hobz Posted June 22, 2010 Share 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? Link to comment Share on other sites More sharing options...
Dave Posted June 22, 2010 Share 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. Link to comment Share on other sites More sharing options...
hobz Posted June 22, 2010 Author Share 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? Link to comment Share on other sites More sharing options...
the tree Posted June 22, 2010 Share 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. Link to comment Share on other sites More sharing options...
hobz Posted June 22, 2010 Author Share Posted June 22, 2010 Ah yes. Of course. Thank you both. Link to comment Share on other sites More sharing options...
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