Question

Let G= a∈ ℚ ∪ {0}. Is a subgroup of (ℚ, +)? Explain

Answer 1

Yes, ( G ) is a subgroup of
`\( (\mathbb{Q}, +) \)`.

Step-by-step explanation:

The set ( G ) is defined as the set of rational numbers ( a ) such that the absolute value of ( a ) is greater than or equal to 1, union {0}. To show that ( G ) is a subgroup of
`\( (\mathbb{Q}, +) \)`, we need to verify the three subgroup criteria: closure under addition, identity element, and inverses.

Firstly, consider any two elements ( a, b ) in ( G ). Since ( |a| geq 1 ) and
`\( |b| \geq 1 \)`, it follows that
`\( |a + b| \geq |a| + |b| \geq 2 \)`. Therefore, ( a + b ) is also in ( G ), satisfying the closure under addition criterion.

Secondly, the identity element in
`\( (\mathbb{Q}, +) \)` is 0, and 0 is included in ( G ). Thus, ( G ) contains the identity element.

Finally, for any ( a ) in ( G ),
`\( |a| \geq 1 \)`, and
`\( |-a| = |a| \geq 1 \)`as well. Therefore, the inverse of ( a ) (which is ( -a )) is also in ( G ), meeting the inverses criterion.

In conclusion, ( G ) satisfies all three subgroup criteria, confirming that it is indeed a subgroup of
`\( (\mathbb{Q}, +) \)`.

Full Question:

Is
`\( G = \{ a \in \mathbb{Q} \mid |a| \geq 1 \} \cup \{0\} \) a subgroup of \( (\mathbb{Q}, +) \)`? Explain.