1 = ?

[Scatty]

@You said two objects are equal if they represent the same thing. And that's very redundant' date=' if you ask me.

[/quote']

 

We're defining equality. If you already have a definition of equality in mind, then stating that definition will come across as a tautology. However, until you have defined equality, the "=" sign is completely meaningless, and thus a definition is required.

 

The definition of equality varies significantly depending on your system and objectives. For example, two cardinalities are equal if the sets to which they correspond are bijective. Equality must be defined to have meaning, even if that definition is intuitive.

 

Nah, I just wanted to get those 6 words off your keyboard. Anyway you twist it, mathematics, at its core, was created for practical purposes, and for something to work, it mustn't be too philosophical.

 

@Twisted: Infinity. Or negative infinity' date=' depends on the situation.

[/quote']

 

No.

 

I have no desire to go into detail here, but you are wrong.

 

Still, please do it.

[CrabHelmet]

Still' date=' please do it.

[/quote']

 

I wouldn't know where to start. The poor definitions of positive infinity, negative infinity, and unsigned infinity? The fact that simply sending everything over 0 to one of the infinities undermines the foundations of calculus?

[Scatty]

Yeah, kinda.

[Zeroshot]

[Scatty]

Meh, let me try this again.

 

Dividing by 1 is equivalent with multiplying by 1, right? In the same way, dividing by 0.1 is equivalent with multiplying by 10, dividing by 0.000001 is equivalent with multiplying by 1000000, and so on, to infinity.

 

Here, we have an increasingly lower number that never reaches 0, which is equivalent to an increasingly higher number that never reaches infinity. Wouldn't that ultimately mean that dividing by 0 is equivalent with multiplying by infinity?

 

Please prove me wrong.

[EHN.]

Calculator stupidity test: 100/3x3

 

A clever calculator would say 100, but a dumb would say 99.99999999999999999999999999999999999999999999999999999999999999999999999999999999

[Scatty]

Calculator stupidity test: if it does not have a battery, it's not stupid, it's just plastic. If it DOES have a battery, it's not stupid, it's still plastic.

[Luna Lovegood]

After weeks of wallowing in my pit of depression' date=' I returned to the bowels - no, not bowels - I returned to the [i']appendix[/i] of the internet to find that my people-are-wrong-about-simple-math senses were tingling.

 

Basically, the problem here is that 0.999... is not just "really really close to 1"; it is exactly equal to 1, in the same way that .5 is exactly equal to .50 and 1/2.

 

Here is an explanation that should work for those who know basic high school algebra: Observe that 0.999... represents a geometric series, a summation equal to 9/10 + 9/100 + 9/1000 + ... and so on forever. This geometric series has first term t = 9/10 and common ratio r = 1/10, which means that, by the standard formula t/(1-r) for the convergent sum of a geometric series - and we want the convergent sum, since we are literally summing every term in this (countably) infinite sequence - produces a value of 1.

 

Unfortunately, YCM is not known for its ability to understand high school math just about anything, so I don't expect my waffling about the convergent sum of a geometric series to be met with anything more than "me brain hurt", so I will need to abandon this path of proof and move not toward the more rigorous (Cauchy sequences) but toward the less rigorous, which, though less mathematically sound, will suffice to explain why this is true. To do this, I will recall a proof from my seventh-grade math textbook.

 

Let's start with 0.999... and call it x.

 

x = 0.999...

 

Now, let's multiply both sides by 10.

 

10x = 10*0.999...

 

Now, we all know how multiplying by 10 works in the decimal system, right?

 

10x = 9.999...

 

Good. Next, let's subtract x (which is 0.999...) from both sides.

 

10x-x = 9.999... - 0.999...

 

The result of the subtraction should be obvious.

 

9x = 9

 

And dividing both sides of the equation by 9, we have:

 

x = 1

 

Thus, 0.999... = 1.

