Quantum Mechanics/Wave Mechanics in Chemistry Beginners Thread

Quote: Originally posted by aga

Is the "Oxygen group" All elements on the O column (#16) of the periodic table?

Also:

For double bonds in C-C bonds, both C atoms hybridise to sp² and the p_z isn't involved. The sp² AOs bond with each other and the H atoms (σ bonds), the two p_z form a π bond (above and below the C-C σ bond).

Will add a figure (not to scale):



All half-filled sp² AOs and σ bonds in the xy plane. Two ungerate half filled p_z (z axis) AOs form a π bond (above and below the C-C σ bond).

Note: no hydrogen atoms shown.


[Edited on 11-12-2015 by blogfast25]

Quote: Originally posted by aga

Now to see if i can apply the teachings ...

Assignment: draw a triple bond: 1 σ + 2 π bonds. The hybridisation is sp.

Take your time.

[Edited on 11-12-2015 by blogfast25]

Tried in ChemSketch and got nowhere.
Ditto MS Pain.
Best i could manage was in Fireworks :-



Blue is the sigma bond, green one pi bond orbital, magenta/red the other pi orbital, with magenta behind, red in front.

[Edited on 12-12-2015 by aga]

Ships. LOL.



I had a bit of trouble with it in ChemSketch too.

Ignore the acetylene and pent-2-yne for a minute.

My second pair of π lobes are in (parallel to) the xy plane and painted as red solid lines. So your structure is correct.

With one H on each of the two sp orbitals we get acetylene (IUPAC: ethyne), which is an interesting case. Due two the high electron density between the to C, both H are very slightly acidic (but less than ethanol, e.g.) and salts of it exist, in particular calcium carbide, CaC₂. With water that generates acetylene because the acetylene is such a weak acid that even water displaces it.

Another thing about alkynes is that the triple bond imparts on the molecule some linearity not seen in alkanes, see the case of pent-2-yne, where the C atoms 1,2,3 and 4 are all on one line! Compare to the 'squiggly' structure of pentane, e.g.

And talking about double/triple bonds, I'm going to present a few more exercises here, shortly.

[Edited on 13-12-2015 by blogfast25]

Using the electrophilic addition mechanism to alkenes and Markovnikov's Rule, predict the reaction products of the following additions:



[Edited on 12-12-2015 by blogfast25]

Not sure about any of these, especially the last one.

It just looked stupid so it got rearranged to look better.

The answers:



Quite a few errors there. Not quite sure why.

The ones with H2SO4 are perhaps understandably harder because you have to realise that HSO₄^- acts here as a Lewis base and latches on to the carbocation. Will add a diagram tomorrow.

Markovnikov says that the carbocation is formed on the C-atom that is already the most connected to other C-atoms.

In order of stability: primary carbocation < secondary carbocation < tertiary carbocation.

[Edited on 15-12-2015 by blogfast25]

Quote: Originally posted by blogfast25

Quite a few errors there. Not quite sure why.

Quote: Originally posted by aga

#5 was just idiocy and not being able to count to 4. Seeing the mechanisms for #2 and #3 would be very helpful. I think i've hit my limit on how much can be absorbed in a given time. Certainly need to revise the past several lessons again (a few times).

#2:

Note that H2SO4 === > H+ + HSO4(-)



Although the hydrogen sulphate ion (bisulphate ion) is a Bronsted Lowry acid (quite a strong one too!) it has unbonded electron pairs on the O atom, one of which bonds to the hard Lewis acid that is the t-butyl carbonium cation. That bisulphate ion is negatively charged helps, of course.

The resulting structure is itself a strong acid by virtue of that dangling -OH group.

In ChemSketch the structure can be easily drawn as follows. Draw your C structure. Add an O atom where you want the sulphate to be. Choose HSO₃ (right hand side of structure menu) and add it to the O.

A well known acid in this category is the following:



It's a strong, non-oxidising acid, often used in OC.

************

Why don't we build in a pause and you can work on your empirical dissertation: the conversion of alpha-pinene to alpha-terpineol (and beyond)?

I've found another potentially doable reaction with alpha-pinene: the preparation of bornyl chloride, so the money spent on a bottle of high quality turpentine would be well spent!

[Edited on 15-12-2015 by blogfast25]

Quote: Originally posted by aga

A Pause before further enlightment rains down would be appreciated. Some practical would be welcome !

Agreed.

