Anyone want to try and tell me which necessary law I left out in proving that 5+2=1
It's a really funny concept to try out on people who don't understand that all important rule which can only be disregarded in the case of limits.
calculus is fun to mess with
Umm.. I would love to critique your proof but I would need to see it first. Where is it? Did you post it somewhere else? 
here is anothe one prove 2+2=5
answer is,
-20 = -20
16 - 36 = 25 - 45
16 - 36 + 81/4 = 25 - 45 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 = 5
2 + 2 = 5
__________________
| kjuvale wrote: |
here is anothe one prove 2+2=5
answer is,
-20 = -20
16 - 36 = 25 - 45
16 - 36 + 81/4 = 25 - 45 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 = 5
2 + 2 = 5
__________________ |
nice. this is of course faulty...but nice.well the rule is that infinity divided by itself does not equal one.
ignoring it
anything plus anything equals 1, unless you are in limits
| tony wrote: |
| kjuvale wrote: | here is anothe one prove 2+2=5
answer is,
-20 = -20
16 - 36 = 25 - 45
16 - 36 + 81/4 = 25 - 45 + 81/4
(4 - 9/2)^2 = (5 - 9/2)^2
4 = 5
2 + 2 = 5
__________________ |
nice. this is of course faulty...but nice. |
the error lies between the third and fourth arguments. When you took the square root of both sides to produce (4-9/2) and (5-9/2) you should have used absolute values. When you look at these numbers simplified they equal (-1) and (1). And so the trick comes out when you reverse the process and square them (-1)^2 = (1)^2. This may not be the best description of the violation but that is the problem.It's very hard to say where the problem is because your argument hasn't enough brackets to make sense.
You are assuming that
16 - 36 + 81 /4 is the same as (16 - 36 + 81) /4.
It isn't.
Here is another and i think is the easiest way to say....
as we know 0^anything =0
Now, 0+0=0
or, 0^2 + 0^2 = 0^5
from the law of indeces....
0^(2+2) = 0^5
or, 2+2 = 5
i though know its wrong....
because there is one basic wrong statement here...
anyone can point that out?
lets see....................
Anyone can explain why x^0=1 ?
| Shin wrote: |
| Anyone can explain why x^0=1 ? |
From the same law of indeces:
x^0 * x^n = x^(0+n) = x^n
so
x^0=1 | garghya wrote: |
Here is another and i think is the easiest way to say....
as we know 0^anything =0
Now, 0+0=0
or, 0^2 + 0^2 = 0^5
from the law of indeces....
0^(2+2) = 0^5
or, 2+2 = 5
i though know its wrong....
because there is one basic wrong statement here...
anyone can point that out?
lets see.................... |
The law only applies to nonzero base. You are doing the same type of stuf as dividing by infinity - create a singularity.Hehe, funny stuff. I'm only in PreCalculus this year, and I nearly failed the section on limits.. Otherwise I'd have more to offer. But for now...
1.9(repeating) = 2.0
How so?
10x - x = 18
x= 1.9(repeating) or 2; it works out both ways. 
I do not think mathmatic games like this are of any sense.I just can not understand why so many people waste so much time on things like this.
it's because we're bored as hell
| osbits wrote: |
| I do not think mathmatic games like this are of any sense.I just can not understand why so many people waste so much time on things like this. |
Like anything that forces you to stretch your brain, it allows you to see further, over the horizon. The first person (I know of) to measure the earth's circumference did so with a stick and some goniometry. | Juparis wrote: |
1.9(repeating) = 2.0
How so?
10x - x = 18
x= 1.9(repeating) or 2; it works out both ways. |
Not quite. Let x+e=2, then:
10(2-e) - (2-e) = 18 - 9e
So the difference between x and 2 is e, while the difference for the other is 9e. Now, if you take the limit then e goes to zero. Both equations have the same limit. 1.999(rep) is not the same as 2.Sorry, but all of that? Flew right over my head.. *imitates a plane flying overhead, sound effects included*
1.9(rep) = 2, as I've been taught (in more places that one 
)
But again, I'm only in PreCalculus and nearly failed the unit on limits, so maybe I'll understand whatever it is you said later on in life. 
| withaar wrote: |
| garghya wrote: | Here is another and i think is the easiest way to say....
as we know 0^anything =0
Now, 0+0=0
or, 0^2 + 0^2 = 0^5
from the law of indeces....
