iso paper sizes: sides are 1:sqrt2 or about 1:1.414
A4 is 210x297mm
A4 folded in half is 148x210 or A5.
A3 is two A4s, or 297x420
go to A0 841x1189 or one square metre.
A-1 (A Minus One) is 2 square metres.
Australia is ~A-43
The visible universe ~A-179
A proton about A139. Quark? A151
A4 is 210x297mm
A4 folded in half is 148x210 or A5.
A3 is two A4s, or 297x420
go to A0 841x1189 or one square metre.
A-1 (A Minus One) is 2 square metres.
Australia is ~A-43
The visible universe ~A-179
A proton about A139. Quark? A151
March 13, 2026 at 9:32 PM
iso paper sizes: sides are 1:sqrt2 or about 1:1.414
A4 is 210x297mm
A4 folded in half is 148x210 or A5.
A3 is two A4s, or 297x420
go to A0 841x1189 or one square metre.
A-1 (A Minus One) is 2 square metres.
Australia is ~A-43
The visible universe ~A-179
A proton about A139. Quark? A151
A4 is 210x297mm
A4 folded in half is 148x210 or A5.
A3 is two A4s, or 297x420
go to A0 841x1189 or one square metre.
A-1 (A Minus One) is 2 square metres.
Australia is ~A-43
The visible universe ~A-179
A proton about A139. Quark? A151
Top of my head, I think we're looking approx 1/sqrt2 along the way so about 70, but it starts at 3 not 1 so that pushes it slightly upwards, 72 or 73?
September 25, 2026 at 8:52 PM
Top of my head, I think we're looking approx 1/sqrt2 along the way so about 70, but it starts at 3 not 1 so that pushes it slightly upwards, 72 or 73?
It's basically 100/sqrt2 with adjustments for two reasons: it's actually 99 not 100, and you start at 3 not 1
September 25, 2026 at 9:34 PM
It's basically 100/sqrt2 with adjustments for two reasons: it's actually 99 not 100, and you start at 3 not 1
not woke:
tau (centrist)
sqrt2 (totally checked out of politics)
rabidly anti woke:
e ("centrist" (fascist))
1 (openly fascist)
infinity (not even a number)
woke:
golden ratio
imaginary unit
zero
tau (centrist)
sqrt2 (totally checked out of politics)
rabidly anti woke:
e ("centrist" (fascist))
1 (openly fascist)
infinity (not even a number)
woke:
golden ratio
imaginary unit
zero
April 8, 2025 at 8:49 PM
not woke:
tau (centrist)
sqrt2 (totally checked out of politics)
rabidly anti woke:
e ("centrist" (fascist))
1 (openly fascist)
infinity (not even a number)
woke:
golden ratio
imaginary unit
zero
tau (centrist)
sqrt2 (totally checked out of politics)
rabidly anti woke:
e ("centrist" (fascist))
1 (openly fascist)
infinity (not even a number)
woke:
golden ratio
imaginary unit
zero
August 23, 2025 at 5:54 PM
@davidkbutler.bsky.social we did this in a tutoring session today.
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
September 25, 2026 at 2:59 AM
@davidkbutler.bsky.social we did this in a tutoring session today.
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
cos60°=cos(45°+15°)
cos60°=cos45°cos15°-sin45°sin15°
1/2=(cos15°-sin15°)/sqrt2
cos60°=cos45°cos15°-sin45°sin15°
1/2=(cos15°-sin15°)/sqrt2
December 6, 2024 at 6:26 PM
cos60°=cos(45°+15°)
cos60°=cos45°cos15°-sin45°sin15°
1/2=(cos15°-sin15°)/sqrt2
cos60°=cos45°cos15°-sin45°sin15°
1/2=(cos15°-sin15°)/sqrt2
The standard schrodinger representation, X=x and P=-id/dx, is isomorphic to a Fock space, by defining A = 1/sqrt2[X+P] and Ω = exp(-x2/2). This is just not generally used unless the dynamics happen to be the QHO.
September 20, 2026 at 8:45 PM
The standard schrodinger representation, X=x and P=-id/dx, is isomorphic to a Fock space, by defining A = 1/sqrt2[X+P] and Ω = exp(-x2/2). This is just not generally used unless the dynamics happen to be the QHO.
But of course, the diagonal pairs will be sqrt2 distance apart, so there has to be a better way
May 29, 2026 at 7:44 AM
But of course, the diagonal pairs will be sqrt2 distance apart, so there has to be a better way
Fucked it up: a = sqrt2^sqrt2, b = sqrt2 in the latter
April 27, 2025 at 6:30 AM
Fucked it up: a = sqrt2^sqrt2, b = sqrt2 in the latter
Proof that a, b can be irrational and a^b rational:
suppose sqrt(2)^sqrt(2), if this is rational a = b = sqrt2, otherwise a = sqrt2, b = sqrt2^sqrt2
suppose sqrt(2)^sqrt(2), if this is rational a = b = sqrt2, otherwise a = sqrt2, b = sqrt2^sqrt2
April 27, 2025 at 6:29 AM
Proof that a, b can be irrational and a^b rational:
suppose sqrt(2)^sqrt(2), if this is rational a = b = sqrt2, otherwise a = sqrt2, b = sqrt2^sqrt2
suppose sqrt(2)^sqrt(2), if this is rational a = b = sqrt2, otherwise a = sqrt2, b = sqrt2^sqrt2
I am unreasonably fond of sqrt2.
March 14, 2025 at 10:43 AM
I am unreasonably fond of sqrt2.