 

The reason this works is that, as Pika said, there most emphatically does not exist a .0~1 difference between 0.999... and 1. To try to call for a 1 after an endless string of zeroes is meaningless - there is no omega-plus-one decimal place because there are only omega decimal places after the decimal point because the set of decimal places after the decimal point has the same Von Neumann order type as the set of positive integers by the very definition of real numbers and here I go talking about set theory when I have previously concluded that my intended audience doesn't even understand high school algebra.

 

I learned it with X= 1.9999, but your way worked to

[Scatty]

Thanks for using up half the page for simply AGREEING with somebody, Luna. That's why I love you so much.

[Lyfe.]

1/3 + 1/3 + 1/3 = 3/3 = 1

 

Decimals...I hate them.

[CrabHelmet]

Meh' date=' let me try this again.

 

Dividing by 1 is equivalent with multiplying by 1, right? In the same way, dividing by 0.1 is equivalent with multiplying by 10, dividing by 0.000001 is equivalent with multiplying by 1000000, and so on, to infinity.

 

Here, we have an increasingly lower number that never reaches 0, which is equivalent to an increasingly higher number that never reaches infinity. Wouldn't that ultimately mean that dividing by 0 is equivalent with multiplying by infinity?

 

Please prove me wrong.

[/quote']

 

As b approaches 0 from the positive direction, a/b indeed becomes arbitrarily large given that a is a fixed positive real number. However, to say that this means dividing something by 0 is the same as multiplying it by infinity is simply ludicrous.

[Raylen]

Can we define division by 0 using limits?

[Guest Welche]

0.99999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999 counts as 1.

[Womi]

^no.

[Scatty]

Meh' date=' let me try this again.

 

Dividing by 1 is equivalent with multiplying by 1, right? In the same way, dividing by 0.1 is equivalent with multiplying by 10, dividing by 0.000001 is equivalent with multiplying by 1000000, and so on, to infinity.

 

Here, we have an increasingly lower number that never reaches 0, which is equivalent to an increasingly higher number that never reaches infinity. Wouldn't that ultimately mean that dividing by 0 is equivalent with multiplying by infinity?

 

Please prove me wrong.

[/quote']

 

As b approaches 0 from the positive direction, a/b indeed becomes arbitrarily large given that a is a fixed positive real number. However, to say that this means dividing something by 0 is the same as multiplying it by infinity is simply ludicrous.

 

Yes but WHY?

[CrabHelmet]

Can we define division by 0 using limits?

 

You can try' date=' but it generally doesn't work out well. If a is fixed, then trying to define a/0 to be a/b as b approaches 0 doesn't work out nicely; if a is not 0, then a/b's limit is different depending on whether b is approached from the positive or negative direction (and thus the overall limit does not exist), and if a is 0, you get the weird result of 0/0 = 0, which may be a convenient definition in some systems for dealing with that case but is entirely arbitrary and doesn't really make any outside sense.

 

Of course, if a is not fixed and varies with b, then something meaningful can be produced. That's one of the foundations of calculus, and is the basis of differentiation.

 

Meh' date=' let me try this again.

 

Dividing by 1 is equivalent with multiplying by 1, right? In the same way, dividing by 0.1 is equivalent with multiplying by 10, dividing by 0.000001 is equivalent with multiplying by 1000000, and so on, to infinity.

 

Here, we have an increasingly lower number that never reaches 0, which is equivalent to an increasingly higher number that never reaches infinity. Wouldn't that ultimately mean that dividing by 0 is equivalent with multiplying by infinity?

 

Please prove me wrong.

[/quote']

 

As b approaches 0 from the positive direction, a/b indeed becomes arbitrarily large given that a is a fixed positive real number. However, to say that this means dividing something by 0 is the same as multiplying it by infinity is simply ludicrous.

 

Yes but WHY?

 

See my answer to Raylen above, but basically, the two bolded phrases combined with a lack of a good definition of infinity cause problems.

[Scatty]

Yeah but if you say that then there is no way you can divide by 0.