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AR15.COM
7/15/2005 8:13:00 PM EDT
I was explaining about the DUm-DUm’s to a relative when the phrase “Irrational Numbers” popped into my brain.

Back in school when I first learned about Irrationals I once had an idea:
Most Irrational Numbers come from getting the square root of a number.  Pi and e being two obvious exceptions.

But I wondered: Is there an Irrational Number where its Square Root is a Rational Number?

Example: Pi is an Irrational Number.  Its Square Root is an Irrational Number.
But what if the Square Root of Pi is a Rational Number?
My theory would be proven.
But can anybody out there prove it?
And how would you prove it?

Can the Arfcom Mathematical geniususususus Hive Mind Prove or Disprove my Theory?

And my Math Teachers often wondered why my eyes often glazed over.
7/15/2005 8:15:02 PM EDT
[#1]
Your theory is flawed.

ANd irrational number can't become rational numbers.

Same with democrats.

SGat1r5
7/15/2005 8:20:20 PM EDT
[#2]

Quoted:
I was explaining about the DUm-DUm’s to a relative when the phrase “Irrational Numbers” popped into my brain.

Back in school when I first learned about Irrationals I once had an idea:
Most Irrational Numbers come from getting the square root of a number.  Pi and e being two obvious exceptions.

But I wondered: Is there an Irrational Number where its Square Root is a Rational Number?

Example: Pi is an Irrational Number.  Its Square Root is an Irrational Number.
But what if the Square Root of Pi is a Rational Number?
My theory would be proven.
But can anybody out there prove it?
And how would you prove it?

Can the Arfcom Mathematical geniususususus Hive Mind Prove or Disprove my Theory?

And my Math Teachers often wondered why my eyes often glazed over.




irrational means it can't be written as a quotient of two integers

proof by contradiction:

assume you have a number like you describe, call it x

then x^(1/2) would be a rational number, so

x^(1/2) = a/b for some integers a and b

then

(x^(1/2))^2 = (a/b)^2

->

x = a^2/b^2

the square of an integer is an integer (not going to try to prove that), so what you have above is that if such an x exists, it can be written as the quotient of two integers, a^2/b^2

but x was assumed to be an irrational number, so this is a contradiction.

so there is no such x

7/15/2005 8:21:57 PM EDT
[#3]

Quoted:

Quoted:
I was explaining about the DUm-DUm’s to a relative when the phrase “Irrational Numbers” popped into my brain.

Back in school when I first learned about Irrationals I once had an idea:
Most Irrational Numbers come from getting the square root of a number.  Pi and e being two obvious exceptions.

But I wondered: Is there an Irrational Number where its Square Root is a Rational Number?

Example: Pi is an Irrational Number.  Its Square Root is an Irrational Number.
But what if the Square Root of Pi is a Rational Number?
My theory would be proven.
But can anybody out there prove it?
And how would you prove it?

Can the Arfcom Mathematical geniususususus Hive Mind Prove or Disprove my Theory?

And my Math Teachers often wondered why my eyes often glazed over.




irrational means it can't be written as a quotient of two integers

proof by contradiction:

assume you have a number like you describe, call it x

then x^(1/2) would be a rational number, so

x^(1/2) = a/b for some integers a and b

then

(x^(1/2))^2 = (a/b)^2

->

x = a^2/b^2

the square of an integer is an integer (not going to try to prove that), so what you have above is that if such an x exists, it can be written as the quotient of two integers, a^2/b^2

but x was assumed to be an irrational number, so this is a contradiction.

so there is no such x




Proof by contradiction brings back horrible memories of Discrete Math in college.

Can you prove it by contraposition?
7/15/2005 8:28:16 PM EDT
[#4]

Quoted:

Quoted:

Quoted:
I was explaining about the DUm-DUm’s to a relative when the phrase “Irrational Numbers” popped into my brain.

Back in school when I first learned about Irrationals I once had an idea:
Most Irrational Numbers come from getting the square root of a number.  Pi and e being two obvious exceptions.

But I wondered: Is there an Irrational Number where its Square Root is a Rational Number?

Example: Pi is an Irrational Number.  Its Square Root is an Irrational Number.
But what if the Square Root of Pi is a Rational Number?
My theory would be proven.
But can anybody out there prove it?
And how would you prove it?

Can the Arfcom Mathematical geniususususus Hive Mind Prove or Disprove my Theory?