0^(2+2) = 0^5
or, 2+2 = 5
i though know its wrong....
because there is one basic wrong statement here...
anyone can point that out?
lets see.................... |
The law only applies to nonzero base. You are doing the same type of stuf as dividing by infinity - create a singularity. |
You don't need to use a zero base, do you?
I mean, 1^n is well-defined for all real n, isn't it? 1^n = 1
So:
1^4 = 1^5
4 = 5
2 + 2 = 5
Or you could use 0 indices.
4^0 = 5^0
Squaring both sides:
4^0 * 4^0 = 5^0 * 5^0
4^1 = 5^1
4 = 5
2 + 2 = 5
If you look at the general case, it gets nuts:
n^0 = 1 for all n
1^n = 1 for all n
So:
x^0 = y^0
(x^0)^a = (y^0)^b
x^a = y^b
for any x, y, a and b
Fun fun until you see the trick ^_^Is this basically Calculus? Because i can't wait until i start doing to this stuff 
. Seems like it can get very complicated when dealing with the theorys of math. | NjRocket wrote: |
| Is this basically Calculus? Because i can't wait until i start doing to this stuff . Seems like it can get very complicated when dealing with the theorys of math. |
I'd say most of what we've seen so far is basic algebra, not calculus.
I don't know of a way to "prove" 2+2=5 with calculus, but I can "prove" 1=2.
Start with x. x can be any number.
Now, x = 1 + 1 + ... + 1 for x number of 1's
For example, if x = 3, then:
x = 1 + 1 + 1
And if x = 4
x = 1 + 1 + 1 + 1
In general:
x = ∑(from i = 0 to x)(1)
(In plain English, x = 1 + 1 + ... + 1 for x number of 1's.)
So:
x = ∑(from i = 0 to x)(1)
Now multiply both sides by x:
x² = ∑(from i = 0 to x)(x)
(or, x² = x + x + ... + x for x number of x's)
Now, differentiate both sides with respect to x:
d(x²)/dx = d(∑(from i = 0 to x)(x))/dx
2x = ∑(from i = 0 to x)(1)
Now, since ∑(from i = 0 to x)(1) = x:
2x = x
Thus, 2 = 1.
(Figuring out the trick in this one is a bitch ^_^)Ok, some ideas, for some of you:
1. Juparis:
1.999(rep) doesn't equal 2. But
lim 2-(1/x) = 2
(x->oo) //infinity
2. garghya and Indi:
Logarithm is undefined with zero and one base. Nice try anyway...
3. Indi:
I liked your 2=1 prove - quite impressive. I think I figured the trick out, but I don't want to ruin the others' fun... 
| Nyizsa wrote: |
Ok, some ideas, for some of you:
1. Juparis:
1.999(rep) doesn't equal 2. But
lim 2-(1/x) = 2
(x->oo) //infinity
2. garghya and Indi:
Logarithm is undefined with zero and one base. Nice try anyway...
3. Indi:
I liked your 2=1 prove - quite impressive. I think I figured the trick out, but I don't want to ruin the others' fun... |
It's my favourite math trick. ^_-
If you want to PM me what you think the trick is, I'll let you know.i think i get it as well...except i hate series, even though this is an easy one. Calc BC AP blows my mind, some of you will understand this. for those who don't, Calc BC AP is basically second calc course in college taken in high school. So yes my head hurts
Ah, but the trick isn't in the series. ^_^
What education level you guys have?? Cause I'm only in 7 grade, and now I'm wondering, will I ever understand what are you talking about | mantasx wrote: |
| What education level you guys have?? Cause I'm only in 7 grade, and now I'm wondering, will I ever understand what are you talking about |
Most of this stuff is taught at least in the calculus/precalculus levels. Most of this isn't too hard (to understand what is being talked about anyways), you just have to know what some functions mean. (Like Sigma (∑), or e)
Some of these are pretty good. 
