Algebra Problem of The Day-9

BigBeachBanana said:
Looks interesting. I haven't thought about that. Wondering what we get...
Click to expand...
Steven G said:
See post 17
Click to expand...
Surely you meant post 16 where it says 1*x=1 for any x, thus the answer would be 1

Some interesting properties that work for any x,y (I'm pretty sure that these aren't useful for finding the answer therefore there's no spoiler)...

1*x = 1 (as per post 16)
(-1)*x = -1
0*x = x
(-x)*(-y) = -(x*y) ...I had to edit this line to correct a mistake!
 
Last edited:
Cubist said:
Surely you meant post 16 where it says 1*x=1 for any x, thus the answer would be 1

Some interesting properties that work for any x,y (I'm pretty sure that these aren't useful for finding the answer therefore there's no spoiler)...
1*x = 1 (as per post 16)
(-1)*x = -1
0*x = x
(-x)*(-y) = x*y
Click to expand...
(-x)*(-y) = -(x+y)/(1+xy) = -(x*y)
 
Cubist said:
Surely you meant post 16 where it says 1*x=1 for any x, thus the answer would be 1

Some interesting properties that work for any x,y (I'm pretty sure that these aren't useful for finding the answer therefore there's no spoiler)...

1*x = 1 (as per post 16)
(-1)*x = -1
0*x = x
(-x)*(-y) = -(x*y) ...I had to edit this line to correct a mistake!
Click to expand...
It's amusing how 0 and 1 swap their properties in this algebra.
 
My answer is below...
(a−1a+1)∗(b−1b+1)=(a−1a+1)+(b−1b+1)1+(a−1a+1)⋅(b−1b+1) by the definition of *=(a−1)(b+1)+(b−1)(a+1)(a+1)(b+1)+(a−1)(b−1)=2ab−22ab+2=ab−1ab+1\left( \dfrac{a-1}{a+1} \right) * \left( \dfrac{b-1}{b+1} \right)\\ = \dfrac{ \left(\dfrac{a-1}{a+1}\right) + \left(\dfrac{b-1}{b+1}\right) }{ 1 + \left(\dfrac{a-1}{a+1}\right) \cdot \left(\dfrac{b-1}{b+1}\right) }\text{ by the definition of *}\\ = \dfrac{ (a-1)(b+1) + (b-1)(a+1) }{ (a+1)(b+1) + (a-1)(b-1) }\\ = \dfrac{2ab-2}{2ab + 2}\\ = \dfrac{ab - 1}{ab + 1}
This extends to any number of terms, since the inputs and outputs are in the same format of (x-1)/(x+1) where x is any number. And because...

y=x−1x+1  ⟹  x=1+y1−yy=\dfrac{x-1}{x+1} \implies x=\dfrac{1 + y}{1 - y}
...then the answer for 6 terms would be...
a∗b∗c∗d∗e∗f=c2−1c2+1where c2=(a+1)(b+1)(c+1)(d+1)(e+1)(f+1)(a−1)(b−1)(c−1)(d−1)(e−1)(f−1)a*b*c*d*e*f=\dfrac{c_2-1}{c_2+1}\\ \text{where }c_2=\dfrac{(a+1)(b+1)(c+1)(d+1)(e+1)(f+1)}{(a-1)(b-1)(c-1)(d-1)(e-1)(f-1)}
Extending this for the original question, the answer would be...

c−1c+1where c=3×4×5×6×⋯×2019×2020×2021×20221×2×3×4×⋯×2017×2018×2019×2020=2021×20222=2043231\dfrac{c-1}{c+1}\\ \text{where }c=\dfrac{\cancel{3} \times \cancel{4} \times \cancel{5}\times \cancel{6}\times \cdots \times \cancel{2019}\times \cancel{2020} \times 2021\times 2022}{1\times 2\times\cancel{3}\times \cancel{4} \times \cdots \times \cancel{2017}\times \cancel{2018}\times \cancel{2019}\times \cancel{2020}} = \dfrac{2021\times 2022}{2}=2043231therefore the answer isc−1c+1=10216151021616 in lowest terms\text{therefore the answer is}\dfrac{c-1}{c+1} = \dfrac{1021615}{1021616} \text{ in lowest terms}This is slightly different to BBB's answer in post#13 :cry: (I still suspect that BBB made a typo ;):D)

