Basic algebra question/observation
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
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?
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
|
|
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?
| 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
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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?
(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?
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or in other words why does (x/y)(z) = (z/y)(x) ?
as division isn't commutative nor associative.
as division isn't commutative nor associative.
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)
(z/y)(x) --> (x)(z/y) --> (x)(z)/(y) --> (z)(x)/(y) --> (z)(x/y) --> (x/y)(z)
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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?
(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?
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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.
Multiplication is distributive, so it doesn't matter.
| or does it mean I have to distribute k/6 over (k+1) first?? or does it matter? |
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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 :)
thanks :)
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)
(k/6)(k+1) = k(1/6)(k+1) = k(k+1)(1/6) = k((k+1)/6)
Give or take the order, it's the last line of my post, @helios.
Not sure how this thread got this long.
Not sure how this thread got this long.
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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.
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