I'll have to print some out to remember later. | withaar wrote: |
| Shin wrote: | | Anyone can explain why x^0=1 ? |
From the same law of indeces:
x^0 * x^n = x^(0+n) = x^n
so
x^0=1 |
Usually I just remember by heart, Now I can explain why it is. cool. Thanks | Shin wrote: |
| withaar wrote: | | Shin wrote: | | Anyone can explain why x^0=1 ? |
From the same law of indeces:
x^0 * x^n = x^(0+n) = x^n
so
x^0=1 |
Usually I just remember by heart, Now I can explain why it is. cool. Thanks |
Well, it is not really because of this. It was made up in order to equations like the above will be fulfilled. | Shin wrote: |
| withaar wrote: | | Shin wrote: | | Anyone can explain why x^0=1 ? |
From the same law of indeces:
x^0 * x^n = x^(0+n) = x^n
so
x^0=1 |
Usually I just remember by heart, Now I can explain why it is. cool. Thanks |
Well, it is not really because of this. It was made up in order to equations like the above will be fulfilled. | Shin wrote: |
| withaar wrote: | | Shin wrote: | | Anyone can explain why x^0=1 ? |
From the same law of indeces:
x^0 * x^n = x^(0+n) = x^n
so
x^0=1 |
Usually I just remember by heart, Now I can explain why it is. cool. Thanks |
Well, it is not really because of this. It was made up in order to equations like the above will be fulfilled. | Indi wrote: |
| NjRocket wrote: | | Is this basically Calculus? Because i can't wait until i start doing to this stuff . Seems like it can get very complicated when dealing with the theorys of math. |
I'd say most of what we've seen so far is basic algebra, not calculus.
I don't know of a way to "prove" 2+2=5 with calculus, but I can "prove" 1=2.
Start with x. x can be any number.
Now, x = 1 + 1 + ... + 1 for x number of 1's
For example, if x = 3, then:
x = 1 + 1 + 1
And if x = 4
x = 1 + 1 + 1 + 1
In general:
x = ∑(from i = 0 to x)(1)
(In plain English, x = 1 + 1 + ... + 1 for x number of 1's.)
So:
x = ∑(from i = 0 to x)(1)
Now multiply both sides by x:
x² = ∑(from i = 0 to x)(x)
(or, x² = x + x + ... + x for x number of x's)
Now, differentiate both sides with respect to x:
d(x²)/dx = d(∑(from i = 0 to x)(x))/dx
2x = ∑(from i = 0 to x)(1)
Now, since ∑(from i = 0 to x)(1) = x:
2x = x
Thus, 2 = 1.
(Figuring out the trick in this one is a bitch ^_^) |
actually 2x = x, would mean x = 0 wouldn't it?? (subtract x from 2x) | squirrelmaster wrote: |
| Indi wrote: | | NjRocket wrote: | | Is this basically Calculus? Because i can't wait until i start doing to this stuff . Seems like it can get very complicated when dealing with the theorys of math. |
I'd say most of what we've seen so far is basic algebra, not calculus.
I don't know of a way to "prove" 2+2=5 with calculus, but I can "prove" 1=2.
Start with x. x can be any number.
Now, x = 1 + 1 + ... + 1 for x number of 1's
For example, if x = 3, then:
x = 1 + 1 + 1
And if x = 4
x = 1 + 1 + 1 + 1
In general:
x = ∑(from i = 0 to x)(1)
(In plain English, x = 1 + 1 + ... + 1 for x number of 1's.)
So:
x = ∑(from i = 0 to x)(1)
Now multiply both sides by x:
x² = ∑(from i = 0 to x)(x)
(or, x² = x + x + ... + x for x number of x's)
Now, differentiate both sides with respect to x:
d(x²)/dx = d(∑(from i = 0 to x)(x))/dx
2x = ∑(from i = 0 to x)(1)
Now, since ∑(from i = 0 to x)(1) = x:
2x = x
Thus, 2 = 1.
(Figuring out the trick in this one is a bitch ^_^) |
actually 2x = x, would mean x = 0 wouldn't it?? (subtract x from 2x) |
Nope, because you know the value of x before you start, and it does not need to equal zero all the time for x² = ∑(from i = 0 to x)(x) to be true. Or to put it another way, you're not solving for x - you already know x, and it's not necessarily 0.
