Get expert insights and community support for your questions on IDNLearn.com. Discover comprehensive answers from knowledgeable members of our community, covering a wide range of topics to meet all your informational needs.
Sagot :
To determine which expression must be an odd integer given that [tex]\(m+I\)[/tex] is an even integer, let us analyze each option step-by-step.
1. Option E: [tex]\(m - 1\)[/tex]:
- Given that [tex]\(m + I\)[/tex] is even, let [tex]\(m + I = 2k\)[/tex] for some integer [tex]\(k\)[/tex].
- Substituting [tex]\(m = 2k - I\)[/tex], we get:
[tex]\[ m - 1 = (2k - I) - 1 \][/tex]
- The parity (odd or even nature) of [tex]\(m - 1\)[/tex] depends on [tex]\(I\)[/tex]. Since [tex]\(I\)[/tex] is not specified, [tex]\(m - 1\)[/tex] can either be odd or even depending on the value of [tex]\(I\)[/tex].
2. Option F: [tex]\(2m - 2\)[/tex]:
- Using [tex]\(m = 2k - I\)[/tex], we calculate:
[tex]\[ 2m - 2 = 2(2k - I) - 2 = 4k - 2I - 2 \][/tex]
- Since [tex]\(4k - 2I - 2\)[/tex] is always even (product and difference of even integers), [tex]\(2m - 2\)[/tex] is always even and therefore cannot be an odd integer.
3. Option G: [tex]\(2m + 1\)[/tex]:
- Substituting [tex]\(m = 2k - I\)[/tex], compute:
[tex]\[ 2m + 1 = 2(2k - I) + 1 = 4k - 2I + 1 \][/tex]
- [tex]\(4k - 2I\)[/tex] is an even number since [tex]\(4k\)[/tex] and [tex]\(-2I\)[/tex] are both even. Adding 1 to an even number gives us an odd number.
- Therefore, [tex]\(2m + 1\)[/tex] is always odd.
4. Option H: [tex]\(2m + 2\)[/tex]:
- Substituting [tex]\(m = 2k - I\)[/tex], we get:
[tex]\[ 2m + 2 = 2(2k - I) + 2 = 4k - 2I + 2 \][/tex]
- Since [tex]\(4k - 2I + 2\)[/tex] is always even (sum of even numbers), [tex]\(2m + 2\)[/tex] is always even and hence cannot be an odd integer.
Among the provided options, only [tex]\(2m + 1\)[/tex] must be an odd integer.
Therefore, the correct answer is:
[tex]\[ \boxed{G} \][/tex]
1. Option E: [tex]\(m - 1\)[/tex]:
- Given that [tex]\(m + I\)[/tex] is even, let [tex]\(m + I = 2k\)[/tex] for some integer [tex]\(k\)[/tex].
- Substituting [tex]\(m = 2k - I\)[/tex], we get:
[tex]\[ m - 1 = (2k - I) - 1 \][/tex]
- The parity (odd or even nature) of [tex]\(m - 1\)[/tex] depends on [tex]\(I\)[/tex]. Since [tex]\(I\)[/tex] is not specified, [tex]\(m - 1\)[/tex] can either be odd or even depending on the value of [tex]\(I\)[/tex].
2. Option F: [tex]\(2m - 2\)[/tex]:
- Using [tex]\(m = 2k - I\)[/tex], we calculate:
[tex]\[ 2m - 2 = 2(2k - I) - 2 = 4k - 2I - 2 \][/tex]
- Since [tex]\(4k - 2I - 2\)[/tex] is always even (product and difference of even integers), [tex]\(2m - 2\)[/tex] is always even and therefore cannot be an odd integer.
3. Option G: [tex]\(2m + 1\)[/tex]:
- Substituting [tex]\(m = 2k - I\)[/tex], compute:
[tex]\[ 2m + 1 = 2(2k - I) + 1 = 4k - 2I + 1 \][/tex]
- [tex]\(4k - 2I\)[/tex] is an even number since [tex]\(4k\)[/tex] and [tex]\(-2I\)[/tex] are both even. Adding 1 to an even number gives us an odd number.
- Therefore, [tex]\(2m + 1\)[/tex] is always odd.
4. Option H: [tex]\(2m + 2\)[/tex]:
- Substituting [tex]\(m = 2k - I\)[/tex], we get:
[tex]\[ 2m + 2 = 2(2k - I) + 2 = 4k - 2I + 2 \][/tex]
- Since [tex]\(4k - 2I + 2\)[/tex] is always even (sum of even numbers), [tex]\(2m + 2\)[/tex] is always even and hence cannot be an odd integer.
Among the provided options, only [tex]\(2m + 1\)[/tex] must be an odd integer.
Therefore, the correct answer is:
[tex]\[ \boxed{G} \][/tex]
Thank you for contributing to our discussion. Don't forget to check back for new answers. Keep asking, answering, and sharing useful information. Thank you for choosing IDNLearn.com. We’re dedicated to providing clear answers, so visit us again for more solutions.