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Exercise 2.8 (Solutions)

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ity element. d- For $a\in {{\mathbb{Q}}^{+}}$, $\dfrac{1}{a}\in {{\mathbb{Q}}^{+}}$ such that $a\times \dfrac{1}{a}=\dfrac{1}{a}\times a=1$. Thus inverse of $a$ is $\dfrac{1}{a}$. Hence ${{\mathbb{Q}}^{+}}$ is group under addi