'Blast from the past':

http://www.falstad.com/qmmo/

To see actual calculated versions of the most common molecular orbitals, use this latest version version of the Java applet. They are for the H₂⁺ molecule (no inter-electronic repulsions).

Only works in IE (NOT Chrome) and you'll probably have to gratis upgrade your Java version.

A π (ungerate) 2p_x molecular orbital:



[Edited on 20-12-2015 by blogfast25]

Fugitive Fermions

Quote: Originally posted by blogfast25

A π (ungerate) 2px molecular orbital

Blog the structure you show is not p-TsOH. p-TsOH forms via electrophilic aromatic substitution as the active electrophile is HSO3/SO3 where the sulphur is attacked by the C=C in the arene.

however the below mechanism would in practice be completed using oleum, rather than H2SO4 as with toluene. This is as the methyl group is electron donating which activates the ring system.

source: http://www.chem.ucalgary.ca/courses/351/Carey5th/Ch12/ch12-4...

Quote: Originally posted by HeYBrO

Blog the structure you show is not p-TsOH. p-TsOH forms via electrophilic aromatic substitution as the active electrophile is HSO3/SO3 where the sulphur is attacked by the C=C in the arene.

Quote: Originally posted by HeYBrO

I gave the correction not so much for you but just so no one becomes confused with your structure, hence gave an explanation of the mechanism. great thread btw

Quote: Originally posted by blogfast25

Quote: Originally posted by HeYBrO   I gave the correction not so much for you but just so no one becomes confused with your structure, hence gave an explanation of the mechanism. great thread btw No worries. That mechanism you showed seems for the sulphonation of benzene with oleum, though, as it starts with an attack on the π-ring by SO3, not by H<sup>+</sup>.

Quote: Originally posted by Magpie

There's a subject I would like to see B&D university cover: activity coefficients and fugacity. I know I can read about these on Wiki but what I'm looking for is how a chemist actually uses them. If you can't look them up of what value are they? Can they be derived? When are they of value to the home chemist?

Quote: Originally posted by Darkstar

no longer force students to carry out their chemical reactions via makeshift condensers and flasks made from copper pipes and pickle jars.

Quote: Originally posted by aga

Speaking of which, exactly how does a tiny amount of water interfere in the chlorination of ethanol ? (anhydrous=rapid formation of white trichloroacetaldehyde precipitate, any water=clear, sweet smelling solution). Glad to see you back Darkstar !

Quote: Originally posted by aga

Speaking of which, exactly how does a tiny amount of water interfere in the chlorination of ethanol ? (anhydrous=rapid formation of white trichloroacetaldehyde precipitate, any water=clear, sweet smelling solution).

Carbon atoms as Lewis bases:

Recapping from what we seen so far, many reactions in OC can be summarised by the following scheme:
$$\text{A}+\text{:B} \to \text{A-B}$$

Where A is a Lewis acid, :B a Lewis base and : an unbonded electron pair and A-B the adduct. Of course this step may be preceded and/or followed up with other steps to make up the full reaction path.

So far, our Lewis acid has invariable been a partly positively charged C-atom or a carbocation.

Is it possible for a C-atom to play the part of a Lewis base, an electron pair donor? Enter organo-metal compounds.

Organo-metal compounds:

For those unfamiliar with this class of chemical compounds, there’s at least one you’ll have heard of, i.e. the now infamous lead-based anti-knocking agent tetraethyl lead:



A more interesting one is methyl lithium, CH₃Li, or LiMe.

Over-simplified (the real structure is more complicated and less than fully ionic), LiMe would be made up of negatively charged CH₃ so-called carbanions and positively charged Li cations. The structure of the carbanion is shown below:



The Me carbanion has a tetrahedral structure with an unpaired, filled MO sitting ‘on top’. This makes the ion a very hard Lewis base.

To the right is a generically substituted carbanion. The substitutions tend to stabilise the ion by charge distribution.

As said, the representation of LiMe as an ionic compound is a simplification but one thing is certain: due to the big difference in electronegativity between Li and C, the Li-C bond is highly polarised, with a strong partial charge residing on the C-atom. This makes LiMe a useful reagent in OC.

Organo-metal compounds like LiMe are particularly useful as reagents towards compounds containing a partly positively charged C-atom, like the C=O carbonyl group. This double bond is highly polarised due to the difference in electronegativity between C and O.