And my Math Teachers often wondered why my eyes often glazed over.




irrational means it can't be written as a quotient of two integers

proof by contradiction:

assume you have a number like you describe, call it x

then x^(1/2) would be a rational number, so

x^(1/2) = a/b for some integers a and b

then

(x^(1/2))^2 = (a/b)^2

->

x = a^2/b^2

the square of an integer is an integer (not going to try to prove that), so what you have above is that if such an x exists, it can be written as the quotient of two integers, a^2/b^2

but x was assumed to be an irrational number, so this is a contradiction.

so there is no such x




Proof by contradiction brings back horrible memories of Discrete Math in college.

Can you prove it by contraposition?



i generally prefer proof by googleization.
7/15/2005 8:29:08 PM EDT
[#5]
aren't all negative values irrational?
7/15/2005 8:29:15 PM EDT
[#6]
fuck this
7/15/2005 8:30:31 PM EDT
[#7]
By definition: An Irrational number is any number that cannot be expressed as some Fraction (ie. Rational Number).

There are 2 types of Irrational Numbers: (a.) Algebraic Irrational Numbers and (b.) Transcendental Irrational Numbers.

Algebraic Irrational Numbers arise from finding the roots of a polynomial (examples are square root of two, cube root of 5, etc.)

Transcendental Irrational Numbers are all Irrational Numbers which cannot arise Algebraically,
Examples are PI, e, etc.

The proof for proving that the square root of 2 is an Irrational Number is quite trivial.
Hint: do a proof by Contridiction.

Similarly,

If I took a Transcental Irrational Number and took its square root, I would still end up with a Irrational Number.
Use the same method for proving the above: ie. proof by contradiction.



7/15/2005 8:35:54 PM EDT
[#8]

Quoted:
aren't all negative values irrational?



Nope, but the square root of all negative values are imaginary.
7/15/2005 8:36:13 PM EDT
[#9]

Quoted:
aren't all negative values irrational?



No.

Only numbers which cannot be represented by a fractional integer  e.g.  2/3   (repeating decimal answer if you carry out the math (.6bar), but still a rational number...)
7/15/2005 8:36:42 PM EDT
[#10]

Quoted:
fuck this



Wait till they try to tell you that 1 = .999999999999999999999999999999999999999999999999999999+

SGatr15
7/15/2005 8:37:53 PM EDT
[#11]

Quoted:

Quoted:
aren't all negative values irrational?



No.

Only numbers which cannot be represented by a fractional integer  e.g.  2/3   (repeating decimal answer if you carry out the math (.6bar), but still a rational number...)



Couldn't you just write an irrational number as the fraction:

irrational number
------------------
             1


?
7/15/2005 8:39:40 PM EDT
[#12]

Quoted:

Quoted:

Quoted:
aren't all negative values irrational?



No.

Only numbers which cannot be represented by a fractional integer  e.g.  2/3   (repeating decimal answer if you carry out the math (.6bar), but still a rational number...)



Couldn't you just write an irrational number as the fraction:

irrational number
------------------
             1


?



Great.

Now we are getting back into the great ".9bar" debate...  
7/15/2005 8:40:32 PM EDT
[#13]

Quoted:

Quoted:

Quoted:
I was explaining about the DUm-DUm’s to a relative when the phrase “Irrational Numbers” popped into my brain.

Back in school when I first learned about Irrationals I once had an idea:
Most Irrational Numbers come from getting the square root of a number.  Pi and e being two obvious exceptions.

But I wondered: Is there an Irrational Number where its Square Root is a Rational Number?

Example: Pi is an Irrational Number.  Its Square Root is an Irrational Number.
But what if the Square Root of Pi is a Rational Number?
My theory would be proven.
But can anybody out there prove it?
And how would you prove it?

Can the Arfcom Mathematical geniususususus Hive Mind Prove or Disprove my Theory?

And my Math Teachers often wondered why my eyes often glazed over.




irrational means it can't be written as a quotient of two integers

proof by contradiction:

assume you have a number like you describe, call it x

then x^(1/2) would be a rational number, so

x^(1/2) = a/b for some integers a and b

then

(x^(1/2))^2 = (a/b)^2

->

x = a^2/b^2

the square of an integer is an integer (not going to try to prove that), so what you have above is that if such an x exists, it can be written as the quotient of two integers, a^2/b^2

but x was assumed to be an irrational number, so this is a contradiction.

so there is no such x




Proof by contradiction brings back horrible memories of Discrete Math in college.

