Extension 2 Inequalities (1 Viewer)

_luke_ · Nov 13, 2015

Hi all,

Just wondering whether for an inequality question, that in proving the inequality, it is sufficient to work backwards to obtain a solution that cannot be refuted (e.g.a square number is greater than or equal to 0).

If not, is there any advice you could give me on how to prove these solutions?

Thanks

Drsoccerball · Nov 13, 2015

_luke_ said:
Hi all,

Just wondering whether for an inequality question, that in proving the inequality, it is sufficient to work backwards to obtain a solution that cannot be refuted (e.g.a square number is greater than or equal to 0).

If not, is there any advice you could give me on how to prove these solutions?

Thanks

Yes you are allowed to work with the inequality backwards and come to an established result.

InteGrand · Nov 13, 2015

_luke_ said:
Hi all,

Just wondering whether for an inequality question, that in proving the inequality, it is sufficient to work backwards to obtain a solution that cannot be refuted (e.g.a square number is greater than or equal to 0).

$This is indeed logically valid, assuming you are using ``if and only if'' steps, and it is good to make this explicit by using the $\Longleftrightarrow$ symbol between lines.$

Trebla · Nov 13, 2015

_luke_ said:
Hi all,

Just wondering whether for an inequality question, that in proving the inequality, it is sufficient to work backwards to obtain a solution that cannot be refuted (e.g.a square number is greater than or equal to 0).

If not, is there any advice you could give me on how to prove these solutions?

Thanks

Ideally you should start from the inequality that 'cannot be refuted' and work your way towards the inequality you are trying to prove because the logical flow makes more sense.

leehuan · Nov 13, 2015

To build on what Trebla said however, you can just work backwards then copy out your steps in the reverse order to look like you're working forwards.

Carrotsticks · Nov 13, 2015

leehuan said:
To build on what Trebla said however, you can just work backwards then copy out your steps in the reverse order to look like you're working forwards.

I teach my kids to work backwards (in the case that they have absolutely no clue and are desperate), but starting from the bottom of the page.

This way, it looks like you're a genius and figured out each step magically, and the work is literally halved compared to the method you outlined.

Carrotsticks · Nov 13, 2015

InteGrand said:
$This is indeed logically valid, assuming you are using ``if and only if'' steps, and it is good to make this explicit by using the $\Longleftrightarrow$ symbol between lines.$

I would be wary about using this, although it is mathematically correct.

HSC markers are absolutely ruthless against students trying any sort of 'backwards' method of inequality proofs. They've been known to give zero the instant they see anything that remotely smells like a backwards proof, and then move on from there without looking back.

Also, unless your teacher is very mathematically sound, I highly doubt they would accept it either for school assessments. I have taught this technique to many of my students and the general report is that the teacher won't accept it, even the ones who acknowledge its correctness.

In a university setting, by all means go for it. But in the flawed high school setting, maybe not.

InteGrand · Nov 13, 2015

Carrotsticks said:
I would be wary about using this, although it is mathematically correct.

HSC markers are absolutely ruthless against students trying any sort of 'backwards' method of inequality proofs. They've been known to give zero the instant they see anything that remotely smells like a backwards proof, and then move on from there without looking back.

Also, unless your teacher is very mathematically sound, I highly doubt they would accept it either for school assessments. I have taught this technique to many of my students and the general report is that the teacher won't accept it, even the ones who acknowledge its correctness.

In a university setting, by all means go for it. But in the flawed high school setting, maybe not.

$Yes, I would not do that in the HSC, despite its validity, because the markers seem to hate it and give 0, as you say.$

$On the topic of proofs in the HSC, if a question asked something like prove $\frac{\sin \theta}{1-\cos \theta}=\frac{1+\cos \theta}{\sin \theta}$, would something like this be allowed in the HSC?:$

$We know $\sin^2 \theta + \cos^2 \theta = 1$ (Pythagorean identity).$

$\begin{align*}\therefore \sin^2 \theta &= 1-\cos^2 \theta \\ \Rightarrow \sin\theta \cdot \sin \theta &= (1+\cos \theta)(1-\cos \theta) \quad (\text{factorising the R.H.S.}) \\ \Rightarrow \frac{\sin \theta}{1-\cos \theta}&=\frac{1+\cos \theta}{\sin \theta} \quad (\text{rearranging}).\,\,\,\, Q.E.D.\end{align*}$

$I remember doing something like that in a school assessment once and I got 0 for doing it haha.$

Drongoski · Nov 13, 2015

InteGrand said:
$Yes, I would not do that in the HSC, despite its validity, because the markers seem to hate it and give 0, as you say.$

$On the topic of proofs in the HSC, if a question asked something like prove $\frac{\sin \theta}{1-\cos \theta}=\frac{1+\cos \theta}{\sin \theta}$, would something like this be allowed in the HSC?:$

$We know $\sin^2 \theta + \cos^2 \theta = 1$ (Pythagorean identity).$

$\begin{align*}\therefore \sin^2 \theta &= 1-\cos^2 \theta \\ \Rightarrow \sin\theta \cdot \sin \theta &= (1+\cos \theta)(1-\cos \theta) \quad (\text{factorising the R.H.S.}) \\ \Rightarrow \frac{\sin \theta}{1-\cos \theta}&=\frac{1+\cos \theta}{\sin \theta} \quad (\text{rearranging}).\,\,\,\, Q.E.D.\end{align*}$

$I remember doing something like that in a school assessment once and I got 0 for doing it haha.$

Nothing wrong with the proof. I find this approach rather cool.