Here’s an example of the reaction between LiMe and acetone (carried out in an ether-type solvent):



In the first step the lone electron pair of the carbanion latches on to the partly positively charged carbonyl C-atom, while the double carbonyl bond opens up to the Li cation. An adduct has been formed.

To complete the synthesis, in the second step some weak acid solution is added, which hydrolyses the adduct to a tertiary alcohol.

Note that in the reaction two carbon atoms have been linked together, it’s a so-called C-C coupling reaction.

LiMe has many drawbacks though: its incompatible with oxygen, CO₂ and reacts violently with water and alcohols to boot.

Safer, tamer organo-metal reagents have therefore been developed.

Next instalment: Grignard organo-metal reagents.


[Edited on 7-3-2016 by blogfast25]

Quote: Originally posted by aga

Hello ? Anybody here ? hello ? ello ? An echo. Good. I must be the first here.

Hmmm... for some time one of the absentees here was you.

Normal service to be resumed very shortly now.

Quote: Originally posted by aga

OC wise, are those two Cl on 1,4 dichlorobenze not rip-offable, e.g. with conc NaOH, perhaps with a little electrolysis stimulation?

Without a strong electron-withdrawing group (EWG) ortho or para to the halogen, the only way you're going to get a chloro group off of a benzene ring with NaOH is by reacting them under extremely harsh conditions (like 300 °C at 200 bar/3,000 psi). Cathodic reduction in an electrochemical cell will remove a halogen from a benzene ring, though, albeit through an entirely different mechanism.

As far as the reaction between aryl chlorides and NaOH goes, if an EWG like -NO₂ is present, either in the ortho- or para-position (or both), NaOH will remove the chlorine by first adding to the benzene ring and breaking aromaticity. The unstable intermediate then collapses back into a much more stable aromatic system by eliminating either the hydroxyl group or the chloro group. If the hydroxyl group gets eliminated, a nucleophilic attack by hydroxide can occur again. If the chloro group gets eliminated, the reaction stops. This kind of reaction is a type of nucleophilic aromatic substitution, and proceeds via the usual S_NAr mechanism:



While not shown in the simplified mechanism above, the intermediate of the second step is stabilized by the nitro group like this:



Notice that not only does the nitro group help to further delocalize the negative charge by spreading it out across more atoms, but the nitrogen also has a positive charge on it, increasing the stability of the adjacent negative charge on the carbon that the nitro group is connected to. And because the nitro group is strongly electron-withdrawing, it also significantly increases the electropositivity of the carbon opposite of it as well, making the chloro group carbon much more susceptible to a nucleophilic attack by hydroxide in the initial step of the substitution.

Also, as mentioned above, at high enough temperatures and pressures it's possible for substitution by NaOH to take place even without an EWG; however, substitution in this manner proceeds through a different mechanism that involves the formation of an extremely reactive benzyne intermediate:

It's actually the other way around. Because the benzene ring's carbons are sp²-hybridized and the C-Cl bond has partial double bond character (due to the partial delocalization of one of chlorine's lone pairs into the aromatic ring), the bond length between chlorine and carbon in chlorobenzene is actually shorter than in alkyl chlorides, making it much stronger and harder to break.

Look at the resonance structures for chlorobenzene:



Notice that three of the structures have a double bond between chlorine and carbon. This means that, in reality, the C-Cl bond isn't a true single bond, and the actual structure of chlorobenzene has a partial double bond between chlorine and carbon.

Depending on the solvent and conditions employed, sodium metal can indeed remove aryl chloro groups. What happens is that the sodium transfers an electron to the substrate to give a radical anion, which quickly decomposes into a chloride anion and a phenyl radical. A second sodium atom then transfers an electron to the phenyl radical to give phenylsodium (an organometallic compound):



Subsequent reaction of this superbase with a proton source will replace the sodium with a hydrogen.

Quote: Originally posted by aga

So OC is just as hard as Calculus then. OK.

Based on this particular chapter I'd say it was the other way around!

Quote: Originally posted by blogfast25

Based on this particular chapter I'd say it was the other way around!

Quote: Originally posted by aga

Is that relatively strong, or strong enough to make All the difference reaction-behaviour-wise ?

Quote: Originally posted by aga

All photos erased.

Quote: Originally posted by aga

It would be awesome if we could do the Theory behind a particular reaction, then do the Practical.

Grignard reagents and reactions:

This post is the follow up to Carbon atoms as Lewis bases, please refresh as what follows might not make a lot of sense otherwise.

