Ive stumble across an impossible eqaution

[megadedhed]

X^n+Y^n=Z^n

 

 

 

now if we do what i like to call substitution

 

 

 

X^n+Y^n-Z^n=0

 

 

 

devide all 3 by 1^n

 

 

 

x+y-z=0

 

 

 

so x+y=z or x-z=-y, or y-z=-x

 

 

 

x and y both = 2, Z equals four, n=1

 

 

 

I don't think that x and y can be equal and n has to be >2

 

 

 

since the original written problem had no limitations written, it would work for just that

[Devilsdragon]

X^n+Y^n=Z^n

 

 

 

now if we do what i like to call substitution

 

 

 

X^n+Y^n-Z^n=0

 

 

 

devide all 3 by 1^n

 

 

 

x+y-z=0

 

 

 

so x+y=z or x-z=-y, or y-z=-x

 

 

 

x and y both = 2, Z equals four, n=1

 

 

 

I don't think that x and y can be equal and n has to be >2

 

 

 

since the original written problem had no limitations written, it would work for just that

 

 

 

What? You didn't read the topic? It's All the numbers have to be intergers and n has to be greater than 2

 

the x and y thing was just speculation

[Kashi]

X^n+Y^n=Z^n

 

 

 

Ha! There, I found something that satisfies all the requirements of your equation.

 

 

 

X = 2

 

Y = 1

 

Z = 3

 

N = 3

 

 

 

Which comes to: 8 + 1 = 9

 

 

 

If there's anything wrong with that answer, please tell me, because I think I may be a genius yes?

[runesmithie]

X^n+Y^n=Z^n

 

 

 

Umm ... well the equation balances if you value;

 

 

 

X = 1

 

Y = 1

 

Z = 2

 

N = 1

 

 

 

There, did I just solve the equation that has baffled mathematicians for hundreds of years?

 

 

 

If you had taken a few minutes to read the topic you would have found that it baffled them for N>2, not N<=2

[Kashi]

If you had taken a few minutes to read the topic you would have found that it baffled them for N>2, not N<=2

 

 

 

Heeheehee, read my above post, I just solved your un-solvable equation.

[runesmithie]

If you had taken a few minutes to read the topic you would have found that it baffled them for N>2, not N<=2

 

 

 

Heeheehee, read my above post, I just solved your un-solvable equation.

 

 

 

EDIT: Ah sheesh, you go and edit it, anyways it'd be

 

 

 

X = 2

 

Y = 1

 

Z = 3

 

N = 3

 

 

 

Which comes to: 8 + 1 = 9

 

 

 

2^3+1^3=3^3

 

 

 

which would be 8+1=27, which it doesn't :P

[Kashi]

Oh yeah .. I was thinking 3*3, not 3^3. Damn, got all excited over nothing ..

[darkmage099]

X^n+Y^n=Z^n

 

 

 

Ha! There, I found something that satisfies all the requirements of your equation.

 

 

 

X = 2

 

Y = 1

 

Z = 3

 

N = 3

 

 

 

Which comes to: 8 + 1 = 9

 

 

 

If there's anything wrong with that answer, please tell me, because I think I may be a genius yes?

 

 

 

oooh kashi gj gj its probably wrong but i dunt see what's wrong with it.

 

 

 

EDIT: yeah wrong

[Devilsdragon]

Oh yeah .. I was thinking 3*3, not 3^3. Damn, got all excited over nothing ..

 

 

 

Lol! Everytime I though about the equation I came up with the same thing, then noticed the same problem that 3^3 isn't 9 :x

[Kashi]

HAHAHA!!

 

 

 

This was bugging me, so I looked it up ..

 

 

 

If N>2, then there are no whole number solutions for X Y Z!

 

 

 

You were right about what you said in the title, given requirements of X Y Z being integers and N>2, there is no solution to this problem. Tell your calculus teacher he's an [wagon].

 

 

 

If N>2 you can solve it if X Y Z aren't whole numbers, but otherwise it really is impossible.

 

 

 

Jerk, you had everyone thinking hard about it for no reason.

[Blipo]

looks like your calculus teacher was having a bit of a laugh with you there. Shame that you spent a long time trying to prove it.

 

 

 

To answer the question above:

 

 

how about

 

x=2

 

y=3

 

z=5

 

n=1

 

 

 

 

this is a triangle with zero area!

 

 

 

Here's one to get your revenge on your calculus teacher with: Prove that 1=2!!

 

(there is a a flaw in this, see if you can figure it out)

 

A=B

 

AB=BB

 

AB-AA=BB-AA

 

A(B-A)=(B-A)(B+A)

 

A=B+A

 

as A=B, then A=2A, therefore 1=2!

 

:D

 

 

 

No, silly rabbit, you're dividing by zero! (At the bolded part)

 

 

 

No, silly rabbit, you're not dividing! (You're simplifying)

 

 

 

A=B

 

AB=BB

 

AB-AA=BB-AA

 

A(B-A)=(B-A)(B+A) 1(1-1)=(1-1)(1+1), when simplified, is 1=2

[yonman14]

Integrate: x*tan(x)dx

 

:lol:

 

 

 

1. We are not here to do your homework for you!

 

 

 

2. Integration and differentiation have nothing to do with this!

 

 

 

While they may not...lemme just do this problem!

 

Integration by parts!

 

u= x dV=tan(x)dx

 

du=1 dx v= -ln(cos(x))

 

 

 

-xln(cos(x))-(inetragl of){-xln(cos(x)}

 

 

 

u = x dv=-ln(cos(x))

 

du = 1 dx v= tan(x)

 

 

 

-xln(cos(x))-tan(x)+(integral of){tan(x)]

 

-xln(cos(x)) - tan(x) - ln(cox(x)) + C

 

 

 

I love calculus.

