# First Year Mathematics B (Integration, Series, Discrete Maths & Modelling) (1 Viewer)

#### leehuan

##### Well-Known Member

I remember doing proof by contradiction. I also remember an interesting discussion on mathstackexchange about proof by contradiction and contrapositive.

But like, if you have a super logical mind, then you'll recognise contrapositive. Consider: https://en.wikipedia.org/wiki/Wason_selection_task
That's nice and all but unfortunately not the entire 1231 cohort because most are there cause they do engineering not maths (and not all of them also do 1081)

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#### seanieg89

##### Well-Known Member

It may just be unfair that we know a method some normal maths students wouldn't know but let's not take the risk.
If a student taking any tertiary maths course doesn't know how to argue by contradiction (of which contrapositive is a special case), that is 100% on them.

It's like not understanding the notion of disproving a non-existence statement by providing a counterexample. These are fundamental notions of proof, and if you do not have decent facility with them you shouldn't expect to be able to answer everything thrown at you.

Some unis offer introductory courses on mathematical thinking that explicitly go through methods of proof, but there are also countless resources online. The onus is on a struggling student to recognise such fundamental gaps in his knowledge and fill them.

#### leehuan

##### Well-Known Member

If a student taking any tertiary maths course doesn't know how to argue by contradiction (of which contrapositive is a special case), that is 100% on them.
So a ton of engineering students are screwed because they're only doing maths because they need the tools and aren't interested in the nature of proof?

#### seanieg89

##### Well-Known Member

So a ton of engineering students are screwed because they're only doing maths because they need the tools and aren't interested in the nature of proof?
They aren't "screwed", but if they don't learn how to prove things (which is completely fine if it is a decision they have made) they shouldn't expect full marks in a typical math course.

You could still pass and do reasonably well in most year math courses regardless, which is probably the goal of an engineering student that doesn't care too much about mathematical rigour.

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#### leehuan

##### Well-Known Member

They aren't "screwed", but if they don't learn how to prove things (which is completely fine if it is a decision they have made) they shouldn't expect full marks in a typical math course.

You could still pass and do reasonably well in most year math courses regardless, which is probably the goal of an engineering student that doesn't care too much about mathematical rigour.
Well I'm not sure why I used the word "screwed" any more because yeah that is a bit excessive. Still if they want to do really well I don't exactly think they should be asked on something they haven't been taught; you're not meant to assume the students will learn it themselves

#### Flop21

##### Well-Known Member

When solving integration questions that involve partial fractions... usually I turn it into something like this:

A/(x-5) + B/(x+1)

3x-8 = A(x+1) + B(x-5)

and find A and B by making x equal something that cancels A/B out.

What do I do when it cannot cancel out? E.g. I get x^2 + x = A(x^2 + 2) + B(x-2). See I cannot make A go away here by making x equal something.

#### iforgotmyname

##### Metallic Oxide

When solving integration questions that involve partial fractions... usually I turn it into something like this:

A/(x-5) + B/(x+1)

3x-8 = A(x+1) + B(x-5)

and find A and B by making x equal something that cancels A/B out.

What do I do when it cannot cancel out? E.g. I get x^2 + x = A(x^2 + 2) + B(x-2). See I cannot make A go away here by making x equal something.
Wut... If it's linear then you can just sub in the numbers like (3x-8) / (x+1)(x-5) you can sub in x=-1 and 5 to get A and B respectively

#### Flop21

##### Well-Known Member

Wut... If it's linear then you can just sub in the numbers like (3x-8) / (x+1)(x-5) you can sub in x=-1 and 5 to get A and B respectively
What do you mean? If you sub x=-1 or 5 in that you will get a math error since the denominator = 0.

#### iforgotmyname

##### Metallic Oxide

What do you mean? If you sub x=-1 or 5 in that you will get a math error since the denominator = 0.
You ignore the side where it is zero and it gives you the factors. The method is inside the purple book

#### leehuan

##### Well-Known Member

When solving integration questions that involve partial fractions... usually I turn it into something like this:

A/(x-5) + B/(x+1)

3x-8 = A(x+1) + B(x-5)

and find A and B by making x equal something that cancels A/B out.

What do I do when it cannot cancel out? E.g. I get x^2 + x = A(x^2 + 2) + B(x-2). See I cannot make A go away here by making x equal something.
Once you make B go away by letting x=2 you have a value for A. Which you can plug back in.

Then, one way of finishing it off is just to sub any random number, say 0. Here, you'd get 0 = 2A - 2B.
So sub in your value of A and it still works out.

