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42

Reading Multi-Factor Decision Making Math and take note of any math I don’t understand.

A typical approach is to score each factor, multiply that by a weighting for how important the factor is, and to add up the results.

Not sure what a weighting is. Is that like a rank of importance relative to all other factors? So like, if you’re considering 10 factors of different importance ranked 1 through 10 and “how comfy the cars seats are” is ranked 2, then to multiplying by its weighting is multiplying its score of e.g 80 by 2?

3x + 8y + 5z

From a mathematical perspective, this cannot be simplified further and the terms can’t be combined.[…]Terms can only be added when every part is identical other than the number.

I don’t know what the term ‘terms’ means in this context. It seems like it’s being used in a math way.

I looked this up and wikipedia says:

In elementary mathematics, a term is either a single number or variable, or the product of several numbers or variables.

Okay I think I’ve got it. In Elliot’s example 3x, 8y, and 5z are terms.

The way fractions work is related. You can only additively combine fractions if the denominators are the same. If they’re different, you have to convert them to the same denominator by multiplying by a conversion factor, which equals one, before you can add them into one term.

I understand the point here about how additively combining fractions is related. But I don’t understand the ‘which equals one’ part. I don’t think you mean that you convert them to the same denominator by multiplying them by one (because afaik that wouldn’t convert them?). But I can’t think of another thing that this would mean.

So reading on, does ‘which equals one’ mean: a conversion factor for a variable which would make it equal one amount of the other variable? So applied to fractions, you convert one fraction’s denominator to another by multiplying by some amount so that it equals one amount of the other denominator?

Equivalently to using substitution to convert, we can use a conversion factor: 3𝑦/𝑥 = 1 (which I got by taking 𝑥 = 3𝑦 and dividing both sides by x). (Note: The conversion factor is just the fraction, not the whole equation, but I want to emphasize that all conversion factors are equal to one.) We can multiply the 3x by this conversion factor to change it from x units to y units, and we’ll get 9y just like when we used the substitution method.

I’m not confident that I understand the conversion factor equalling one concept and could use it easily. I didn’t have a problem with the substitution concept.

“which I got by taking x = 3y and dividing both sides by x” okay I understand how the math for this worked but not why you did it.

“We can multiply the 3x by this conversion factor to change it from x units to y units, and we’ll get 9y just like when we used the substitution method.”

I just tried to simplify the equation 3x * 3y/x and I couldn’t do it.

I’m going to stop there for today and continue reading the article tomorrow.

Data:
Start: 26 Apr 2024 at 1:14:24 PM
Stop: 26 Apr 2024 at 2:44:08 PM
Had some 5 min breaks.

43

The function approach has been very helpful. For one, it’s made it easier to check my own work. Combined with the realisation that participles and not just gerunds can end in -ing, I think I have made some progress.

I read many things explaining how smoking is bad

read(I, things(many, explaining(how(is(smoking, bad)))))

  • read = action-verb(subject, object): finite clause
    • I = subject
    • things = noun(modifier, modifier): noun = object
      • many = modifier
      • explaining = participle(modifier): modifier
        • how = subordinator(finite-clause): modifier
          • is = linking-verb(subject, complement): finite-clause
            • smoking = gerund(): noun = subject
            • bad = modifier = complement

Participles return modifiers, unlike gerunds which return nouns. So as a participle I can have ‘explaining’ in this sentence modifying ‘things’. I think this is the key to fixing some problems with my earlier tree that had ‘including’ which I’ll fix up and explain tomorrow.

44

Great. And your smoking sentence analysis is fine.

45

You read enough for now.

Find some prealgebra resources and see if you can learn to do this.

46

Okay, I can see how this was simplified to 9y now.

3x \times \frac {3y}{x} \\ =3x \times 3y \times \frac {1}{x} \\ \\ =3y \times \frac {3x}{x} \\ \\ =3y \times 3 \times \frac {x}{x} \\ \\ =3y \times 3 \times 1 \\ \\ = 9y
47

Do you know what you had trouble with before?

