Use the ideas of sets and the definition of the order relation for the whole numbers to justify the transitive property of "less than." That is, if a, b, and c are whole numbers seltistying $$ a

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Let $$ a, b, c $$ be whole numbers. One way to define the order relation is as follows: we say that
$$
a < b
$$
if there exist sets $$ A $$ and $$ B $$ for which 
$$
n(A) = a, \quad n(B) = b,
$$
and such that
$$
A \subset B.
$$
Similarly, if
$$
b < c,
$$
then there exist sets $$ B $$ and $$ C $$ with 
$$
n(B) = b, \quad n(C) = c,
$$
and 
$$
B \subset C.
$$

Now, if we choose sets $$ A, B, C $$ satisfying 
$$
A \subset B \subset C,
$$
then by the transitivity of set inclusion, it follows that
$$
A \subset C.
$$
Since the combination of $$ A \subset C $$ with 
$$
n(A) = a \quad \text{and} \quad n(C) = c,
$$
gives the definition $$ a < c $$, the transitive property is justified.

Thus, the correct answer is:

D. Since $$ aThe transitive property of "less than" is justified by the following: - If $$ a < b $$ and $$ b < c $$, then by choosing sets $$ A, B, C $$ such that $$ A \subset B \subset C $$, it follows that $$ a < c $$. Therefore, the correct answer is: D. Since $$ a < b $$ and $$ b < c $$, choose sets $$ A, B, C $$ satisfying $$ A \subset B \subset C $$ and $$ n(A) = a $$, $$ n(B) = b $$, and $$ n(C) = c $$. Since $$ A \subset B \subset C $$, then $$ a < c $$.

Use the ideas of sets and the definition of the order relation for the whole numbers to justify the transitive property of "less than." That is, if a, b, and c are whole numbers seltistying \( a

Ask by Salinas Gray. in the United States

Mar 16,2025

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Use the ideas of sets and the definition of the order relation for the whole numbers to justify the transitive property of "less than." That is, if a, b, and c are whole numbers seltistying \( a Ask by Salinas Gray. in the United States Mar 16,2025