@davidkbutler.bsky.social we did this in a tutoring session today.
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
September 25, 2026 at 2:57 AM
@davidkbutler.bsky.social we did this in a tutoring session today.
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
Built 1st quadrant unit circle from special right triangles, remembered +/- of coordinates in each quadrant from algebra1, & talked about multiples of fractions for radians (pi/6, 2pi/6, 3pi/6 etc.)
The golden ratio! 1:sqrt2
May 28, 2024 at 1:16 PM
The golden ratio! 1:sqrt2
I like ℚ(sqrt2), then ℂ, then ℝ.
October 31, 2024 at 7:41 AM
I like ℚ(sqrt2), then ℂ, then ℝ.
It's not 100% accurate, as sqrt2 is 1.41412... but yeah moving minis via a ruler is about as accurate.
And you don't even need to multiply anything. Just count as you go, every 2nd diagonal is 2 squares of movement. Easy.
And you don't even need to multiply anything. Just count as you go, every 2nd diagonal is 2 squares of movement. Easy.
August 10, 2025 at 1:05 AM
It's not 100% accurate, as sqrt2 is 1.41412... but yeah moving minis via a ruler is about as accurate.
And you don't even need to multiply anything. Just count as you go, every 2nd diagonal is 2 squares of movement. Easy.
And you don't even need to multiply anything. Just count as you go, every 2nd diagonal is 2 squares of movement. Easy.
When you say that it's for the same reason that complex roots come in conjugate pairs, what do you see as being the reason? (The reason that I have in my head doesn't extend very neatly to the sqrt2 case)
April 24, 2025 at 8:36 PM
When you say that it's for the same reason that complex roots come in conjugate pairs, what do you see as being the reason? (The reason that I have in my head doesn't extend very neatly to the sqrt2 case)
Which does indeed follow.
But in q7, 3+sqrt2 is not a root...
But in q7, 3+sqrt2 is not a root...
April 24, 2025 at 6:12 PM
Which does indeed follow.
But in q7, 3+sqrt2 is not a root...
But in q7, 3+sqrt2 is not a root...
Sqrt2 --> sqrt5 which is 1.58 x bigger
Sqrt5 --> sqrt17, 1.84 x bigger
Sqrt5 --> sqrt8, 1.26 x bigger
Sqrt5 --> sqrt17, 1.84 x bigger
Sqrt5 --> sqrt8, 1.26 x bigger
April 5, 2025 at 1:02 PM
Sqrt2 --> sqrt5 which is 1.58 x bigger
Sqrt5 --> sqrt17, 1.84 x bigger
Sqrt5 --> sqrt8, 1.26 x bigger
Sqrt5 --> sqrt17, 1.84 x bigger
Sqrt5 --> sqrt8, 1.26 x bigger
This here is why i specify my hex table is 4' edge to edge, because corner to corner is a bit smaller.
I normally measure edge to edge, like a 1" square is 1" edge to edge (and sqrt2 corner to corner?)
Same deal.
I normally measure edge to edge, like a 1" square is 1" edge to edge (and sqrt2 corner to corner?)
Same deal.
July 17, 2024 at 4:19 PM
This here is why i specify my hex table is 4' edge to edge, because corner to corner is a bit smaller.
I normally measure edge to edge, like a 1" square is 1" edge to edge (and sqrt2 corner to corner?)
Same deal.
I normally measure edge to edge, like a 1" square is 1" edge to edge (and sqrt2 corner to corner?)
Same deal.
I think I'd go for expanding (x-3+sqrt2)(x^2+ax+b) and equating coefficients to find a and b
April 24, 2025 at 8:21 PM
I think I'd go for expanding (x-3+sqrt2)(x^2+ax+b) and equating coefficients to find a and b
I didn't intend to leave out the square root of 2 but Bsky can only handle four images per post. Love you, sqrt2
February 15, 2025 at 5:32 AM
I didn't intend to leave out the square root of 2 but Bsky can only handle four images per post. Love you, sqrt2
Of course you know exactly which book 😉
I found that chapter while getting a presentation together for Pi Day called "Pi And Friends: Appreciating Special Numbers in Math".
{we had done a Pi webinar the prior year & we were out of pi ideas 😂🤪 so we added in 0, e, golden ratio, sqrt2, i etc)
I found that chapter while getting a presentation together for Pi Day called "Pi And Friends: Appreciating Special Numbers in Math".
{we had done a Pi webinar the prior year & we were out of pi ideas 😂🤪 so we added in 0, e, golden ratio, sqrt2, i etc)
January 16, 2025 at 1:49 PM
Of course you know exactly which book 😉
I found that chapter while getting a presentation together for Pi Day called "Pi And Friends: Appreciating Special Numbers in Math".
{we had done a Pi webinar the prior year & we were out of pi ideas 😂🤪 so we added in 0, e, golden ratio, sqrt2, i etc)
I found that chapter while getting a presentation together for Pi Day called "Pi And Friends: Appreciating Special Numbers in Math".
{we had done a Pi webinar the prior year & we were out of pi ideas 😂🤪 so we added in 0, e, golden ratio, sqrt2, i etc)
This one stumped them; I had to demonstrate a method in the end, using similar triangles. sqrt2 and sqrt3 were easily dealt with by most, and then all with the hint 'Pythagoras'.
March 4, 2025 at 2:24 PM
This one stumped them; I had to demonstrate a method in the end, using similar triangles. sqrt2 and sqrt3 were easily dealt with by most, and then all with the hint 'Pythagoras'.