If x = 3 then ∑(from i = 0 to x)(x) = x + x + x
So:
x² = ∑(from i = 0 to x)(x)
x² = x + x + x
d(x²)/dx = d(x + x + x)/dx
2x = 1 + 1 + 1
2x = 3
But x = 3, so:
2x = x
The only way 2x = x for x = 3 is if 2 = 1, thus, 2 = 1.
Same if x = 5:
∑(from i = 0 to x)(x) = x + x + x + x + x
So:
x² = ∑(from i = 0 to x)(x)
x² = x + x + x + x + x
d(x²)/dx = d(x + x + x + x + x)/dx
2x = 1 + 1 + 1 + 1 + 1
2x = 5
But x = 5, so:
2x = x
The only way 2x = x for x = 5 is if 2 = 1, thus, 2 = 1. | Indi wrote: |
Start with x. x can be any number.
Now, x = 1 + 1 + ... + 1 for x number of 1's
For example, if x = 3, then:
x = 1 + 1 + 1
And if x = 4
x = 1 + 1 + 1 + 1
In general:
x = ∑(from i = 0 to x)(1)
(In plain English, x = 1 + 1 + ... + 1 for x number of 1's.)
So:
x = ∑(from i = 0 to x)(1)
Now multiply both sides by x:
x² = ∑(from i = 0 to x)(x)
(or, x² = x + x + ... + x for x number of x's)
Now, differentiate both sides with respect to x:
d(x²)/dx = d(∑(from i = 0 to x)(x))/dx
2x = ∑(from i = 0 to x)(1)
Now, since ∑(from i = 0 to x)(1) = x:
2x = x
Thus, 2 = 1.
(Figuring out the trick in this one is a bitch ^_^) |
Well, I've figured out what's wrong but I can't prove why.
The problem is actually to prove that we cannot differentiate a sum that way, when the number of substracts depend on the variable x. To be honest, I'm stuck thinking what is, and what is not obvious. What we have to make sure is that d(∑(from i = 0 to x)(1))/dx = 1 but it could be simplified into finding the evidence that ∑(from i = 0 to x)(1) = x, which is absurd.
EDIT: What is absurd is not the equotation before, but searching evidence for it | the_mariska wrote: |
Well, I've figured out what's wrong but I can't prove why.
The problem is actually to prove that we cannot differentiate a sum that way, when the number of substracts depend on the variable x. To be honest, I'm stuck thinking what is, and what is not obvious. What we have to make sure is that d(∑(from i = 0 to x)(1))/dx = 1 but it could be simplified into finding the evidence that ∑(from i = 0 to x)(1) = x, which is absurd.
EDIT: What is absurd is not the equotation before, but searching evidence for it |
You pretty much have the answer without the why right there. ^_^
The problem isn't actually the sum itself, the problem is - as you figured - that the number of terms in the sum is dependent on x.
The trick is the nature of differentiation. For d(f(x))/dx to be valid for some interval, f(x) has to be continuous over that interval. So is x² = ∑(from i = 0 to x)(x) continuous?
Nope. ^_^ x² = ∑(from i = 0 to x)(x) is only valid when x is a positive integer, because you can't have a sum with a fractional (or negative, for that matter) number of terms. If you were to draw the graph of x² = ∑(from i = 0 to x)(x) using x vs. x², it would be a series of dots when x = 0, 1, 2, ... Since it's a bunch of dots and not a line, it's not continuous anywhere over its range, and thus not differentiable.
Even when you write it as:
x² = x + x + ... + x (for x number of x's)
You're still implicitly saying that x ∈ ℕ₀, because you can't have a negative or fractional number of terms.
So the expression x² = ∑(from i = 0 to x)(x) is not differentiable, and the differentiation step is undefined, which means everything after that is nonsense.OK, you've beaten me . I guess my maths teacher would kill me if he saw that I forgot about this again. Hope I'll remember about this at my maths secondary school exam that I'm having the day after tomorrow 
Thanks | the_mariska wrote: |
| OK, you've beaten me . I guess my maths teacher would kill me if he saw that I forgot about this again. Hope I'll remember about this at my maths secondary school exam that I'm having the day after tomorrow Thanks |
No one ever remembers that. ^_^ That's why it works so well. Try it on your maths teacher, and it will probably get him, too.If you havent noticed ppl you are proving that 2x = x. Even though x may = 1.
