Basic algebra question/observation

Feb 18, 2021 at 5:28pm
Hi guys,

SO I've come to ask a simple math question, the answer doesn't perplex me but rather surprises me slightly.

so (k/6)(k+1)(2k+1) = (k+1/6)(k)(2k+1)
let's call (k/6)(k+1)(2k+1) A and (k+1/6)(k)(2k+1) B for simplicity.

let's just focus are concern on (k/6)(k+1) as (2k+1) will be the constant in this scenario or maybe redundant would be more precise.

What surprises ms is that (k/6)(k+1) = (k+1/6)(k).

I know that multiplication is obviously both associative and commutative, so
 
(2 * 3) * 6 == 2 * (3 * 6) and also (2)(3)(8) == (3)(8)(2) == (8)(2)(3) and so on



but division isn't neither commutative or associative so how does A == B

let's try 2 test cases, let's set k = 2 and 5 respectively.

let's first test A with k = 2,

(2/6)(2+1)(2(2)+1)
(1/3)(3)(5)
(1)(5) = 5

let's now test B with k = 2,

((2+1)/6)(2)(2(2)+1)
(3/6)(2)(5)
(0.5)(2)(5)
(1)(5) = 5

Now let's set k = 5,

(5/6)(5+1)(2(5)+1)
(5/6)(6)(11)
(5)(11) = 55

now B

((5+1)/6)(5)(2(5)+1)
(6/6)(5)(11)
(5)(11) = 55

so they both clearly equal one another but what law or rule allows this to hold true?
Feb 18, 2021 at 5:38pm
What surprises ms is that (k/6)(k+1) = (k+1/6)(k).


That would surprise me too.

If I put k=12 then:
the LHS is 2 * 13, which is 26
the RHS is (12+1/6) * 12, which is 146

So they aren't equal.


You need to put brackets in the correct places. You have omitted a rather crucial pair.
(k/6)(k+1) = ((k+1)/6) * k
Last edited on Feb 18, 2021 at 5:40pm
Feb 18, 2021 at 5:59pm
True, bracket error but my observation is still true I tried k = 12

(12/6)(12+1)(2(12) + 1)
(2)(13)(25) = 650

( (12+1)/6) (12)(25) = 650

again they both equal each other, I can't seem to figure out the reasoning to this as division isn't commutative nor associative , I would for example understand (12/6)(6)(2) = (6)(12/6)(2) because of the associative property, but how come we can swap k+1 with k in this scenario?
Last edited on Feb 18, 2021 at 6:02pm
Feb 18, 2021 at 6:29pm
or in other words why does (x/y)(z) = (z/y)(x) ?

as division isn't commutative nor associative.
Feb 18, 2021 at 6:57pm
Division is left-associative, but that doesn't prevent you from commuting the multiplication around.

(z/y)(x) --> (x)(z/y) --> (x)(z)/(y) --> (z)(x)/(y) --> (z)(x/y) --> (x/y)(z)
Last edited on Feb 18, 2021 at 6:59pm
Feb 18, 2021 at 6:57pm
Probably answering my own question, please correct me if I'm wrong

(k/6)(k+1) = ((k+1)/6)(k)

(k/6)(k+1) = (k/6)((k+1) / 1) bottom denominator will always equal 6 as 6 x 1 is 6, and since multiplication is commutative we can swap k and k+1 so (k/6)((k+1)/1) = ((k+1)/6)(k/1)

*Edit thanks Ganado :), finally got it haha but still have another question in regards to the same expression

but staying on the same topic, when it comes to precedence regarding order of operations

does (k/6)(k+1)(2k+1) = k/6(k+1)(2k+1)?? the actual question didn't have k/6 in brackets

so does this mean I have to multiply what's in brackets by each other first? (k+1)(2k+1)

or does it mean I have to distribute k/6 over (k+1) first?? or does it matter?
Last edited on Feb 18, 2021 at 6:59pm
Feb 18, 2021 at 7:04pm
Generally the author should avoid potential ambiguities in notation like that, but yes if we go by standard precedence rules that calculators and such use, the k/6 would be done first regardless of the parentheses around it.

or does it mean I have to distribute k/6 over (k+1) first?? or does it matter?
Multiplication is distributive, so it doesn't matter.
Last edited on Feb 18, 2021 at 7:05pm
Feb 18, 2021 at 7:09pm
That makes sense, I wrote out a couple of examples and yes it seems order doesn't matter but yeah those brackets are fairly ambiguous, would make more sense to put the brackets around each term.

thanks :)
Feb 18, 2021 at 11:32pm
I'm surprised no one has mentioned that
(k/6)(k+1) = k(1/6)(k+1) = k(k+1)(1/6) = k((k+1)/6)
Feb 19, 2021 at 6:19am
Give or take the order, it's the last line of my post, @helios.

Not sure how this thread got this long.
Last edited on Feb 19, 2021 at 6:20am
Feb 19, 2021 at 2:12pm
You moved the division, yes, but the way you did it doesn't emphasize how a division is really just a multiplication, which I feel i part of what's tripping up adam2016.
Last edited on Feb 19, 2021 at 2:12pm
Topic archived. No new replies allowed.