backwards reasoning working out (1 Viewer)

ilovemangos · Nov 16, 2023

hi, just checking that this is a valid method for proof. also how would you structure it? would you just say "therefore by backwards reasoning ___ is true" or draw arrows going back in your working out? thank you

cossine · Nov 16, 2023

ilovemangos said:
hi, just checking that this is a valid method for proof. also how would you structure it? would you just say "therefore by backwards reasoning ___ is true" or draw arrows going back in your working out? thank you

You can work from the RHS and show it is equal to the LHS.

Sometimes yes it can be useful forwards/backwards however your final proof should be in one direction only. i.e. you can't work left to right than start working right-to-left. Don't ask me why. Weird quirk of the HSC

Trebla · Nov 16, 2023

ilovemangos said:
hi, just checking that this is a valid method for proof. also how would you structure it? would you just say "therefore by backwards reasoning ___ is true" or draw arrows going back in your working out? thank you

Generally I would avoid this type of structure because it is susceptible to logical flaws and it is not really a well argued “proof” if you start with the statement you are trying to prove in the first place. In fact, some arguments in the converse direction may not actually be true.

A classic example would be to prove definite integral inequalities. It is not correct to start with the integral inequality then differentiate both sides to arrive at something that is true. This is because differentiating both sides of an inequality in general does not necessarily preserve the inequality sign (nor does it necessarily flip the inequality sign). It may be a useful hack to find the starting point of the proof (i.e here is a true inequality and then take definite integral of both sides to get the result) but it is not a valid proof itself.

ilovemangos · Nov 16, 2023

yeah true, i was introduced to this method in a number proof question: Prove that the second digit after the decimal point of √2 is 1 without a calculator.
and the working out started with this:
1.41 < √2 < 1.42
then it squared everything.
then it evaluated 1.41^2 and 1.42^2 and showed that the inequality was true.

is there an alternate method possible for this question?

but yeah some questions seem to just be unlocked with this method, kind of like reverse engineering the question

Luukas.2 · Nov 21, 2023

ilovemangos said:
hi, just checking that this is a valid method for proof. also how would you structure it? would you just say "therefore by backwards reasoning ___ is true" or draw arrows going back in your working out? thank you

Sometimes a better approach would be to use a proof by contradiction.

For example, if $a$ and $b$ are unequal and positive reals, then $a+b > 2\sqrt{ab}$ can easily be proven forwards starting from

$\begin{align*} \left(\sqrt{a} - \sqrt{b}\right)^2 &> 0 \\ \left(\sqrt{a}\right)^2 - 2\sqrt{ab} + \left(\sqrt{b}\right)^2 &> 0 \\ a - 2\sqrt{ab} + b &> 0 \\ a + b &> 2\sqrt{ab} \qquad \qquad \text{as required} \end{align*}$

But equally, rather than working from the result backwards to find the starting point, and then having to re-write forwards, a proof by contradiction can be constructed based on the reasoning used to find the starting point (if it was not already known):

Proof - By Contradiction:

Assume that the theorem is false. Thus, there must exist some real and unequal values $a$ and $b$ such that $a+b \leq 2\sqrt{ab}$ . It follows that:

$\begin{align*} 0 < a + b &\leq 2\sqrt{ab} \qquad \qquad \text{as both $a,\,b \in \mathbb{R}^+$} \\ 0^2 < \left(a + b\right)^2 &\leq \left(2\sqrt{ab}\right)^2 \\ a^2 + 2ab + b^2 &\leq 4ab \\ a^2 - 2ab + b^2 &\leq 0 \\ (a - b)^2 &\leq 0 \end{align*}$

This cannot be true as $a \neq b$ and thus $a - b \in \mathbb{R}^\pm$ and so must have a positive square.

This contradiction inevitably follows from the assumption, and so the assumption must be false and the theorem must be true.

PS: @jimmysmith560, maybe this belongs in the current MX2 forum, rather than here in the 2018 archive?

