I am not quite sure what you want me to do with post # 14.
Arithmetic is about learning mechanical procedures and which procedure to use in what situations. Justifying those procedures is frequently more easily done by considering problems using algebra as Dr. Peterson said almost at the start of this thread.
[MATH]a = \text {number of rupees going to Person A.}[/MATH]
[MATH]b = \text {number of rupees going to Person B.}[/MATH]
According to the problem, we are dividing 12 rupees between A and B and the ratio is 4:2.
[MATH]\therefore a + b = 12 \text { and } \dfrac{a}{b} = \dfrac{4}{2}.[/MATH]
That translates the problem into algebraic language.
[MATH]\dfrac{a}{b} = \dfrac{4}{2} \implies a = b * \dfrac{4}{2} \implies 2a = 4b \implies a = 2b.[/MATH]
[MATH]\therefore a + b = 12 \implies 2b + b = 12 \implies 3b = 12 \implies b = 4 \implies a = 8.[/MATH]
Notice that 8/4 = 4/2 and 8 + 4 = 12.
Notice that [MATH]b = 4 \text { because } b = \dfrac{3b}{3} = \dfrac{12}{3} = 2 * \dfrac{12}{6}.[/MATH]
There is no philosophy involved. A very easy problem in algebra can be also be solved in arithmetic by using a procedure that is usually taught by rote because justifying it in general is hard to do without mathematics more advanced than arithmetic. In this case, the procedure is made even more mysterious by failing to state the ratio in lowest terms.
It is, however, possible to show BY ALGEBRA and MATHEMATICAL INDUCTION that the arithmetic procedure will always give the answer easily demonstrated by algebra.
Arithmetic is about learning mechanical procedures and which procedure to use in what situations. Justifying those procedures is frequently more easily done by considering problems using algebra as Dr. Peterson said almost at the start of this thread.
[MATH]a = \text {number of rupees going to Person A.}[/MATH]
[MATH]b = \text {number of rupees going to Person B.}[/MATH]
According to the problem, we are dividing 12 rupees between A and B and the ratio is 4:2.
[MATH]\therefore a + b = 12 \text { and } \dfrac{a}{b} = \dfrac{4}{2}.[/MATH]
That translates the problem into algebraic language.
[MATH]\dfrac{a}{b} = \dfrac{4}{2} \implies a = b * \dfrac{4}{2} \implies 2a = 4b \implies a = 2b.[/MATH]
[MATH]\therefore a + b = 12 \implies 2b + b = 12 \implies 3b = 12 \implies b = 4 \implies a = 8.[/MATH]
Notice that 8/4 = 4/2 and 8 + 4 = 12.
Notice that [MATH]b = 4 \text { because } b = \dfrac{3b}{3} = \dfrac{12}{3} = 2 * \dfrac{12}{6}.[/MATH]
There is no philosophy involved. A very easy problem in algebra can be also be solved in arithmetic by using a procedure that is usually taught by rote because justifying it in general is hard to do without mathematics more advanced than arithmetic. In this case, the procedure is made even more mysterious by failing to state the ratio in lowest terms.
It is, however, possible to show BY ALGEBRA and MATHEMATICAL INDUCTION that the arithmetic procedure will always give the answer easily demonstrated by algebra.