Grignard reactions make use of organo-metallic compounds known as Grignard reagents, obtained by reacting an organo-halide with magnesium metal. The resulting complex R-CH₂-Mg-X carries a partial negative charge on the Mg bonded C atom due to the strong difference in electronegativity between C and Mg, making it a strong nucleophile (Lewis base).

Below’s an example of the full sequence (three distinct steps) of a Grignard reaction between 1-bromopropane, Mg and pentan-2-one:



Step 1: formation of the Grignard reagent.

1-bromopropane is slowly added to a suspension of Mg shavings in super-dry ether of THF. The organo-magnesium complex forms.

Step 2: formation of the Adduct.

pentan-2-one is added to the solution of Grignard reagent. The partially negatively charged C-Mg atom bonds to the partially positively charged C atom of the carbonyl >C=O group. The partially positively charged Mg atom bonds to the O atom in the carbonyl group.

The reaction between two carbon atoms (one as Lewis base, one as Lewis acid) is often called a C-C coupling reaction or chain extending reaction.

Step 3: hydrolysis of the Adduct.

The adduct is rarely the desired end-product. Usually and equivalent of dilute HCl solution is added, which converts the adduct to alcohol by electrophylic attack of H⁺ on the Mg-O bond.

After work up, the end product 4-methylheptan-4-ol is then obtained.

Carbonyl groups are very suited as Grignard substrates because of the strong polarisation of the C=O bond, as shown here.

As regards the organo-halide starting point, suitability depends on choice of halogen. In order of suitability: I>Br>Cl>F.

Water is the enemy of this reaction, as water hydrolyses the Grignard reagent. All reagents used must therefore be of suitable dryness.

[Edited on 2-5-2016 by blogfast25]

Quote: Originally posted by aga

[...] in that the group are electron deficient. Reading up (in this thread) why the C ends up being a strong Lewis Base instead of the Mg or Br ...

It's simpler to merely consider that the electron density (ψ² is the light green 'bell', remember that? ) of the Mg-C bond is skewed toward the C-atom (the picture is a schematic and not to any scale):



The C-atom therefore carries a partial negative charge.

[Edited on 2-5-2016 by blogfast25]

Quote: Originally posted by aga

Inescapable is the Br's superior electronegativity, which is what makes me wonder why It is not the 'active' component instead of the C.

Good question but easily answered. Br's electronegativity means that it won't share another of its three remaining unbonded MOs with any Lewis acid. So structures like -Mg-Br-C- can never arise. Br^- is an extremely weak Lewis base.

See also HBr:



One could imagine a [H-Br-H]⁺ ('bromonium cation') structure but in reality that doesn't exist (bar perhaps in extreme acid conditions).

Quote: Originally posted by aga

Perhaps a week was being optimistic ...

Formation of Grignard reagents are thought to proceed through a SET (single-electron transfer) mechanism that takes place on the surface of the magnesium metal. For the sake of simplicity, you can envision this reaction as consisting of three steps: oxidation of the magnesium metal, reduction of the organic halide, and formation of the R-MgX complex:



For the purposes of preparing a Grignard reagent as part of your next practical, the above explanation should be more than sufficient; however, if you would like a slightly more advanced understanding of what is going on, we can further break the reaction down to elementary steps and intermediates. While the true mechanism for the formation of Grignard reagents isn't actually known (particularly the mechanism of the halide reduction step), the reaction most likely involves an initial electron transfer from the magnesium metal to the organic halide (R-X) to give a radical anion (R−X^−•) and a magnesium(I) cation (Mg⁺).

The unstable radical anion then collapses into an organic radical (R•) and a halide anion (X⁻), the latter of which serves as the new counterion for the magnesium(I) ion (Mg⁺X⁻). The magnesium(I) ion then transfers a second electron to the organic radical to give a carbanion (R⁻) and a magnesium(II) ion (Mg²⁺), which together with the halide anion make up the organometallic complex known as the Grignard reagent (R-MgX).

The mechanism described above can be conceptualized as follows:



Recall that arrows with a "fishhook" on the end depict the transfer of a single electron, while the more common "full" arrowhead represents the transfer of a pair of electrons. Also, remember that a radical is where an atom has an unpaired valence electron. In other words, one of the occupied valence orbitals only has one electron in it (called a singly occupied molecular orbital, or "SOMO" for short). This typically makes radical species extremely reactive. But don't beat yourself up if you don't completely understand everything discussed above, especially the SET stuff, as it isn't exactly introductory-level material.

