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cupoosta [38]
3 years ago
15

Explain whether the following given sets are closed under addition:

Mathematics
1 answer:
Ivan3 years ago
7 0

Answer:

Step-by-step explanation:

a) Yes.

\forall\ a,b\ \in\ B,\ a+b\in B\ (since\ 0+0=0)

b) Yes

W=\{0,1,2,3,..,\}\ (whole\ numbers)\\\forall\ a,b\ \in\ 3W,\ a+b\in 3W\\since:\\a\ \in\ 3W \Longrightarrow\ \exists\ k_1\in W | a=3*k_1\\\\b\ \in\ 3W \Longrightarrow\ \exists\ k_2\in W | b=3*k_2\\\\a+b=3*k_1+3*k_2=3*(k_1+k_2) \in 3W\\

c) Yes

\forall n\ \in\ \mathbb{N}: \Longrightarrow\ n+1 \in\ \mathbb{N}\\

d) No

3\in V , 3+3=6\notin V\\

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\text{ Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }

<em><u>Solution:</u></em>

From given question,

Number of pounds Jake carry = 6\frac{1}{4} \text{ pounds }

Number of pounds his father carry is 1\frac{5}{7} times as much as jake

To find: Number of pounds Jake father can carry

<em><u>Convert the mixed fractions to improper fractions</u></em>

Multiply the whole number part by the fraction's denominator.

Add that to the numerator.

Then write the result on top of the denominator

\rightarrow 6\frac{1}{4} = \frac{4 \times 6+1}{4} = \frac{25}{4}\\\\\rightarrow 1\frac{5}{7} = \frac{7 \times 1+5}{7} = \frac{12}{7}

<em><u>Then according to question,</u></em>

\text{Pounds father carry } = \frac{12}{7} \text{ times the pounds jake carry }\\\\\text{Pounds father carry } = \frac{12}{7} \times \frac{25}{4}\\\\\text{Pounds father carry } = \frac{75}{7}\\\\\text{In terms of mixed fractions, }\\\\\text{Pounds father carry } = \frac{75}{7} = 10\frac{5}{7}

\text{ Thus, Jake father can carry } 10\frac{5}{7} \text{ or } \frac{75}{7} \text{ pounds }

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