if x = 1 and x = 2x then 2x = 1
and
therefore 2x = x is completely void or you could divide by x and 2 = 0
this is not possible to use a variable to describe this function.
explain it with theorms and explanations not simple algebra
| Nyizsa wrote: |
Ok, some ideas, for some of you:
1. Juparis:
1.999(rep) doesn't equal 2. But
|
weird, my computer programming teacher said it does. He explained that one day.Computer programming is different from math. If the computer is set to only go to a certain place value, i'm pretty sure it just rounds. However, in math 1.99999999999999999999999999999999999999999...does not equal to, un less it is:
lim x->infinity 2-(1/x)
which does infact equal 2
| Indi wrote: |
| No one ever remembers that. ^_^ That's why it works so well. Try it on your maths teacher, and it will probably get him, too. |
I hope so He caught me at this once, and it seems I haven't learned anything since then
| Quote: |
Juparis:
1.999(rep) doesn't equal 2.
weird, my computer programming teacher said it does. He explained that one day |
It does equal. Here's a proof.
Assume:
1.99999..(infinite number of 9s) = x / multiply both sides times 10
19.999999... = 10x / subtract 1.999999.... from both sides
19.99999... - 1.99999... = 10x - 1.99999....
18.0 = 10x - 1.999999...
But we know that 1.999999... = x, so we replace it with x. We get
18 = 10x - x
18 = 9x
x = 2
Clear? (This is a real proof, you don't have to look for a trick here )
| Quote: |
However, in math 1.99999999999999999999999999999999999999999...does not equal to, un less it is:
lim x->infinity 2-(1/x)
which does infact equal 2 |
This equotation above has nothing to do with limes in infinity. And limes is only limes, 2-(1/x) will never equal exactly 2, it will only get closer and closer. But recurring decimals like this above do equal normal fractions, like 0.333333.... = 1/3; 0.6666666... = 2/3; 0.999999.... = 3/3 = 1
| Quote: |
If you havent noticed ppl you are proving that 2x = x. Even though x may = 1.
if x = 1 and x = 2x then 2x = 1
and
therefore 2x = x is completely void or you could divide by x and 2 = 0
this is not possible to use a variable to describe this function.
explain it with theorms and explanations not simple algebra |
I guess you don't get the point. We know that the answer is wrong. We were just trying to figure out the trick where was the mistake in the 'evidence'. And there is no real theorem that 2=1. It was just a trick using some attribute of differentation that no one really remembers of. Heh, I don't think you understand what I'm saying, so I'll leave that matter I still say that 1.(infinite no. of 9s) DOESN'T equal 2, but it is very close to it. So close that the difference is negligible, so we may say that they are equal, if you insist. But according to the properties of the set os rational numbers, only 2=2 and no other rational number!
If you plot them on the line, 1.999... will be very close to 2. But what if you add some more 9s? You get a number that is even closer to 2. So between your original number and 2 there are several (actually infinite) numbers. And you can do this forever.
| Nyizsa wrote: |
| If you plot them on the line, 1.999... will be very close to 2. But what if you add some more 9s? You get a number that is even closer to 2. So between your original number and 2 there are several (actually infinite) numbers. And you can do this forever. |
1.99999.. means that after this 1 there is infinite number of 9s. I know that it's hard to imagine, but this is the way that infinite numbers behave . and you simply cannot add more 9s because there are already infitely many of them. It's the same problem as adding anything to the infinity. Infinity plus any number will still be infinity, no matter how much you add.
If it's hard for you to believe this, try starting from the equotation 1/3 = 0.333333.....
The evidence is simple and even kids in primary school can show that:
1 : 3 = 0.333...
10
- 9
__
. 10
. - 9
. __
. . 10
. . - 9
. . __
. . . 1....
Yeah, there are exactly infinitely many 3s after that comma, but it's exactly equal to 1/3. | the_mariska wrote: |
| Infinity plus any number will still be infinity, no matter how much you add. |
That's what I have just said. You can add as many 9s as you want.
| the_mariska wrote: |
If it's hard for you to believe this, try starting from the equotation 1/3 = 0.333333.....
The evidence is simple and even kids in primary school can show that:
1 : 3 = 0.333...
10
- 9
__
. 10
. - 9
. __
. . 10
. . - 9
. . __
. . . 1....
