hobz Posted June 22, 2010 Posted June 22, 2010 To my knowledge: [math] 3 \cdot 5 = 5 + 5 + 5 [/math] However [math] 3 \, \mathrm{m} \cdot 5 \,\mathrm{m} = 15 \, \mathrm{m}^2 \neq 5 \,\mathrm{m} + 5 \,\mathrm{m} + 5 \,\mathrm{m} [/math] What is wrong?
Dave Posted June 22, 2010 Posted June 22, 2010 Simply that you've not done the multiplication right. [math] 3 \, \mathrm{m} \cdot 5 \,\mathrm{m} = (3\cdot 5) \mathrm{m}\cdot \mathrm{m} = (3\cdot 5) \mathrm{m}^2 [/math] and the rest follows.
hobz Posted June 22, 2010 Author Posted June 22, 2010 I don't think I have made my self clear. The problem is that if multiplication is repeated addition, how do I add the units to make them square?
the tree Posted June 22, 2010 Posted June 22, 2010 Well, crudely speaking: [math]3m \cdot 5m = (3\cdot 5) (m \cdot m) = \underbrace{(5 + 5 + 5)}_{ 3 \; \mbox{times} } \underbrace{(m + m \dots + m)}_{ m \; \mbox{times} } = 15m^2 [/math] Like Dave said, it's mostly just a case of doing the multiplication right.
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