring and \( I \) a subset of \( R \). Then \( I \) is an ideal if and only if
1. \( 0 \in I \);
2. \( i - j \in I \) for all \( i, j \in I \); and
3. \( ir, ri \in I \) for all \( i \in I, r \in R \).
**Example 9.1.** Let \( n \) be any integer. Then \( n\mathbb{Z} \) is an ideal of \( \mathbb{Z} \). Indeed, we already know that it is a subring. But also, if \( nk \in n\mathbb{Z} \), then for any integer \( r, r(nk) = n(rk) \in n\mathbb{Z} \).
**Example 9.2.** Let \( I \) be the set of all polynomials \( f(x) \in \mathbb{R}[x] \) such that \( f(0) = 0 \). We claim that \( I \) is an ideal in \( \mathbb{R}[x] \). Certainly \( I \) contains the zero polynomial. Also, if \( f(0) = g(0) = 0 \), then \( (f-g)(0) = f(0) - g(0) = 0 \), hence \( f(x) - g(x) \in I \). Also, if \( f(0) = 0 \) and \( h(x) \in \mathbb{R}[x] \), then \( h(0)f(0) = h(0)0 = 0 \). Hence, \( h(x)f(x) \in I \).
**Example 9.3.** Let \( I \) be the set of all polynomials in \( \mathbb{Z}[x] \)
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