@@Nick-the-foxthat's what he said. In base 10 the prime factors happen to be 2 and 5. But in base 12, for example, the prime factors are 3 and 2, so 1/3 has a terminating decimal expansion in base 12
This is important when dealing with computers. Computers use binary so only fractions with twos in their prime factorialization terminate. Especially 1/100 does not terminate. This is important when dealing with money. On successive calculations errors may cumulate and change the cents. There are several ways to solve this. One can store the values in cents so they will be integers (mathematically). One can round the results. It is not necessary to do it to cents nor after every calculation, just often enough. One can also use a statem that uses binary codes decimal numbers. They behave like decimals and are more predictable. They still lose precision on things like 1/3.
it always bugs me when people say "irrational numbers dont repeat in their decimal expansion". well of course they can repeat sometimes. for example you might have "888" appear somewhere, or "753753" or something. really what i think people want to say is "the expansion never falls into an infinitely repeating tail of digits". of course, in context it makes sense, but i feel like people throw around that saying out of context and it can be confusing. for example, i remember when i first learned that the digits of pi go on forever, and someone told me they never repeat, and i remember being blown away because i thought they meant it in terms of some kind of penrose tiling thing where the digits somehow conspire to avoid having repeating strings. of course, thats impossible if you think about it for very long, but it did confuse me for awhile.
0:34 my first tought on this: Actually, 1/10 also repeats. 1/10=0.1000000... with 0 repeating (or 0.0999999... with 9 repeating) But there are some numbers that don't repeat. For example: decimal expansion for pi doesn't repeat. This is because pi is irrational. Rational numbers decimal expansion repeat, irrarional numbers do not.
You don't mention it; but of course this means when you use different base systems you get different sets of prime factors which changes which fractions terminate and which fractions repeat. More generally you could say: The expansion of any fraction with a GCF of 1 in any base system will terminate if the prime factors of the denominator are shared with the prime factors of the base, and will repeat if the denominator has prime factors not shared with the base. At least with integer bases.
for some rational number P/Q where P and Q share no factors, its expansion terminates in base B if and only if the diference of the Q and B's prime factors is empty. Basically the denominator of the number in simplest form cannot have prime factors which the base does not have
Question. When you explain the fact that in order for the decimal expansion to terminate the denominator of the fraction has to fully cancel out with some 10^n, that makes me wonder: is it possible to have a situation where b > 10^n (of course assuming that b also has only 2s and 5s in its factorization)? In that case, the 10^n will cancel out, but it will still leave some factors of b in the denominator. Can this situation ever happen? And if so, what happens?
It happens all the time, when you have inconsistent powers of 2 and 5 in the denominator. Consider 1/40. The number 40 has 3 copies of 2 and only 1 copy of 5 in its prime factorization. Multiply by 10^1, and we remove a single copy of each 2 and 5 in the denominator, leaving us with 1/4. The 10^1 will cancel, and we still have two copies of 2 in the denominator's prime factorization. Of course, this all can be corrected for, by picking a different value of n. Instead of picking n = 1, pick n = 3. 1000/40 = 25. Now we've eliminated all copies of 2 and 5, by picking the larger of the two exponents in the prime factorization, rather than the smaller of the two exponents. For all fractions of the form 1/(2^m * 5^n), it will always be possible to either multiply by 10^m or 10^n, whichever is larger, to eliminate the denominator completely, and thus the original fraction has an ending decimal, rather than a repeating decimal.
@@carultch Thank you! That makes complete sense. In the video he started the proof by assuming that the decimal expansion has n digits to begin with, and therefore he multiplied by 10^n to eliminate the denominator, but in an application you wouldn't know in advance how many digits the expansion has, unless you factorize the denominator first and retroactively choose an appropriate value for n. That's what confused me. In your example, therefore, by picking n=3, which is the smallest n for which the denominator is eliminated completely, we can see that the decimal expansion of 1/40 is 0.025 and it indeed has 3 digits after the decimal point, and terminates as expected. Now it's all clear!
my theory. don't know if it makes sense. For rational numbers, i saw in another way. If dividing a number by y, decimals repeat and go on forever, then it's trying to yield a infinite amount of 9s when multiplied by y. for example, for 1/7, in base 10, it finds it hard to yield exactly 1 without the decimals going on forever. it asks, what will yield 9s when multiplied by 7, and it has 142857. now, 0.142857 will yield 0.999999 when multiplied by 7. but when 142857 repeats infinitely, creating a repeating decimal that goes on forever, it can converge to infinite amount of 9s when multiplied by 7, which is mathematically equal to 1. same goes for others. even other bases.
Well, then the video is not directed at you. You sound like the type of person I wouldn't wanna be around. Okay you know it, so what? There are people who don't and that's why we need videos like this one so people that aren't so versed in the math field to feel welcomed to enter into it
Couple early episodes out now for channel members: ua-cam.com/play/PLztBpqftvzxXQDmPmSOwXSU9vOHgty1RO.html
I thought irrational nums dont have repeating deciaml expansons
It's because a terminating decimal is a fraction whose denominator uses a prime factor of the base.
@@Nick-the-foxthat's what he said. In base 10 the prime factors happen to be 2 and 5. But in base 12, for example, the prime factors are 3 and 2, so 1/3 has a terminating decimal expansion in base 12
This is important when dealing with computers. Computers use binary so only fractions with twos in their prime factorialization terminate. Especially 1/100 does not terminate. This is important when dealing with money. On successive calculations errors may cumulate and change the cents.