@BigBeachBanana my post#14 spoiler does actually contain an answer, but it's inside a spoiler within a spoiler. The outer spoiler says "without answer" because I thought this would enable people to click on the outer spoiler more readily. My English language "skills" sometimes let me down. (This post proves the conjecture of my post#14)
 
Last edited:
BBB,
Please hold off on supplying your solution as I think I may have a nice one. I just need to proof a conjecture.
Thanks,
Steven
 
Here is what I got. I will state two theorems which I did prove (by induction)
Theorem #1: (2)(2+2*1)(2+2*2)...(2+4n) = (2n+2)/(2n+1).
Theorem #2: (1+2*1)(1+2*2)...(1+4n) = n/(n+1)

2021 = 1+4n implies n = 505
So (3)(5)(7)...(2021) = 505/506

2018=2+4n implies n=504
So (2)(4)...(2018) = 1010/1009

Now (2)(3)...(2021) = (1010/1009)(2020)(505/506) = (1010/1011)(505/506) = 1021615/1021616

I am retiring from these problems.
 
Last edited:
Making observations:
2∗3=57, (2∗3)∗4=119,[(2∗3)∗4]∗5=1415,{[(2∗3)∗4]∗5}∗6=22202*3=\frac{5}{7},\, (2*3)*4 = \frac{11}{9}, [(2*3)*4]*5=\frac{14}{15}, \{[(2*3)*4]*5\}*6=\frac{22}{20}
The conjecture:
Let a3=57, a4=119,...\text{Let } a_3 =\frac{5}{7},\, a_4=\frac{11}{9},...then:
an={n(n+1)/2−1n(n+1)/2+1for odd nn(n+1)/2+1n(n+1)/2−1for even nor an={n(n+1)−2n(n+1)+2for odd nn(n+1)+2n(n+1)−2for even na_n = \begin{cases} \frac{n(n+1)/2-1}{n(n+1)/2+1} &\text{for odd } n \\ \\ \frac{n(n+1)/2+1}{n(n+1)/2-1} &\text{for even } n \end{cases} \text{or } a_n = \begin{cases} \frac{n(n+1)-2}{n(n+1)+2} &\text{for odd } n \\ \\ \frac{n(n+1)+2}{n(n+1)-2} &\text{for even } n \end{cases}
Therefore,
a2021=2021(2022)−22021(2022)+2=1 021 6151 021 616a_{2021}=\frac{2021(2022)-2}{2021(2022)+2}=\frac{1\,021\,615}{1\,021\,616}
The proof below was optional. If you noticed the pattern then it was sufficient.

Proof (by induction) for the odd case:
The base case (n=3) has shown to be true above in the observation.
The induction case:
ak+1=ak∗(k+1)=k(k+1)−2k(k+1)−2∗(k+1)a_{k+1}=a_k*(k+1)=\frac{k(k+1)-2}{k(k+1)-2}*(k+1)−some algebra steps−-\text{some algebra steps}-k(k+1)+2+(k(k+1)−2)(k+1)k(k+1)−2+(k(k+1)+2)(k+1)\frac{k(k+1)+2+(k(k+1)-2)(k+1)}{k(k+1)-2+(k(k+1)+2)(k+1)}(k+1)(k+2)+2(k+1)(k+2)−2\frac{(k+1)(k+2)+2}{(k+1)(k+2)-2}
The even case can be proven in a similar fashion.
 
I looked at (2*4)(6*8)... and noticed that the 1st entry in each bracket went up by 4. That was how I based my answer.
I think that I looked early on at what you did and then for some reason I did not see a pattern.

This is a nice algebra to look at. We noted how the usual role of 0 and 1 got interchanged. I am thinking about what the range of this would be.
Lets try this here. Let k be any real number.
(x+y)/(1+xy) = k
k(1+xy) = x+y
k+kxy = x+y
k-y = x +kxy = x(1+ky)
x = (k-y)/(1+ky)
Since y is free, we can be sure that 1+ky≠0
Hence the range is all reals.
I actually thought proving this would be hard.
Now we can talk about classes for k
 
You must log in or register to reply here.
Top