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Here's another way:

It got taken out because determinants and matrices got taken out. So when they went, so did this little area formula. It had nothing to do with validity.

It was taken out (along with many other things) because the course was changed from a 2-year course to a 1-year course: http://community.boredofstudies.org/2220688-post2.html

Dare I say use it at Ian Woodhouse's risk. It IS a superior method and therefore I do recommend using it, despite Ian Woodhouse (or perhaps more poignantly - BECAUSE of Ian Woodhouse........)

I assure you it has been seen since 1980 and it should resonate with people today because it's all over the internet: eg., http://mathworld.wolfram.com/TriangleArea.html

It used to be in the old Level 1 course: http://community.boredofstudies.org/2220688-post2.html It was taken out of the syllabus in 1980. But it's still valid. So I'd say still use it. But Ian Woodhouse would say don't because it's not in the current syllabus, as he said last Saturday...

No need. It is already at the 2 unit resources page http://www.boredofstudies.org/view.php?course=1

Good. Thanks for that. So &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;))=4.000000303.... > 4 (by calculator) and therefore &pi;⁴+&pi;⁵ < e⁶. But if the idea is to avoid calculators altogether, won't you need to prove &alpha;=(ln(&pi; +1) -2)/(1 - ln(&pi;)) is...

I tried it a similar way to brett86 and got &pi; < 3.14159266 & 2.718281828 < e. So &pi;⁴+&pi;⁵ < 3.14159266⁴+3.14159266⁵ = 403.4287797363725532846657585016733756588576 (exactly) <...

I disagree. "e=2.7... . Therefore e < 3" is a perfectly valid proof.

I just think it's sad that no_arg still thinks &pi;⁴+&pi;⁵=e⁶ exactly. It doesn't. &pi;⁴+&pi;⁵=403.42877... and e⁶=403.42879... Hence &pi;⁴+&pi;⁵&ne;e⁶ It's quite simple, really.

I was trying to edit a post in the <a href="http://community.boredofstudies.org/showthread.php?t=81019">pi and e thread</a> when McLake closed it. The - should have been a + in my last post in that thread at the bottom. So here it is again. Here's another proof using only integers...

Here's another proof using only integers. e*10⁹ > 2718281828 and &pi;*10⁸ < 314159266 Hence (e⁶-&pi;⁴-&pi;⁵).10⁵⁴ > 2718281828⁶-314159266⁴*10²²-314159266⁵*10¹⁴...

Unfortunately, they are both offline at the moment, so we'll have to wait till they come back later.

Exactly! And the reference to Castellanos is Castellanos, D. "The Ubiquitous Pi. Part I." Math. Mag. 61, 67-98, 1988. no_arg has no publication refuting Castellanos's claim, and if he ever submits such a thing for publication, it would get rejected!

Well, you're the one claiming &pi;⁴+&pi;⁵=e⁶ exactly, and your invective does not constitute proof. I've now proved &pi;⁴+&pi;⁵&ne;e⁶ two ways, and several others on this thread have also proved it. You're still claiming...

How about this. &pi; < 3.14159266 & 2.718281828 < e. So &pi;⁴+&pi;⁵ < 3.14159266⁴+3.14159266⁵ = 403.4287797363725532846657585016733756588576 (exactly) < 403.428793083965001476126676589903866851868865301397704704 =...

1989 four unit HSC Question 8(b)(i)(&alpha;) "The difference between a real number r and the greatest integer less than or equal to r is called the fractional part of r, F(r). Thus F(3.45) = 0.45. Note that for all real numbers r, 0 < F(r) < 1. Let a = 2136 log₁₀2. Given that...

Yeah. Anyway at least we know that &pi;⁴+&pi;⁵ is not equal to e⁶! no_arg said they are equal, so I guess you could just deduce from this that they aren't! Really? Well how about this. If f(x)=sin(x), then f''(x)=-sin(x), so f''(0)=0, and f'''(x)=-cos(x)...