48

Yes, apart from my simplifying skills being limited and in general being rusty (I hadn’t done this in earnest since high school), I also didn’t understand how to simplify a fraction. So once I understood that:

\frac {x}{x} = x \times \frac {1}{x} = \frac{(x \times 1)}{x}

I understood how to do it.

I’ve been doing some practise today ~2hrs on combining terms, solving equations, and simplifying.

49

Do some practice problems for arithmetic with fractions, without variables. Including converting between mixed numbers and improper fractions in both directions.

50

Cool. I’ve done about ~3.5 hours today doing worksheets, reading, and watching videos. Had 4 or 5 5min breaks.

One of the things i’ve been reading through is this pdf:

http://maths.mq.edu.au/numeracy/consolidated_mums/module1/module1.pdf

It has some of the info i needed. But it’s not great. For example it introduces operations on fractions in 1.3 sections 3 & 4, after sections 1 & 2 where it uses operations on fractions in its examples of simplifying fractions and mixed numbers and improper fractions, but that wasn’t a big deal.

Some of the worksheets I did were here and here.

I didn’t run into any problems that I couldn’t figure out with these. I haven’t gone over division of fractions in enough depth just yet but I think I’ve done all I can on this today. I’ll go over that tomorrow.

One problem that took me a little while to get my head around conceptually, was why:

\frac {x \times n}{y \times n} = \frac {x}{y} \times \frac {n}{n} = \frac {x}{y}

Like, everything I found on it said that they were equivalent, but didn’t offer much in the way of conceptual explanation. One way I understand it is that the ratio of x to y stays the same when you multiply both x and y by the same amount n. It just changes the amounts in which the ratio x to y is expressed. In other words xn/yn and x/y are equivalent fractions. When simplifying the fraction, the factors in common between the denominator and the numerator simplify to 1. So you have

\frac {x \times 1}{y \times 1}

which is the same as

\frac {x}{y}

This makes sense, but it still doesn’t feel super concrete to me. It’s tempting to just learn the algorithms to use in these situations but I think that’s going to make progress down the line a little harder.

51

Do you understand why x*y*n = n*x*y? Why you can reorder multiplications?

52

Yeah, i do kind of intuitively, and I do know that I can do that. Three variables like that make me think of a 3 dimensional object, and that its volume is the same regardless of what order you gives its dimensions in. I don’t think I could explain well why that is the case though.

53

Do you understand how subtraction is a form of addition, and how you can rewrite equations with no subtractions? (Instead of subtracting 4, you add negative 4.)

54

Yup, I do.

55

Do you understand how division is a form of multiplication, and how you can rewrite equations with no divisions? (Instead of dividing by 4, you multiply by one quarter.)

56

Yes

57

so xn/yn is 4 things multiplied together, in any order: x, n, 1/y, 1/n. (for this, look at 1/y and 1/n as ways to write particular real numbers, not as indicating a division operation. they can be written in other ways without division, like 1/n might be the number 0.5 or 0.7 or some other number).

so you can take any two of the things you want to deal with, like n and 1/n, and multiply them first since order doesn’t matter.

hopefully that gives you some leads. i’m logging off now.

1 Like
58

One of the next things you should do is prime factorization practice. You should practice enough problems that all primes below 10 become memorized.

After that, one of your next steps will be using prime factorization to simplify fractions. After you’ve practiced that, I’d like you to remember all primes below 20.

Try to learn the low primes as a natural result of practicing factorizing, not by directly memorizing them.

1 Like
59

This is an update on my perspective on the problems I encountered in the article.

Using a conversion factor that equals one lets you add together two fractions that don’t have the same denominator. Because you are multiplying by one you aren’t changing the number, so you can convert it to an equivalent fraction with the same denominator. You did say ~this in the article, but I couldn’t see why. When converting mixed numbers to improper fractions I used this method and I can see how it works now.

And so I understand better why you did this (to make a conversion factor equal to one):

60

This makes sense to me. Great!

61

Spent ~1hr on prime factorisation. Included some reading on composite numbers, prime numbers, and units.

Did ~40 prime factorisations using trees.

Used prime factorisation to simplify 18 fractions with 100% accuracy.

primes under 10 from memory are: 2, 3, 5, 7

I had no trouble with any of this.

1 Like