Yeah, there are exactly infinitely many 3s after that comma, but it's exactly equal to 1/3. |
This is true, because these are the same numbers, just in two different forms. (Fraction and decimal, if they are called so in English.)
But if we follow the same analogy:
2 : 1 = 2
0
| the_mariska wrote: |
| ...even kids in primary school can show that. |
And one last thing: 2 is an integer, and 1.999... is NOT. | Nyizsa wrote: |
And one last thing: 2 is an integer, and 1.999... is NOT. |
I know I am damn stubborn, but here I have something else. Here is a quote from Wikipedia:
| Wikipedia wrote: |
The case of 0.99999...
The method of calculating fractions from repeated decimals, especially the case of 1 = 0.99999..., is sometimes contested by the mathematically naive:
x = 0.99999...
10x = 9.9999...
10x − x = 9.9999... − 0.99999...
9x = 9
x = 1
Some argue that, in the second step of the equation given above, 10x is 9.999...0 and not 9.999... but this is not the case: the right-hand side does not terminate (it is recurring) and so there is no end to which a zero can be appended.
Generalising this, any nonzero number with a finite decimal expression (a decimal fraction) can be written in a second way as a recurring decimal.
For example 3/4 = 0.75 = 0.750000000... = 0.74999999 ...
Recurring decimals can be applied to cryptography.
|
I'm not going to say anything more about that matter, I promise But if you still don't believe me click here to see a few more proofs It's just like the halfway problem, you place a point, no mass no space, a certain distance away from the door say 2 ft. If you half the distance an infinite number of times, you will never reach the door. you will never travel two feet. you will travel 1.99999repeating ft. and be .0repeating1 ft. away from the door in the end.
| s43ros wrote: |
| It's just like the halfway problem, you place a point, no mass no space, a certain distance away from the door say 2 ft. If you half the distance an infinite number of times, you will never reach the door. you will never travel two feet. you will travel 1.99999repeating ft. and be .0repeating1 ft. away from the door in the end. |
But this problem has already been solved, haven't you know? The evidence based on the theory of continuum and infinitely little numbers, but unfortunately I cannot do this myself I think we were saying the same thing all the time. As I took a look at the Wiki page you provided, I realized that in some of the proofs they actually talk about the limes. And I totally agree with that. It is an "infinitely small" difference anyway - let's not waste any more word on it.
| Nyizsa wrote: |
| I think we were saying the same thing all the time. As I took a look at the Wiki page you provided, I realized that in some of the proofs they actually talk about the limes. And I totally agree with that. It is an "infinitely small" difference anyway - let's not waste any more word on it. |
No problem Anyway, for me it was a good training before my final exam in maths. Luckily I did it quite well. I hope that they'll accept me at IT studies (as it depends on my exam's result) and I'll be able to better explain my mathematical thoughts. Peace Hey, congrats for the exam! Mine was 5 or 6 years ago...
what exactly is your point with the last thing? | s43ros wrote: |
what exactly is your point with the last thing? |
Oops that wasn't quite right, forget about that okay 
.A more well-known version is Belgium is the following. Let's assume a and b are both real numbers, which are equal.
a = b
<=> a*a = a*b
<=> a^2 + a^2 - 2*a*b = a*b + a^2 - 2*ab
<=> 2(a^2 - a*b) = a^2 - a*b
<=> 2 = 1
(No need for heavy calculus to solve this one ) | tlambert wrote: |
A more well-known version is Belgium is the following. Let's assume a and b are both real numbers, which are equal.
a = b
<=> a*a = a*b
<=> a^2 + a^2 - 2*a*b = a*b + a^2 - 2*ab
<=> 2(a^2 - a*b) = a^2 - a*b
<=> 2 = 1
(No need for heavy calculus to solve this one ) |
Yeah, but it's also easy to spot because the first thing everyone looks for is a divide by zero. ^_^
You know, I'm surprised that no-one's mentioned the old adage: "2 + 2 = 5, but only for large values of 2."
2.4 ≅ 2 (because 2.4 rounds to 2)
4.8 ≅ 5 (because 4.8 rounds to 5)
2.4 + 2.4 = 4.8
Thus:
2 + 2 = 5
^_^
