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Exercise 1.1 (Solutions) @fsc-part1-ptb:sol:ch01
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\hline \end{array} \] As $0+0=0 \in \{0\}$. This implies $\{0\}$ has closure property w.r.t. '+'. Multipl... \end{array} \] As $0\times 0=0 \in \{0\}$ This implies $\{0\}$ has closure property w.r.t. '$\times$'. ... ne \end{array} \] As $1+1=2 \notin \{1\}$. This implies $\{1\}$ does not satisfy closure property w.r.t. ... e \end{array} \] As $1\times 1=1 \in \{1\}$. This implies $\{1\}$ has closure property w.r.t. '$\times$' *
Definitions: FSc Part 1 (Mathematics): PTB by Aurang Zaib
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A = \{x \ | \ x \) is any natural number\}. This implies that \( A \) consists of all natural numbers. ==