[Devilsdragon]

looks like your calculus teacher was having a bit of a laugh with you there. Shame that you spent a long time trying to prove it.

 

 

 

To answer the question above:

 

 

how about

 

x=2

 

y=3

 

z=5

 

n=1

 

 

 

 

this is a triangle with zero area!

 

 

 

Here's one to get your revenge on your calculus teacher with: Prove that 1=2!!

 

(there is a a flaw in this, see if you can figure it out)

 

A=B

 

AB=BB

 

AB-AA=BB-AA

 

A(B-A)=(B-A)(B+A)

 

A=B+A

 

as A=B, then A=2A, therefore 1=2!

 

:D

 

 

 

No, silly rabbit, you're dividing by zero! (At the bolded part)

 

 

 

No, silly rabbit, you're not dividing! (You're simplifying)

 

 

 

A=B

 

AB=BB

 

AB-AA=BB-AA

 

A(B-A)=(B-A)(B+A) 1(1-1)=(1-1)(1+1), when simplified, is 1=2

 

 

 

What are you [developmentally delayed]ed? 1-1=0 so 1*0=0 and 2*0=0

 

so what you proved is that 0=0 not 1=2

[Diminished2b]

http://en.wikipedia.org/wiki/Fermat%27s_last_theorem

[Hobgoblin11]

 

 

 

Here's one to get your revenge on your calculus teacher with: Prove that 1=2!!

 

(there is a a flaw in this, see if you can figure it out)

 

A=B

 

AB=BB

 

AB-AA=BB-AA

 

A(B-A)=(B-A)(B+A)

 

A=B+A

 

as A=B, then A=2A, therefore 1=2!

 

:D

 

 

 

Take it back to the very beginning: A=B

 

 

 

Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P :

 

 

 

BB=BB

 

BB-BB=BB-BB

 

B(B-B)=(B-B)(B-B)

 

B=(B-B)

 

If B=2, this equation now reads 2=2-2=0

 

 

 

2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted.

[menot]

 

Take it back to the very beginning: A=B

 

 

 

Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P :

 

 

 

BB=BB

 

BB-BB=BB-BB

 

B(B-B)=(B-B)(B-B)

 

B=(B-B)

 

If B=2, this equation now reads 2=2-2=0

 

 

 

2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted.

 

 

 

3rd line you have BB-BB = 2BB-2BB

 

is 0=0

 

you then multiply both sides by 0 (B-B)

 

so the next line should also read 0=0 :-)

[darkmage099]

 

 

 

Here's one to get your revenge on your calculus teacher with: Prove that 1=2!!

 

(there is a a flaw in this, see if you can figure it out)

 

A=B

 

AB=BB

 

AB-AA=BB-AA

 

A(B-A)=(B-A)(B+A)

 

A=B+A

 

as A=B, then A=2A, therefore 1=2!

 

:D

 

 

 

Take it back to the very beginning: A=B

 

 

 

Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P :

 

 

 

BB=BB

 

BB-BB=BB-BB

 

B(B-B)=(B-B)(B-B)

 

B=(B-B)

 

If B=2, this equation now reads 2=2-2=0

 

 

 

2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted.

 

 

 

I can tell it's wrongbut Idk how to explain it

 

 

 

 

 

 

 

 

 

Integrate: x*tan(x)dx

 

:lol:

 

 

 

1. We are not here to do your homework for you!

 

 

 

2. Integration and differentiation have nothing to do with this!

 

 

 

While they may not...lemme just do this problem!

 

Integration by parts!

 

u= x dV=tan(x)dx

 

du=1 dx v= -ln(cos(x))

 

 

 

-xln(cos(x))-(inetragl of){-xln(cos(x)}

 

 

 

u = x dv=-ln(cos(x))

 

du = 1 dx v= tan(x)

 

 

 

-xln(cos(x))-tan(x)+(integral of){tan(x)]

 

-xln(cos(x)) - tan(x) - ln(cox(x)) + C

 

 

 

I love calculus.

 

 

 

Dude, that's not math. IT has like 3 digits in it. That's more like 1st grade English. :roll:

[Hobgoblin11]

 

Take it back to the very beginning: A=B

 

 

 

Everything on from here is void, as you could substitute in B for every A. However, something else slightly odd happens when you do this :P :

 

 

 

BB=BB

 

BB-BB=BB-BB

 

B(B-B)=(B-B)(B-B)

 

B=(B-B)

 

If B=2, this equation now reads 2=2-2=0

 

 

 

2 quite clearly does not equal 0! See if you can spot the bit where it goes twisted.

 

 

 

 

 

3rd line you have BB-BB = 2BB-2BB

 

is 0=0

 

you then multiply both sides by 0 (B-B)

 

so the next line should also read 0=0 :-)

 

 

 

Lol, correct. Its actually the second line, BB-BB=BB-BB. If you convert that to numbers, so B = 2, you have (4-4)=(4-4)=0.

 

 

 

=D>

[Death_By_Pod]

 

xln(cos(x))-(inetragl of){-xln(cos(x)}

 

 

 

 

this line is wrong:

 

xln(cos(x))-(inetragl of){-xln(cos(x)}

 

it should be:

 

-xln(cos(x))-(integral of){-ln(cos(x))}

 

and as you don't know the integral of -ln(cos(x)) you can't continue.

 

 

 

The integral of x*tan(x) doesn't have a real solution thats why I posted it in this topic. It's impossible to do with the normal methods of integration you are taught.

[bull912000]

I don't get it. There are no common variables and it is completely simplified. I can't see what else you can do with it. And I don't see what were trying to prove.