If d = 0, 1, 4, 5, 6, or 9, then d^3 ends in d.
If d = 2, 3, 7, or 8, then d^3 ends in 10-d.
(2/10)
How to mentally compute the cube root of the cube of a two digit number.
Thread (1/10)
If d = 0, 1, 4, 5, 6, or 9, then d^3 ends in d.
If d = 2, 3, 7, or 8, then d^3 ends in 10-d.
(2/10)
For example, if 50653 is the cube of an integer, it's the cube of a number ending in 7.
(3/10)
Then 10a <= n <= 10(a+1), and so n^3 is between (10a)^3 and (10(a+1))^3, i.e. between 1000 a^3 and 1000 (a+1)^3.
(4/10)
Chop off the last three digits of x.
Then a, the first digit of n, is the largest number whose cube is no greater than what's left of x.
(5/10)
(6/10)
Lopping off the last three digits gives 50.
3^3 < 50 < 4^3, so the first digit of the cube root of 50653 is 3.
(7/10)
(8/10)
More from Maths
Loops and their multiplication groups
A thread in 15 parts
(0/15)
Recall that a quasigroup (Q,*) is a set Q with a binary operation * such that for each a,b in Q, the equations a*x=b and y*a=b have unique solutions x,y. Groups are quasigroups and this property is usually one of the first things proved in elementary group theory.
(1/15)
Note that we don't assume associativity of *!
A loop is a quasigroup with an identity element. The story of why they are called loops is an interesting one and may even be true, but I will save it for another day. I am going to focus on loops in this thread.
(2/15)
Natural examples of nonassociative loops:
- The nonzero octonions under multiplication
- The sphere S^7 under octonion multiplication
- I have discussed other examples
For each x in a loop Q, define the left & right translations L_x, R_x : Q->Q by L_x(y)=xy and R_x(y)=yx. These mappings are permutations of Q. The composition L_x L_y of two left translations is not necessarily a left translation because Q is not necessarily associative.
(4/15)
A thread in 15 parts
(0/15)
Recall that a quasigroup (Q,*) is a set Q with a binary operation * such that for each a,b in Q, the equations a*x=b and y*a=b have unique solutions x,y. Groups are quasigroups and this property is usually one of the first things proved in elementary group theory.
(1/15)
Note that we don't assume associativity of *!
A loop is a quasigroup with an identity element. The story of why they are called loops is an interesting one and may even be true, but I will save it for another day. I am going to focus on loops in this thread.
(2/15)
Natural examples of nonassociative loops:
- The nonzero octonions under multiplication
- The sphere S^7 under octonion multiplication
- I have discussed other examples
Rethinking Vector Addition
— Michael Kinyon (@ProfKinyon) December 1, 2020
or
How I Learned to Stop Worrying and Love Nonassociativity
A thread in 29 tweets
(0/28)
For each x in a loop Q, define the left & right translations L_x, R_x : Q->Q by L_x(y)=xy and R_x(y)=yx. These mappings are permutations of Q. The composition L_x L_y of two left translations is not necessarily a left translation because Q is not necessarily associative.
(4/15)
OK, I may be guilty of a DoS attack attempt on mathematicians' brains here, so lest anyone waste too much precious brain time decoding this deliberately cryptic statement, let me do it for you. •1/15
First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15
Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15
Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15
… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15
\u2200x.\u2200y.((\u2200z.((z\u2208x) \u21d2 ((\u2200t.((t\u2208x) \u21d2 (t\u2208z) \u21d2 (t\u2208y)))) \u21d2 (z\u2208y))) \u21d2 (\u2200z.((z\u2208x) \u21d2 (z\u2208y))))
— Gro-Tsen (@gro_tsen) February 12, 2021
First, as some asked, it is to be parenthesized as: “∀x.∀y.((∀z.((z∈x) ⇒ (((∀t.((t∈x) ⇒ ((t∈z) ⇒ (t∈y))))) ⇒ (z∈y)))) ⇒ (∀z.((z∈x) ⇒ (z∈y))))” (the convention is that ‘⇒’ is right-associative: “P⇒Q⇒R” means “P⇒(Q⇒R)”), but this doesn't clarify much. •2/15
Maybe we can make it a tad less abstruse by using guarded quantifiers (“∀u∈x.(…)” stands for “∀u.((u∈x)⇒(…))”): it is then “∀x.∀y.((∀z∈x.(((∀t∈x.((t∈z) ⇒ (t∈y)))) ⇒ (z∈y))) ⇒ (∀z∈x.(z∈y)))”. •3/15
Maybe a tad clearer again by writing “P(u)” for “u∈y” and leaving out the quantifier on y, viꝫ: “∀x.((∀z∈x.(((∀t∈x.((t∈z) ⇒ P(t)))) ⇒ P(z))) ⇒ (∀z∈x.P(z)))” [✯]. Now it appears as an induction principle: namely, … •4/15
… “in order to prove P(z) for all z∈x, we can assume, when proving P(z), that P(t) is already known for all t∈z∩x” (n.b.: “(∀z.(Q(z)⇒P(z)))⇒(∀z.P(z))” can be read “in order to prove P(z) for all z, we can assume Q(z) known when proving P(z)”). •5/15