Carrotsticks · Nov 13, 2015

InteGrand said:
$Yes, I would not do that in the HSC, despite its validity, because the markers seem to hate it and give 0, as you say.$

$On the topic of proofs in the HSC, if a question asked something like prove $\frac{\sin \theta}{1-\cos \theta}=\frac{1+\cos \theta}{\sin \theta}$, would something like this be allowed in the HSC?:$

$We know $\sin^2 \theta + \cos^2 \theta = 1$ (Pythagorean identity).$

$\begin{align*}\therefore \sin^2 \theta &= 1-\cos^2 \theta \\ \Rightarrow \sin\theta \cdot \sin \theta &= (1+\cos \theta)(1-\cos \theta) \quad (\text{factorising the R.H.S.}) \\ \Rightarrow \frac{\sin \theta}{1-\cos \theta}&=\frac{1+\cos \theta}{\sin \theta} \quad (\text{rearranging}).\,\,\,\, Q.E.D.\end{align*}$

$I remember doing something like that in a school assessment once and I got 0 for doing it haha.$

This is perfectly fine and any teacher wishing to not award you full marks for this should seriously consider a new profession.

Trebla · Nov 14, 2015

Just also be wary that the whole 'backwards approach' only works for if and only if logical pathways (i.e. the converse must also be true).

Many inequality questions in the HSC are of the form of "if A is true then prove inequality B". It is possible that if inequality B holds then condition A does NOT hold (i.e. the converse statement is not true) so you can't really go backwards. A trivial example of this is to show that when x > 1 then x² - 1 > 0. If you decide to go backwards with x² - 1 > 0 then you cannot then say that therefore x > 1.

Generally speaking, the "forward" way (i.e. starting from a known inequality rather than the result) works as a logically valid approach for a broader case of inequalities you cover in the HSC, whereas the "backward" way would only be a logically valid approach for a subset of such inequalities.

InteGrand · Nov 14, 2015

Trebla said:
Just also be wary that the whole 'backwards approach' only works for if and only if logical pathways (i.e. the converse must also be true).

Many inequality questions in the HSC are of the form of "if A is true then prove inequality B". It is possible that if inequality B holds then condition A does NOT hold (i.e. the converse statement is not true) so you can't really go backwards. A trivial example of this is to show that when x > 1 then x² - 1 > 0. If you decide to go backwards with x² - 1 > 0 then you cannot then say that therefore x > 1.

Generally speaking, the "forward" way (i.e. starting from a known inequality rather than the result) works as a logically valid approach for a broader case of inequalities you cover in the HSC, whereas the "backward" way would only be a logically valid approach for a subset of such inequalities.

$This is why if one wants to use the ``backward way'', it is a good idea to use the $\Longleftrightarrow$ symbol between lines of working, to make explicit the biconditional nature of the statements (assuming they are indeed biconditional; if they're not, this method won't be valid!).$

Carrotsticks · Nov 14, 2015

Trebla said:
Just also be wary that the whole 'backwards approach' only works for if and only if logical pathways (i.e. the converse must also be true).

Many inequality questions in the HSC are of the form of "if A is true then prove inequality B". It is possible that if inequality B holds then condition A does NOT hold (i.e. the converse statement is not true) so you can't really go backwards. A trivial example of this is to show that when x > 1 then x² - 1 > 0. If you decide to go backwards with x² - 1 > 0 then you cannot then say that therefore x > 1.

Generally speaking, the "forward" way (i.e. starting from a known inequality rather than the result) works as a logically valid approach for a broader case of inequalities you cover in the HSC, whereas the "backward" way would only be a logically valid approach for a subset of such inequalities.

I think this was addressed already.

InteGrand said:
$This is indeed logically valid, assuming you are using ``if and only if'' steps, and it is good to make this explicit by using the $\Longleftrightarrow$ symbol between lines.$

Sy123 · Nov 14, 2015

For those who are curious, the arithmetical manipulations preserve the truth of the inequality if and only if the function being applied is monotone increasing across the domain of the inequality. If it was monotone decreasing on the other hand, the inequality sign 'flips', in fact this is why the inequality sign flips when you multiply by -1, or if you reciprocate both sides (if both sides are the same sign), precisely because f(x) = -x is monotone decreasing, and so is f(x) = 1/x if both sides of the inequality are the same sign.

The reason for this is clear from the definition of an increasing function, that is:

$\\ $Function$ \ f \ $is monotone increasing if and only if, for all$ \ a < b \ $then$ \ f(a) < f(b)$

In fact, all the exceptions to the manipulation rule of inequalities are because the functions being applied are not monotone across the domain.

For instance, 1 > -2, but (1)^2 > (-2)^2 is not true. Here, the function $f(x) = x^2$ was applied across the domain $-2 \leq x \leq 1$ , but we know that $f(x) = x^2$ is not monotone, and thus the inequality is not necessarily preserved.

Extension 2 Inequalities (1 Viewer)

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