Avoiding division absurdities.

[boit]

How do people end with .9r?
Apart from trying to derive square root of numbers using Newton's method, the biggest culprit is how we employ the authorised order of operation in mathematics i.e. BODMAS, PENDMAS e.t.c. Specificaly where we place the quantities.
I saw an equation being used to 'reduce' .9r into 1 (or is it to collapse it into 1?) that went thus
x=.9r
10x=9.9r
subtracting the first equation from the second and we get
9x=9
dividing each side by 9 we find x=1 proving .9r is actually (equal to) 1.
Here is the problem. If we decide to start with division before multiplication, we end up in the same hole that we're trying to get ourselves out of.
Check this. (1/9)x9=?
Now let's divide 1 by 9, what do we end up with? .111 . . . .i.e. 0.1r
multiply that by 9 and we end up with .9r
If we could only have re-arranged the equation to be 1x(9/9)
That way we will avoid unnecessary controversies (but again just like in tabloids, maths seems to thrive in controversies).

 

[pwsnafu]

I think what you are trying to say is:

$1 = \frac{9}{9} = \frac{1 \times 9}{9} = \frac{1}{9} \times 9 = 0.\overline{1} \times 9 = 0.\overline{9}$

Of course when students see this they ask "but 1 divided by 9 isn't really 0.1 recurring, it's just an approximation" and we are back to square one.

That way we will avoid unnecessary controversies (but again just like in tabloids, maths seems to thrive in controversies).

I don't know what you are saying here. Students need a proper understanding of the reals. Just look at this thread, especially from post #34 onwards. As Hurkly wrote in that thread

Did you know there are a number of people who think numbers aren't "exact"? That people get the idea that there is no such number as "one third", with reasoning that decimal can only provide an approximate value of such a thing? Non-terminating decimals really need some amount of coverage, to properly connect decimal notation to arithmetic and algebraic knowledge.

Your solution doesn't address this, and suffers the same problems that the original solution you posted suffers from.

 

[Vorde]

First of all. .9999... is = to 1.

Second of all, these 'absurdities' like PEMDAS are crucial to mathematics. There are the rules math follows.

To say they are arbitrary and wrong is like saying the dictionary definition for the word 'the' is wrong because 'the' can mean anything you want. Sure, you're right in a sense. But you're right in a useless way which gets you no where.

We define an order of operations so that our math works. Without it, math can do nothing.

Remember, all of math is made up, so we can define whatever we want (as long as the system is consistent). It's just a testament to the power of the things we 'define arbitrarily' that we can use math to so accurately describe the real world.

 

[pwsnafu]

Second of all, these 'absurdities' like PEMDAS are crucial to mathematics. There are the rules math follows.

Hardly. Order of operations is "needed" only because frankly we can't be bothered writing the parenthesis $(3 \times 4) + 5$ every time we have more than two operations in an expression. It's a notational convention, and has nothing to do with the underlying mathematics. The only part that is "crucial" is the parenthesis.

Just look at any non-associative algebraic structure. Yes it annoying in that we need to write down the parenthesis, but that's it. It's an annoyance.

 

[arildno]

Hardly. Order of operations is "needed" only because frankly we can't be bothered writing the parenthesis $(3 \times 4) + 5$ every time we have more than two operations in an expression. It's a notational convention, and has nothing to do with the underlying mathematics. The only part that is "crucial" is the parenthesis.

Just look at any non-associative algebraic structure. Yes it annoying in that we need to write down the parenthesis, but that's it. It's an annoyance.

Indeed.
Auxiliary rules are meant to..auxiliate, not have independent insights.

We MIGHT write 3+4*5+5*(2+1) as A(A(3,M(4,5)),M(5,A(2,1)),

where A(a,b) is the addition operator, and M(a,b) the multiplication operator, but it ought to visually intuitive why we do NOT, usually, write A(A(3,M(4,5)),M(5,A(2,1))

 

[Vorde]

I understand your point, of course, but I don't agree with the conclusion. I do believe I know less math than the two of you, but is there a good reason $3 \times 4 +5$ implies $(3 \times 4) + 5$ rather than $3 \times (4 + 5)$?

Isn't it that we've decided that with a lack of parenthesis multiplication goes first? Based on that, if that is correct, I would say something like PEMDAS is very much a rule.

 

[boit]

We wouldn't be in this quagmire if the Indians or Arabs didn't invent the decimal fraction. I guess we have to live with it.

 

[Number Nine]

We wouldn't be in this quagmire if the Indians or Arabs didn't invent the decimal fraction. I guess we have to live with it.

What quagmire, exactly? I'm not entirely sure what the problem is supposed to be.

 

[micromass]

We wouldn't be in this quagmire if the Indians or Arabs didn't invent the decimal fraction. I guess we have to live with it.

You are seeing the entire 1=0.999999... as a huge problem, when it really isn't a problem at all. It's just a consequence of the axioms and definitions we put forward. It really doesn't change math or make it more difficult.

 

[HallsofIvy]

We wouldn't be in this quagmire if the Indians or Arabs didn't invent the decimal fraction. I guess we have to live with it.

That's right. We would, instead, still be doing arithmetic with counters. However, this entire "0.999..." vs "1.0" "quagmire" is, at worst, a mudpuddle that most people step over without ever noticing it.

 

[johnqwertyful]

How do people end with .9r?
Apart from trying to derive square root of numbers using Newton's method, the biggest culprit is how we employ the authorised order of operation in mathematics i.e. BODMAS, PENDMAS e.t.c. Specificaly where we place the quantities.
I saw an equation being used to 'reduce' .9r into 1 (or is it to collapse it into 1?) that went thus
x=.9r
10x=9.9r
subtracting the first equation from the second and we get
9x=9
dividing each side by 9 we find x=1 proving .9r is actually (equal to) 1.
Here is the problem. If we decide to start with division before multiplication, we end up in the same hole that we're trying to get ourselves out of.
Check this. (1/9)x9=?
Now let's divide 1 by 9, what do we end up with? .111 . . . .i.e. 0.1r
multiply that by 9 and we end up with .9r
If we could only have re-arranged the equation to be 1x(9/9)
That way we will avoid unnecessary controversies (but again just like in tabloids, maths seems to thrive in controversies).

.999999...=1.

PEMDAS is a convention.

There are controversies in math. These however aren't among them.

 

[boit]

.999999...=1.

PEMDAS is a convention.

There are controversies in math. These however aren't among them.

To the uninitiated (like me) it is.

 

[Vorde]

To the uninitiated (like me) it is.

No it is not a controversy, to you it is a confusion.

 

[Whovian]

Yep, if we divide by 9 instead of multiplying by 10, we end up deeper in the same hole.

But why do that if we can easily grab the vine of multiplying by 10, and then we just need to pull to get out of the hole?