As blogfast already pointed out, most of the electron density in the C-Mg bond is shifted towards the carbon. But something else to consider, which hasn't been mentioned yet, is the leaving group you'd get if Br⁻ attacked the carbonyl group. Think about it: why is it that when you add concentrated HBr or a bromide salt to a solution of acetone, you don't end up with this (at least not in any significant quantity):



Surely the nucleophilic bromide ions will try to attack the electrophilic carbonyl carbon, right? And if the solution is strongly acidic, the oxygen on acetone will also be protonated, making the carbonyl carbon even more electrophilic than it normally would be. So why don't we get the molecule shown above?

The reason is primarily due to the unusual instability of geminal halohydrins. Because the carbon is bound to two extremely electronegative groups, it is highly electron-deficient and will immediately attract any nearby electrons. Since the oxygen has an extra pair of electrons that it can form bonds with, and the bromide ion makes a rather good leaving group (i.e. it isn't very strongly attracted back to the carbon as it leaves), the oxygen will just donate a pair of electrons to the electron-deficient carbon and kick out bromine, returning the unfavorable geminal halohydrin back to a much more favorable carbonyl configuration.

Also, if the media is strongly basic, like in Grignard reactions, the oxygen will be deprotonated and carry a negative charge, meaning TWO free lone pairs available to donate to carbon. This makes the elimination of bromine in order to return to a neutral and more stable configuration even MORE favorable. With carbon as the attacking nucleophile, however, elimination in this manner cannot happen, as the leaving group (a negatively-charged carbon atom) is way too strongly attracted back to the electrophilic carbon to leave.

Now you've gone and opened Pandora's box. While Grignard reagents are usually taught at the introductory level as simply being this R-MgX complex floating around in some ethereal solvent (usually diethyl ether or THF), the truth is that they actually exist as tetrahedral adducts with the solvent (which is why they dissolve in it). The reason ethers are so commonly used as solvents in Grignard reactions is because not only are they unreactive towards the Grignard reagent, but they also stabilize the R-MgX complex by coordinating with the magnesium atom and giving it a full octet:



Notice that the magnesium is surrounded by electronegative atoms, which provides a sort of stabilizing effect. Similarly, the Grignard reagent also coordinates with the carbonyl compound to form a six-membered cyclic ring, which may also help to explain why bromine isn't the active attacker. It is really through this structure that most Grignard reactions are thought to actually proceed, though the mechanism is often simplified when first introduced at the lower levels. I haven't done a great deal of research regarding the specifics of these Grignard reaction mechanisms, but my guess is that the electrons move like this during the attack on the carbonyl carbon:



Anyway, I hope this has been of at least some help to you. Don't be discouraged if some of this seems a little overwhelming. The level of depth gone into here far exceeds what is typically taught in introductory-level courses, so don't be put off of trying to learn organic chemistry because of it. And don't be afraid to continue to ask questions if you're still not clear on anything.

Seeing as the Dark Lord is back, how about an actual do-able OC synth that demonstrates an OC mechanism/principle ?

Got loads of reagents now

Edit:

And stacks of 250ml beakers for some reason.

[Edited on 3-9-2016 by aga]

Quote: Originally posted by aga

Seeing as the Dark Lord is back, how about an actual do-able OC synth that demonstrates an OC mechanism/principle ? Got loads of reagents now

Quote: Originally posted by aga

What other reagents needed ?

Complex numbers (recap)

Complex numbers (aka imaginary numbers) arise because when we ask "What's the square root of a (a Real number)?", then if the answer is b, then:

$$\sqrt{a}=b \iff b^2=a\:\text{if } (a\in\mathrm{R})$$

As the square of a Real number is always positive, we can't take the square root of a negative Real number.

1. The imaginary number i:

We define:

$$i=\sqrt{-1}$$
Then, with a a Real number:
$$\sqrt{-a^2}=\sqrt{a^2\times -1}=a\sqrt{-1}=ai$$
And:

$$(ai)^2=a^2i^2=a^2\times (\sqrt{-1})^2=a^2\times(-1)=-a^2$$

2. Composite complex numbers:

Most (but not all) complex numbers have a Real and a Complex part, where Re is the Real part and Im is the Complex part:

$$c=Re+Im\times i=x+yi$$

For example:
$$-5+3i$$

We say that c is a member ("an element") of the group of Complex numbers:

$$c\in\mathrm{C}$$

3. Representation in the Complex space:

The Complex number as a vector:



