Your are viewing a read-only archive of the old DiS boards. Please hit the Community button above to engage with the DiS !
I got 1 out of 10.
Eat my dust.
Some obvious fuckups when reading the answers.
2 I couldn't be arsed to work out, but the speed/grandmas house one I don't get at all.
Therefore spends twice as long at 2 mph. Average speed = (2 + 2 + 4)/3 = 8/3 mph
and oh yeah, it was the start of the summer
but i wasnt trying
Was only the last question that screwed me as I couldn't for the life of me remember how to find the area of a rectangle.
The explanation for the solution is pretty half-arsed. There is no way to do it except by trial and error, and his suggestion that you'd guess the correct answer quickly is a bit tenuous. That said it is exactly what I did.
doing them all in my head.
The last one is a bit cruel in that it's more-or-less trial and error/instinct, once you've worked out the whole square-on-the-hypotenuse thing.
would have got 9 (I worked out the size of the rectangle) but didn't read the bloody question properly.
Consider this statement in the explanation:
" But almost all fractions have decimals that are "recurring"
but I'm far from confident of my answer.
i.e. recurring decimals are fractions rather than as they stated.
i guess what they're trying to get at is that fractions are rational maybe, so they will all have some kind of repetitive structure in decimal representation
Some fractions are not recurring. And he's implying that there are more recurring fractions than non-recurring ones.
My own line of thought is that the number of non-recurring fractions is exactly equal to the number of recurring ones. I think both sets are countably infinite and that as a result there is a one to one mapping between the two sets.
erm.... i thought all fractions either terminate or are recurring? maybe get a pure mathematician in here to clear things up though
what fractions can you write that don't recur in decimal representation?
if they didn't recur then they would be irrational so then they wouldn't be representable as fractions, right?
= 0.5000000..... technically i guess
i get what you mean now. are the number that terminate equal to the number that don't ?
seems by intuition that there would be more that don't terminate, dunno how you would prove that though
and I think he's wrong.
The set of all fractions (aka the set of all rational numbers, aka the set of all decimals that either terminate or recur) is known to be countably infinite, ie each element of the set can be mapped with a bijection to the set of natural numbers.
I think that the set of terminating decimals is also countably infinite. It's certainly infinite (0.1, 0.01, 0.001 ...) and since it is a subset of the rational numbers I don't believe it can be uncountably infinite (citation needed). Therefore it is countably infinite. And hence the "number" of fractions with a recurring decimal representation is equal to the "number" of fractions with a terminating decimal representation. And so his statement that "almost all" fractions have a recurring decimal representation is false.
ok what about something like this
a) if a number terminates then it will be possible to write in as a/10, a/100, a/1000 etc. this is because it can be written in the form
where s is some string of numbers and a is an integer
b) for this to hold then it must also be possible to write it in the form b/2 or b/5, where b is again an integer
c) you can get many more fractions from the same b, by writing b/3, b/7, b/11 etc (i.e. any prime that is not 2 or 5). these will all recur (as they don't terminate)
d) hence there must be many more recurring numbers
i think because there is an infinite number of primes then mapping in this way implies that set of all fractions is higher cardinality than the set of terminating fractions
this falls apart at part a, because you're suggesting multiple decimal representations for the same actual number.
If that's "allowed" then his statement is just trivially obvious, because every terminating decimal would have its own infinite set of non-terminating equivalents.
But it isn't allowed. The decimal representation of a number is the one that is expressed in the simplest terms. So no trailing zeros and no 9 recurrings. That way each number has one unique decimal representation.
that's just to intuit that it's possible to express all terminating numbers in that form. it's right
put it a different way:
if a decimal terminates then it can written as
where s is a string of length l. then this as a fraction will be
should be "string of length L"
don't start arguing with me using the creaky method.
"Some real numbers have two infinite decimal representations. For example, the number 1 may be equally represented by 1.000... as by 0.999... (where the infinite sequences of digits 0 and 9, respectively, are represented by "..."). Conventionally, the version with zero digits is preferred; by omitting the infinite sequence of zero digits, removing any final zero digits and a possible final decimal point, a normalized finite decimal representation is obtained."
i expressed it badly. the post below that explains it better maybe. i'm not sure if it's right though
as even numbers?
the idea of there being more or fewer of any two countably-infinite sets is redundant.
I knew someone would have the correct phrasing. This stuff always did my head in.
So what's your take on the original statement I took issue with?
I have DiS open in a small window at work atm..
" But almost all fractions have decimals that are "recurring""
fundamental number-theory untruth of it.
Basically, the only fractions that DON'T recur in base 10 are ones which, as you noted above, can be written in the form "a/(10^n)".
But infinite-set cardinality is counter-intuitive: The sole reason for the cardinality of the real line being greater than that of integers is down to transcendental numbers, which are difficult to calculate in spite of being numerically abundant.
Thus, INTUITIVELY you would say that there are lots more fractions that recur than ones that of the "a/(10^n)" form. When you're doing year 10 level maths, you're unlikely to be particularly familiar with number theory and the cardinality of infinite sets.
So yeah, whilst I'd agree that fundamentally he's incorrect, for the purposes of what you'd be teaching year 10s he's right enough.
so that's my little side-argument dealt with. Both sets are N0 cardinality, and the concept of "more than" doesn't really make sense in the context he's using it in.
But taking things a bit more generally and less rigorously I'm still intrigued by his implication.
I would argue that the most commonly found fractions (by a year 10 or anyone else) are half, third, ..., tenth. Of which only four out of nine are recurring decimals. So even on that level I think he's talking out of his arse.
And this is a reader in maths we're talking about. He really shouldn't be saying stuff that's wrong at all, even if it is intuitive.
this whole decimal nonsense is trash.
I guess the idea is to make recurring fractions less frightening - once you've worked out the whole x=0.123123123123... means 1000x-x=123=> x=41/333 or whatever, then you'r sorted.
But if you don't, then the number 0.142856142856... just seems weird as fuck.
he's only trying to get people just to recognise multiples of 1/9. So you're right.
I'm more annoyed by his pythagorean triples argument to be honest.
It would have been more intuitive to say numbers with recurring decimals can generally be stated as fractions rather than the other way around.
His point was that recurring decimals are very common, not that they can be expressed as fractions.
without even doing them in my head
got the secret bonus question right too
it said you lose a point if you tell others about it
i was on 12/10 before
is cleverer than saps
Because I didn't and only got 6.
Couldn't get my head around the last one. Seems badly written if you ask me.
you just have to know your squares of numbers and then have to exeriment with them a bit...unless you can 'see' I didn't see it immediately, I just experimented with squares ending in 1 so it took me 3 trials to get the answer
that the 20,21,29 pythagorean triple is an algebraic derivative of the 3,4,5 triple (because 20=4x5 and 21=3x7). But I can't work out the algebra :-(.
If you started by looking at squares that end in 1, why did you not take the simpler route and start by looking at squares that end in 0 (which you'd need to match up with your 1)?
shit shit shit shit shit
I put 3mph
I didn't bother to think about it :(