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I completely disagree. First of all, at the time children learn the multiplication table, they definitely don't know the concept of exponentiation. Secondly, 7*

I completely disagree. First of all, at the time children learn the multiplication table, they definitely don't know the concept of exponentiation. Secondly, 7*2*2*2 is not some immediately obvious shortcut. Also, learning multiplication with numbers higher than 10 still relies on knowing the multiplication table. 17*8 is 7*8=56, hold the 5, 1*8 + 5 = 13, so 136.

LouisSayers 2y ago
> Also, learning multiplication with numbers higher than 10 still relies on knowing the multiplication table You've actually just proved my point - you used a method of breaking down the problem into a different problem and then solving it rather than simply memorising. If you give the same question to multiple people there will be numerous ways different people use to go about solving it. As an example, I might solve this by doing 20*8 = 160 3*8 = 24 160 - 24 = 136 Or 10*8 = 80 7*8 = 56 80+56 = 136 And I might apply different tools like the one I originally mentioned within these calculations. I know that 80+20 is 100 and so "borrow" 20 from 56, so that I can easily add 100 and 36 together. These ways of calculating happen in your mind very quickly if this is how you get used to calculating.
- tsimionescu 2y ago
  Sure, but all of those work for numbers higher than 10, and all assume you know the multiplication table by heart. The multiplication table (the result of multiplying every number between 1 and 10 with each other) is something you have to memorize. You can get away with memorizing only some of these results and computing the others based on them, but it's basically impossible to do any more complex arithmetic if you don't know most of it by rote memorization.
lelanthran 2y ago
> Also, learning multiplication with numbers higher than 10 still relies on knowing the multiplication table. 17*8 is 7*8=56, hold the 5, 1*8 + 5 = 13, so 136. Interestingly, I do those less efficiently: 17 * 8 = (10 * 8) + (7 * 8) = (80) + (56) = (80) + (50 + 6) = 130 + 6 I think the reason I do it this way is because I get an approximation sooner when the numbers are very large i.e. I get the most significant digit first, and can stop calculating when I get the precision I require.*

LouisSayers 2y ago

> Also, learning multiplication with numbers higher than 10 still relies on knowing the multiplication table You've actually just proved my point - you used a method of breaking down the problem into a different problem and then solving it rather than simply memorising. If you give the same question to multiple people there will be numerous ways different people use to go about solving it. As an example, I might solve this by doing 20*8 = 160 3*8 = 24 160 - 24 = 136 Or 10*8 = 80 7*8 = 56 80+56 = 136 And I might apply different tools like the one I originally mentioned within these calculations. I know that 80+20 is 100 and so "borrow" 20 from 56, so that I can easily add 100 and 36 together. These ways of calculating happen in your mind very quickly if this is how you get used to calculating.

tsimionescu 2y ago
Sure, but all of those work for numbers higher than 10, and all assume you know the multiplication table by heart. The multiplication table (the result of multiplying every number between 1 and 10 with each other) is something you have to memorize. You can get away with memorizing only some of these results and computing the others based on them, but it's basically impossible to do any more complex arithmetic if you don't know most of it by rote memorization.

lelanthran 2y ago

> Also, learning multiplication with numbers higher than 10 still relies on knowing the multiplication table. 17*8 is 7*8=56, hold the 5, 1*8 + 5 = 13, so 136. Interestingly, I do those less efficiently: 17 * 8 = (10 * 8) + (7 * 8) = (80) + (56) = (80) + (50 + 6) = 130 + 6 I think the reason I do it this way is because I get an approximation sooner when the numbers are very large i.e. I get the most significant digit first, and can stop calculating when I get the precision I require.*