4 ms·

> Back in school, I had to remember a few radicals to use them as needed. √13 was one of them. I have forgotten them by now. We also had to learn the algorithm

> Back in school, I had to remember a few radicals to use them as needed. √13 was one of them. I have forgotten them by now. We also had to learn the algorithm to compute any √ as needed That’s very interesting. In Poland we also had problems of this type, but we were taught to solve them by simply bringing everything under the square root and comparing the numbers. Definitely resulted in much bigger numbers but meant less memorization x)

georgecmu 3y ago
The comment for that particular problem also caught my eye. There's absolutely no need to remember radicals to solve it; just square both sides and compare.
paganel 3y ago
I'm about seven or eight years older compared to the poster (am also Romanian) and most certainly we didn't need to memorise √13, this is actually the first time I hear about it. I think (and now that I've checked I'm 100% sure) that we knew by heart the first two decimals of square root of 2, meaning 1.41, and possibly Euler's number (2.71) too, even though I'm not so sure about that anymore (this was 25+ years ago).
crq-yml 3y ago
That problem can be solved with estimation of the roots on a number line. It's not the most precise way of calculating it but when visualized you can clearly see that the root of 13 will have a fractional part above .5, while the root of 2 will have a fraction below .5.
- lupire 3y ago
  Did you actually solve it? 5 times <1.5 is <7.5, and 2 times >3.5 = >7 so "more or less than .5" is insufficient for this comparison.
  - lupire 3y ago
    You can solve it, however, by knowing that 5√2 < 5 * 1.45 = 7.2 (either because √2 = 1.414... is well known, or 1.45^2 = 2.125 is easy because 1.45^2 = 1.41.5 + .0025) 2√13 > 2√12.96 = 2 * 3.6 = 7.2 (36^2 = 6^4 = 1296 is a common fact from dice probability.) P.S: 0.45^2 = .2025 and 1.45 = 2.125 is an amusing coincidence.

georgecmu 3y ago

The comment for that particular problem also caught my eye. There's absolutely no need to remember radicals to solve it; just square both sides and compare.

paganel 3y ago

I'm about seven or eight years older compared to the poster (am also Romanian) and most certainly we didn't need to memorise √13, this is actually the first time I hear about it. I think (and now that I've checked I'm 100% sure) that we knew by heart the first two decimals of square root of 2, meaning 1.41, and possibly Euler's number (2.71) too, even though I'm not so sure about that anymore (this was 25+ years ago).

crq-yml 3y ago

That problem can be solved with estimation of the roots on a number line. It's not the most precise way of calculating it but when visualized you can clearly see that the root of 13 will have a fractional part above .5, while the root of 2 will have a fraction below .5.

lupire 3y ago
Did you actually solve it? 5 times <1.5 is <7.5, and 2 times >3.5 = >7 so "more or less than .5" is insufficient for this comparison.
- lupire 3y ago
  You can solve it, however, by knowing that 5√2 < 5 * 1.45 = 7.2 (either because √2 = 1.414... is well known, or 1.45^2 = 2.125 is easy because 1.45^2 = 1.41.5 + .0025) 2√13 > 2√12.96 = 2 * 3.6 = 7.2 (36^2 = 6^4 = 1296 is a common fact from dice probability.) P.S: 0.45^2 = .2025 and 1.45 = 2.125 is an amusing coincidence.