Luukas.2 · Nov 21, 2023

ilovemangos said:
yeah true, i was introduced to this method in a number proof question: Prove that the second digit after the decimal point of √2 is 1 without a calculator.
and the working out started with this:
1.41 < √2 < 1.42
then it squared everything.
then it evaluated 1.41^2 and 1.42^2 and showed that the inequality was true.

is there an alternate method possible for this question?

but yeah some questions seem to just be unlocked with this method, kind of like reverse engineering the question

This method is valid as, for any positive reals,

$0 < a < b \implies 0 < a^2 < b^2$

because

$b^2 - a^2 = (b + a)(b - a) > 0 \implies b - a > 0$ .

It suffices as a method to establish that the decimal expansion begins

$\sqrt{2} = 1.41 ...$

as the function $f(x) = x^2 - 2$ is continuous throughout its domain and changes sign on the open interval $x \in (1.41,\,1.42)$ , and so there must be at least one solution of $f(x) = 0$ within that interval - and since there are only two solutions, $x = \pm\sqrt{2}$ , it must be the positive value that satisfies

$1.41 < \sqrt{2} < 1.42$

Note, however, that you would need to show that the root is within $x \in (1.41,\,1.415)$ to establish that $\sqrt{2} \approx 1.41$ to three significant figures.

A method for finding roots called Newton's method would provide an alternative approach to this problem, but is no longer included in the MX1 or MX2 syllabi.

For example, starting with an initial approximation of $x_0 = 1.5$ , the sequence of approximations $\{x_n\}$ would be:

$\begin{align*} \text{The exact decimal} \qquad \qquad \sqrt{2} &= 1.414\ 213\ 562... \\ \\ \text{The first approximation} \qquad \qquad x_0 &= \frac{3}{2} = 1.5 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \text{is correct to 0 decimal places} \\ \text{The second approximation} \qquad \qquad x_1 &= \frac{3}{2} - \cfrac{\left(\frac{3}{2}\right)^2 - 2}{2\times\frac{3}{2}} = \frac{17}{12} = 1.416\ 666\ 666... \qquad \qquad \text{is correct to 2 decimal places} \\ \text{The third approximation} \qquad \qquad x_2 &= \frac{17}{12} - \cfrac{\left(\frac{17}{12}\right)^2 - 2}{2\times\frac{17}{12}} = \frac{577}{408} = 1.414\ 215\ 686... \qquad \qquad \text{is correct to 5 decimal places} \end{align*}$

Starting from a better first approximation, $x_0 = 1.45$ , the approximations converge to $\sqrt{2}$ more rapidly:

$\begin{align*} \text{The exact decimal} \qquad \qquad \sqrt{2} &= 1.414\ 213\ 562... \\ \\ \text{The first approximation} \qquad \qquad x_0 &= \frac{29}{20} = 1.45 \qquad \qquad \qquad \qquad \text{is correct to 1 decimal place} \\ \text{The second approximation} \qquad \qquad x_1 &= \frac{1641}{1160} = 1.414\ 655\ 17... \qquad \qquad \text{is correct to 3 decimal places} \\ \text{The third approximation} \qquad \qquad x_2 &= \frac{5\ 384\ 081}{3\ 807\ 120} = 1.414\ 213\ 63... \qquad \qquad \text{is correct to 5 decimal places} \end{align*}$

... though I do admit that calculating the decimal form of some of these fractions isn't easy without a calculator!

Under the old syllabus, there was a final question on an MX2 paper (1988, if I recall correctly) that showed that using Newton's Method and starting from the first approximation $\sqrt[3]{2} \approx 2$ , the 12th approximation to $\sqrt[3]{2}$ was correct to at least 267 decimal places.

tywebb · Nov 21, 2023

Luukas.2 said:
1988, if I recall correctly

Yeah 1988 Q8b part iv

ilovemangos · Nov 21, 2023

thank you so much, yep i realised that i could just do a contradiction proof most of the time instead of actually doing it backwards and then having to rewrite it.

Luukas.2 · Nov 21, 2023

tywebb said:
Yeah 1988 Q8b part iv

Thanks, that's it... I thought I was doing well remembering that it was $\sqrt[3]{2}$ and that it was 267 decimal places!

backwards reasoning working out (1 Viewer)

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