• Flop21

#### Flop21

##### Well-Known Member

Once you make B go away by letting x=2 you have a value for A. Which you can plug back in.

Then, one way of finishing it off is just to sub any random number, say 0. Here, you'd get 0 = 2A - 2B.
So sub in your value of A and it still works out.
Ohhhhhhh, nice, works out fine thanks.

I see in the purple book it has methods for doing all sorts of things, using C, not just A and B. And like Bx+C. I'll probably have to look over that anyway for the more complicated ones.

#### leehuan

##### Well-Known Member

Ohhhhhhh, nice, works out fine thanks.

I see in the purple book it has methods for doing all sorts of things, using C, not just A and B. And like Bx+C. I'll probably have to look over that anyway for the more complicated ones.
Yeah for Bx+C you might just want to sub in a random value.

(Also, note that 0 was just convenient for the example you provided. Sometimes you may need to sub like say 1 or -1 or something else.)
(And it isn't the only way to do it, it's just the easiest to remember)

• Flop21

#### seanieg89

##### Well-Known Member

Well I'm not sure why I used the word "screwed" any more because yeah that is a bit excessive. Still if they want to do really well I don't exactly think they should be asked on something they haven't been taught; you're not meant to assume the students will learn it themselves
If HSC maths questions (eg irrationality of e or pi) can involve proof by contradiction, why is it so obscene that questions from a first year uni course should? I am sure lots of high school students aren't formally taught proof by contradiction either, but this does not stop it from being in exams that are far more rigidly constrained by syllabus than any uni course.

In any case, they are not being "asked on" proof by contradiction, that is just the most convenient/natural way to phrase the argument. Another possible argument is as follows:

T: Span(S)->Span(R) is a linear map between vector spaces.

By rank-nullity:

dim(span(S))=dim(ker(T))+dim(im(T)) >= dim(im(T)) = dim(R)= 3 since the three vectors in R are linearly independent.

We also have dim(span(S)) =< 3, since it is the span of 3 vectors. Hence dim(span(S))=3 and so the spanning set {v1,v2,v3} must be linearly independent.

This is far less elegant than the previous argument, but it is still an easy proof that a first year could construct which is not a proof by contradiction.

#### leehuan

##### Well-Known Member

Actually, whilst I can't deny the reduced difficulty of the HSC exams over the years, if it were to appear now I would be thoroughly displeased at that as well.

I don't know what the HSC was like in 2003 to speak for back then

Thanks for that proof though; that I'll show him as well.

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#### Flop21

##### Well-Known Member

Once you make B go away by letting x=2 you have a value for A. Which you can plug back in.

Then, one way of finishing it off is just to sub any random number, say 0. Here, you'd get 0 = 2A - 2B.
So sub in your value of A and it still works out.
I've become stuck on this again

[3x + 2]/[(x^2+2x+2)(x-1)] = 1/(x-1) + (bx+c)/(x^2+2x+2).

So I go 3x+2 = A(x^2+2x+2) + (bx+c)(x-1)... right?

A is 1 like they already give us. But I can't seem to find an x value that will allow me to find bx+c?

#### InteGrand

##### Well-Known Member

I've become stuck on this again

[3x + 2]/[(x^2+2x+2)(x-1)] = 1/(x-1) + (bx+c)/(x^2+2x+2).

So I go 3x+2 = A(x^2+2x+2) + (bx+c)(x-1)... right?

A is 1 like they already give us. But I can't seem to find an x value that will allow me to find bx+c?
Sub. x = 0, which easily lets us find C. Then we can find B by subbing (say) x = 2.

• Flop21

#### Flop21

##### Well-Known Member

When doing first order linear differential equations...

How did this turn into this? #### Flop21

##### Well-Known Member

Sub. x = 0, which easily lets us find C. Then we can find B by subbing (say) x = 2.
Oh thanks, silly me was getting confused when I was getting c = 0 for some reason. But that's the answer.

#### InteGrand

##### Well-Known Member

When doing first order linear differential equations...

How did this turn into this? Product rule.

#### InteGrand

##### Well-Known Member

Actually, whilst I can't deny the reduced difficulty of the HSC exams over the years, if it were to appear now I would be thoroughly displeased at that as well.

I don't know what the HSC was like in 2003 to speak for back then

Thanks for that proof though; that I'll show him as well.
Can you recall whether there have been any contradiction proofs in the HSC more recently than 2003? Surely there should've been (though I can't remember any off the top of my head)?

• kawaiipotato