Can you prove it by contraposition?

Like I said; My Math Teachers often wondered why my eyes often glazed over.
7/15/2005 8:40:42 PM EDT
[#14]

Quoted:

Quoted:

Quoted:
aren't all negative values irrational?



No.

Only numbers which cannot be represented by a fractional integer  e.g.  2/3   (repeating decimal answer if you carry out the math (.6bar), but still a rational number...)



Couldn't you just write an irrational number as the fraction:

irrational number
------------------
             1


?




no, rational means it is the ratio of two integers (i.e. two "whole" numbers positive or negative)
7/15/2005 8:44:43 PM EDT
[#15]

Quoted:

Quoted:

Quoted:

Quoted:
aren't all negative values irrational?



No.

Only numbers which cannot be represented by a fractional integer  e.g.  2/3   (repeating decimal answer if you carry out the math (.6bar), but still a rational number...)



Couldn't you just write an irrational number as the fraction:

irrational number
------------------
             1


?




no, rational means it is the ratio of two integers (i.e. two "whole" numbers positive or negative)



+ .9bar  
7/15/2005 9:18:35 PM EDT
[#16]
told ya.

SGat1r5
7/15/2005 9:23:05 PM EDT
[#17]

Quoted:
told ya.

SGat1r5

But they proved it; you didn't.

Now go make me a samich.
7/15/2005 9:26:20 PM EDT
[#18]

Quoted:

Quoted:
fuck this



Wait till they try to tell you that 1 = .999999999999999999999999999999999999999999999999999999+

SGatr15



It is.
7/15/2005 9:34:09 PM EDT
[#19]

Quoted:

Quoted:
told ya.

SGat1r5

But they proved it; you didn't.

Now go make me a samich.



Hell yeah, you just lived up to your name GiggleSmith. No offense sgtar15.
7/16/2005 12:03:59 AM EDT
[#20]

Quoted:

Quoted:
I was explaining about the DUm-DUm’s to a relative when the phrase “Irrational Numbers” popped into my brain.

Back in school when I first learned about Irrationals I once had an idea:
Most Irrational Numbers come from getting the square root of a number.  Pi and e being two obvious exceptions.

But I wondered: Is there an Irrational Number where its Square Root is a Rational Number?

Example: Pi is an Irrational Number.  Its Square Root is an Irrational Number.
But what if the Square Root of Pi is a Rational Number?
My theory would be proven.
But can anybody out there prove it?
And how would you prove it?

Can the Arfcom Mathematical geniususususus Hive Mind Prove or Disprove my Theory?

And my Math Teachers often wondered why my eyes often glazed over.




irrational means it can't be written as a quotient of two integers

proof by contradiction:

assume you have a number like you describe, call it x

then x^(1/2) would be a rational number, so

x^(1/2) = a/b for some integers a and b

then

(x^(1/2))^2 = (a/b)^2

->

x = a^2/b^2

the square of an integer is an integer (not going to try to prove that), so what you have above is that if such an x exists, it can be written as the quotient of two integers, a^2/b^2

but x was assumed to be an irrational number, so this is a contradiction.

so there is no such x



As I have only recently really fully learned the proper mathematical proof methods (somehow lacking in my undergrad math preparation), I was very happy to read your proof and understand exactly what you were doing, and why.

And to think I used to hate math.  Now I wish I had double-majored...

Jim
7/16/2005 12:06:14 AM EDT
[#21]

Quoted:
fuck this



Liberal arts major?
7/16/2005 12:07:10 AM EDT
[#22]

Quoted:

Quoted:

Quoted:
told ya.

SGat1r5

But they proved it; you didn't.

Now go make me a samich.



Hell yeah, you just lived up to your name GiggleSmith. No offense sgtar15.



No they didn't.

The decimal point is still on opisite sides.

PERIOD!

Sgat1r5


PS  I'll take ham & cheese...provolone.
7/16/2005 12:15:39 AM EDT
[#23]

Quoted:

Quoted:
fuck this



Liberal arts major?



actually...dual major of mechanical and aerospace engineering with a minor in physics.  I like to keep math out of my non school life
7/16/2005 5:27:50 AM EDT
[#24]
When I took Real Analysis...one of the first proofs we did was proving that the Square Root of 2 was a Irrational Number.

If you like doing proofs, both Real Analysis and Abstract Algebra are were you first really learn how to do a proof.