There are several ways to solve this. One can store the values in cents so they will be integers (mathematically). One can round the results. It is not necessary to do it to cents nor after every calculation, just often enough. One can also use a statem that uses binary codes decimal numbers. They behave like decimals and are more predictable. They still lose precision on things like 1/3.
it always bugs me when people say "irrational numbers dont repeat in their decimal expansion". well of course they can repeat sometimes. for example you might have "888" appear somewhere, or "753753" or something. really what i think people want to say is "the expansion never falls into an infinitely repeating tail of digits". of course, in context it makes sense, but i feel like people throw around that saying out of context and it can be confusing.
for example, i remember when i first learned that the digits of pi go on forever, and someone told me they never repeat, and i remember being blown away because i thought they meant it in terms of some kind of penrose tiling thing where the digits somehow conspire to avoid having repeating strings. of course, thats impossible if you think about it for very long, but it did confuse me for awhile.
Good point. Sounds pedantic at first, but then pedantry is essential in math.
Decimal expansion: infinite numbers
True, hope I didn't confuse anybody with that! Context is everything, but yeah, obviously they do repeat very often.
11:11 Another proof that .9 repeating is one. q = .9repeating = 9/10¹-1 = 9/9 = 1
0:34 my first tought on this:
Actually, 1/10 also repeats. 1/10=0.1000000... with 0 repeating (or 0.0999999... with 9 repeating)
But there are some numbers that don't repeat. For example: decimal expansion for pi doesn't repeat.
This is because pi is irrational. Rational numbers decimal expansion repeat, irrarional numbers do not.
You don't mention it; but of course this means when you use different base systems you get different sets of prime factors which changes which fractions terminate and which fractions repeat.
More generally you could say: The expansion of any fraction with a GCF of 1 in any base system will terminate if the prime factors of the denominator are shared with the prime factors of the base, and will repeat if the denominator has prime factors not shared with the base.
At least with integer bases.
Okay I learned something new today, thanks! I'd never even stopped to wonder why this was.
That's awesome, thanks for watching!
for some rational number P/Q where P and Q share no factors, its expansion terminates in base B if and only if the diference of the Q and B's prime factors is empty.
Basically the denominator of the number in simplest form cannot have prime factors which the base does not have
That's very interesting! Thanks for the vid!
not always, we only have names for about 0% of all irrational numbers from m-n where n=!=m
For Software developers: try writing 0.1 in binary, it's as impossible as writing 1/3 in decimal.
0.00011001100110011…
Question. When you explain the fact that in order for the decimal expansion to terminate the denominator of the fraction has to fully cancel out with some 10^n, that makes me wonder: is it possible to have a situation where b > 10^n (of course assuming that b also has only 2s and 5s in its factorization)? In that case, the 10^n will cancel out, but it will still leave some factors of b in the denominator. Can this situation ever happen? And if so, what happens?
It happens all the time, when you have inconsistent powers of 2 and 5 in the denominator. Consider 1/40. The number 40 has 3 copies of 2 and only 1 copy of 5 in its prime factorization. Multiply by 10^1, and we remove a single copy of each 2 and 5 in the denominator, leaving us with 1/4. The 10^1 will cancel, and we still have two copies of 2 in the denominator's prime factorization.
Of course, this all can be corrected for, by picking a different value of n. Instead of picking n = 1, pick n = 3. 1000/40 = 25. Now we've eliminated all copies of 2 and 5, by picking the larger of the two exponents in the prime factorization, rather than the smaller of the two exponents.
For all fractions of the form 1/(2^m * 5^n), it will always be possible to either multiply by 10^m or 10^n, whichever is larger, to eliminate the denominator completely, and thus the original fraction has an ending decimal, rather than a repeating decimal.
@@carultch Thank you! That makes complete sense. In the video he started the proof by assuming that the decimal expansion has n digits to begin with, and therefore he multiplied by 10^n to eliminate the denominator, but in an application you wouldn't know in advance how many digits the expansion has, unless you factorize the denominator first and retroactively choose an appropriate value for n. That's what confused me.
In your example, therefore, by picking n=3, which is the smallest n for which the denominator is eliminated completely, we can see that the decimal expansion of 1/40 is 0.025 and it indeed has 3 digits after the decimal point, and terminates as expected. Now it's all clear!
Hey, really appreciate the work you are doing 👏
Could you please make a video on set operations and cantors theory, thanks
Sure thing, thank you!
my theory. don't know if it makes sense.
For rational numbers, i saw in another way.
If dividing a number by y, decimals repeat and go on forever,
then it's trying to yield a infinite amount of 9s when multiplied by y.
for example, for 1/7, in base 10, it finds it hard to yield exactly 1 without the decimals going on forever. it asks, what will yield 9s when multiplied by 7, and it has 142857. now, 0.142857 will yield 0.999999 when multiplied by 7. but when 142857 repeats infinitely, creating a repeating decimal that goes on forever, it can converge to infinite amount of 9s when multiplied by 7, which is mathematically equal to 1.
same goes for others. even other bases.
0:47 its like saying 1/5 / by 2 ≈ 0.~~
(Approx.
This is why I don’t divide by 3 unless the numerator is a multiple of 3
Very interesting
Agreed!
0:48 Real
Its also fun how some non repeating numbers in decimal repeat in binary
Yeah, i think there's no small amount of overlap between base-12 fans and fraction haters, based on the nice terminating expansions in base 12
12:34 ITS PRONOUNCED [FAY] (y as in yogurt and A as in Ukrainian А and F as Ph)
I have known this since I was in 1st grade!
want a cookie?
Me too! ❤
Well, then the video is not directed at you. You sound like the type of person I wouldn't wanna be around. Okay you know it, so what? There are people who don't and that's why we need videos like this one so people that aren't so versed in the math field to feel welcomed to enter into it
why everyone so mean bruh
they just said they knew it, not that they’re better than everyone who didnt know it before
You are certainly disliked here, Peter ;)
I thought this was trivial lol, all primes who are not part of the composite numbers will repeat forever on their reciprocal