This is basically geometric progression in which the common multiple r has value less than 1 so show the formula to get sum of infinity a/(1-r) can be used ,its the same thing. Btw a is the first term in the above formula
The areas may be equal in measure, but there is a point in the center of the square that is never shaded. One way you can think of this is that the limiting difference between 0.888..._9 and 1.0 is an infinitesimal.
@@JoaoGabriel-ew4ic In base b such that b != 1 and b is a natural number and for n such that n = b-1; 0.nnn.... = n/10 + n/100 + n/1000 +... 0.nnn... = n/10 + n/(10^2) + n/(10^3) +... 0.nnn... = n(1/10 + 1/(10^2) + 1/(10^3) +...) 0.nnn... = n * sum 1/(10^k) k->infinity 1/(10^k) = (1/10)^k 0.nnn... = n * sum (1/10)^k k->infinity sum r^k k-> infinity = r/(1-r) 0.nnn... = n * (1/10)/(1-(1/10)) 1 - 1/10 in base b = (b-1)/10 0.nnn... = n * (1/10)/(n/10) 0.nnn... = n * (1/10) * (10/n) 0.nnn... = n * 10/10n 0.nnn... = n * 1/n 0.nnn... = n/n = 1 QED
Or we can be perfectly happy knowing that the area of the square is one square and go about the rest of our days without obsessing over tic tac toe boards. 😅
To be clear, the series 8/9 + 8/9^2 + 8/9^3 + 8/9^4 + ... has the sequence of partial sums: 8/9, 8/9 + 8/9^2, 8/9 + 8/9^2 + 8/9^3, .... The sum of the series is the limit of the seqeunce of the partial sums, and that is (using the geometric series formula) (8/9) / ( 1 - 1/9) = 1. If you prefer [the sum of] 0.888... (base 9) = lim n->oo 0.888...8 (base 9) (n 8s) = lim n->oo 1 - 1/9^n = 1
Then it's 25% true and 75% false that 0 = 1 in base 1. But If one of those hits 0, that means it's true or false. But There's a probability that one of those gets to 0%.
@Naej7 i0, and maybe j0...? I see that surely i0=0 at first glance, but does it at somekind of infinity or the like? After all, set theory is entirely base upon the empty set as the starting notion, then counting the number of empty sets under consideration to reach higher numbers and sets of numbers and things. Right?
Math in school: 🥱
Math in internet: 🤩
Bro is so tru
This is basically geometric progression in which the common multiple r has value less than 1 so show the formula to get sum of infinity a/(1-r) can be used ,its the same thing. Btw a is the first term in the above formula
So basically for any n, 0.nnn repeating will be equal to 1 in base n+1
For n = integer and less than or equal to 9.
@@dougr.2398Why would the pattern stop for n>9 ?
@@raph-ko1706 because the decimal representation is base 10
Base 10 only uses digits 0 through 9 in its decimal representation
@@dougr.2398 Fair answer thanks 🙂
1. Let 0.(n)=x
2. Multiply both sides by 10
n.(n)=10x
3. Substract x from both sides
n.(n)-0.(n)=10x-x
n=9x
x=n/9
4. Recall 0.(n)=x
0.(n)=n/9
This only works in base 10
I mean yeah you can do this with any fraction
He does
Nooooooo. Really?
Only with fractions >= 1/2
@@otinajNo, any works, 3/4, 9/10, 1/7, anything
@@raysye4775 It works for >
Beautiful ❤️
The areas may be equal in measure, but there is a point in the center of the square that is never shaded. One way you can think of this is that the limiting difference between 0.888..._9 and 1.0 is an infinitesimal.
Infinite space is only reached in infinite time
Isn't x-1 over x + x-1 over x² + x-1 over x³ ... always 1?
It should work for X>1
No it works for x > 1 and x < 1
@@jamesmarlowebito8982 i think you are right
I've just realised that this videos of proving that something is 1 in base X is not as impressive since it's just like 0.999... is 1 in base 10 lol
0.nnnnnn is 1 in base n+1
Prove it
7/8 + 7/8² + 7/8³ + 7/8⁴ + ... + 7/8^n = 1
That means
(0.7777777...) in base 8 is 1
@@JoaoGabriel-ew4ic
In base b such that b != 1 and b is a natural number and for n such that n = b-1;
0.nnn.... = n/10 + n/100 + n/1000 +...
0.nnn... = n/10 + n/(10^2) + n/(10^3) +...
0.nnn... = n(1/10 + 1/(10^2) + 1/(10^3) +...)
0.nnn... = n * sum 1/(10^k) k->infinity
1/(10^k) = (1/10)^k
0.nnn... = n * sum (1/10)^k k->infinity
sum r^k k-> infinity = r/(1-r)
0.nnn... = n * (1/10)/(1-(1/10))
1 - 1/10 in base b = (b-1)/10
0.nnn... = n * (1/10)/(n/10)
0.nnn... = n * (1/10) * (10/n)
0.nnn... = n * 10/10n
0.nnn... = n * 1/n
0.nnn... = n/n = 1 QED
How proof shoe r.@@JoaoGabriel-ew4ic
@@JoaoGabriel-ew4ic for proof show the video xD
How many series are left with these proofs?
Wanted one for each base from 2 to 10. Just need base 6 at this point.
Or we can be perfectly happy knowing that the area of the square is one square and go about the rest of our days without obsessing over tic tac toe boards. 😅
0.99...=1 in base 10
You could also do 0.eee = 1 in hexadecimal.
I was trying to get one for each base from 2 to 10. But maybe I’ll work on that one too :)
@@MathVisualProofs Since 16 is square, a 4x4 would make a nice visual.
That's pentadecimal, 0.FFF... = 1 in base 16.
@@narfharder F is 16 in hexadecimal. The repeating decimal is one less than the base, so 0.eee is for hexadecimal.
@@scottabroughtonNo, F in hex is 15 in dec
Good example how 0.99999... is equal to 1 (in base 10).
8/9=0,(8)≠1
Asymptoticism should be a word because this would be Antiasymtoticism
Wouldn't it be more accurate to say it approaches 1 instead of equals 1?
The sequence of finite sums approaches 1 for sure. But the infinite sum IS the limit of the sequence of finite sums so the infinite sum is 1.
Wrong
@@budderman3rd You are 0 years old but you know nothing.
@@jasonhilliker492 We put the equal sign when we speak about limits, but sure it means by approaching
To be clear, the series 8/9 + 8/9^2 + 8/9^3 + 8/9^4 + ... has the sequence of partial sums: 8/9, 8/9 + 8/9^2, 8/9 + 8/9^2 + 8/9^3, .... The sum of the series is the limit of the seqeunce of the partial sums, and that is (using the geometric series formula) (8/9) / ( 1 - 1/9) = 1.
If you prefer [the sum of] 0.888... (base 9) = lim n->oo 0.888...8 (base 9) (n 8s) = lim n->oo 1 - 1/9^n = 1
0.0000000... = 1 in base 1
No
@@Naej7Yes and no. Only 0 exist in base 1. But base 1 has no fixed numerical values.
Well, there is at least the very singular imaginary number system in base 1, unless somehow there isn't.
Then it's 25% true and 75% false that 0 = 1 in base 1.
But
If one of those hits 0, that means it's true or false.
But
There's a probability that one of those gets to 0%.
@Naej7 i0, and maybe j0...? I see that surely i0=0 at first glance, but does it at somekind of infinity or the like? After all, set theory is entirely base upon the empty set as the starting notion, then counting the number of empty sets under consideration to reach higher numbers and sets of numbers and things. Right?
well 0.99999999...=1 base 10
(0.88888...)9=1