Algebraically:

$$c=r\cos\varphi +ri\sin\varphi $$
$$(r\cos\varphi)^2 +(r\sin\varphi)^2=r^2\cos^2\varphi + r^2\sin^2\varphi$$
And because, always:
$$\cos^2\varphi + \sin^2\varphi=1$$
$$\implies (r\cos\varphi)^2 +(r\sin\varphi)^2=r^2(\cos^2\varphi + \sin^2\varphi)=r^2$$
And because:

$$Re=r\cos\varphi\:\text{and }Im=r\sin\varphi$$
$$\implies r^2=Re^2+Im^2=x^2+y^2$$

The value r:

$$r=\sqrt{x^2+y^2}$$

... is often called the modulus of the complex number.

Exercise:

If:

$$r^2=\frac12$$

And:

$$\varphi=\frac{\pi}{4}$$

Then write the complex number as a + bi.

Next up: complex conjugation > basic operations with complex numbers > ...



[Edited on 1-11-2016 by blogfast25]

Quote: Originally posted by aga

Gulp.

We've seen all this above, it's just presented more beautifully.

So if earlier you were only physically present or were really daydreaming about girls with really big t*ts, now's your chance!

Quote: Originally posted by aga

to run off and find the Mafs book, then frantically revise)

Get over yourself! You know enough to revise this quite quickly!

Chop, chop...

Quote: Originally posted by j_sum1

(I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.)

Even more "fun" are things like:

$$\sqrt [4]{-6}$$

But we're not going there. "Lite" here means addition, substraction, multiplication and division only!

[Edited on 2-11-2016 by blogfast25]

Quote: Originally posted by aga

The car insurance got complicated, a bit like where i'm heading on this first (presumably easy) question. So far i got some pythagoras going on to give :- $$ y = \sqrt { \frac {1}{2} } \sin{ \frac {\pi}{4} } $$ Is this going in remotely the right direction, or is it an overthunk, drunk and sunk direction ? [Edited on 2-11-2016 by aga]

Quote: Originally posted by aga

$$ \sqrt { \frac {1}{2} } \sin{ \Big ( 90 - \frac {\pi}{4} \Big ) } + i \sqrt { \frac {1}{2} } \sin{ \Big ( \frac {\pi}{4} \Big ) } $$ The LaTeX is coming back, slowly ...

Quote: Originally posted by blogfast25

Quote: Originally posted by j_sum1   (I distinctly remember asking my maths teacher the value of 2i and it taking him a long time to figure it out.) Even more "fun" are things like: $$\sqrt [4]{-6}$$ But we're not going there. "Lite" here means addition, substraction, multiplication and division only! [Edited on 2-11-2016 by blogfast25]

Quote: Originally posted by aga

Why was the expression 'strange' and where does 'c' some into it ? Edit: The notion of Radians did not enter my head, at all. I recall that Raidans exist, but am not aware of what they are, apart from being an alternate way of expressing Degrees. Gradians as well. Same story.

Quote: Originally posted by Maroboduus

Gradians are 400ths of a full circle. They were an option on various calculators in the 1980s. At the time I heard they were used by some engineers. Your memory is not faulty here.

G*d yeah, I vaguely remember that now (not that I want to!)

Complex numbers: complex conjugation

Quote: Originally posted by aga

To be sure, what does the sigma-like symbol mean ? $$c\in\:\mathrm{R}$$ I guessed at 'in the realm of' but then it all went odd.

Quote: Originally posted by aga

If even Numbers have limits, they cannot be a linear sequence, which, in turn, makes even Numbers meaningless The status of Zero has always bothered me. Ponderment required.

Zero took a long time to get accepted: the other numbers didn't want to know it! But accepting it was a true revolution!

The first part of your ponderment is cryptic. Care to explain?

Quote: Originally posted by blogfast25

The first part of your ponderment is cryptic. Care to explain?

Quote: Originally posted by aga

Quote: Originally posted by blogfast25   The first part of your ponderment is cryptic. Care to explain? Sorry, i missed the question earlier. It was some random drunken notion more than likely. Something about Numbers spanning an infinity, yet allowing an infinity between each number. 0 to 1 (or -1) being an infinity, equal to, er, infinity, making 0 the only True number. Every other number is infinitely distant from 0, and infinitely distant from the next number, making them all basically the same number.

Complex numbers basic operations: