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Science Forum Index » Logic Forum » please help verify this proof
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| Guest |
 Posted: Wed Apr 30, 2008 6:31 pm |
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Prove that if x^3 is irrational, x is irrational.
Proving by contradiction:
suppose that if x^3 is rational, x is irrational
x^3 = a/b definition of rational
x=(a/b)^3 cubing, just simple math
x=(a^3/b^3) rewriting rules of exponent allows this
x=(a/b) This can be because a and b are any arbitrary integer not
equal to zero. Just substitution.
thus a contradiction since we claim x is irrational, which by
definition are numbers that cannot be written as fraction of two
integers.
QED
or by Proof by Contrapositive
1.-x-->-x^3 premise (x is a rational implies x^3 is rational)
conclude: -x^3
2.-x=a/b definition a of -x, a rational number
3.-x^3=a/b definition for rational number
4.-x= (a/b)^3 arithmetic simplification line 3
5.-x= (a^3/b^3) arithmetic rules of exponent line 4
6.-x = (a/b) arithmetic substitution of line 5 since a and b are any
two integers
7.-x^3 6,1 modus ponens
QED
correct? |
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| smn |
| Posted: Thu May 01, 2008 12:22 am |
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Guest
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On Apr 30, 9:31 pm, chalon...@gmail.com wrote:
Quote: Prove that if x^3 is irrational, x is irrational.
Proving by contradiction:
You want to show that x is irrational ! So assume as a premise that x
is rational and try for a contradiction .The contradiction is by your
second proof showing that
x^3 is rational which is a contradiction of your
hypothesis.Contradicting the hypothesis is also called proof by
contrapositive .Your first proof is wrong.
Quote: suppose that if x^3 is rational, x is irrational
There is no point assuming this and anyway it could be true : x=2^1/3
is not rational (its harder to prove this -a little number theory is
required) but x^3 =2 so you could never get a contradiction .
Quote: x^3 = a/b definition of rational
x=(a/b)^3 cubing, just simple math
Here is your mistake, It should say x=(a^1/3)/(b^1/3) which isn't
helpful
Quote: x=(a^3/b^3) rewriting rules of exponent allows this
x=(a/b) This can be because a and b are any arbitrary integer not
equal to zero. Just substitution.
thus a contradiction since we claim x is irrational, which by
definition are numbers that cannot be written as fraction of two
integers.
QED
So above is wrong ,below is ok. Regards,smn
or by Proof by Contrapositive
1.-x-->-x^3 premise (x is a rational implies x^3 is rational)
conclude: -x^3
2.-x=a/b definition a of -x, a rational number
3.-x^3=a/b definition for rational number
4.-x= (a/b)^3 arithmetic simplification line 3
5.-x= (a^3/b^3) arithmetic rules of exponent line 4
6.-x = (a/b) arithmetic substitution of line 5 since a and b are any
two integers
7.-x^3 6,1 modus ponens
QED
correct? |
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| David C. Ullrich |
| Posted: Thu May 01, 2008 7:08 am |
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Guest
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On Wed, 30 Apr 2008 21:31:07 -0700 (PDT), chalong08@gmail.com wrote:
Quote: Prove that if x^3 is irrational, x is irrational.
Proving by contradiction:
suppose that if x^3 is rational, x is irrational
Wrong already. You should be assuming that x^3
is irrational and x is rational.
Quote: x^3 = a/b definition of rational
x=(a/b)^3 cubing, just simple math
This is also backwards.
Quote: x=(a^3/b^3) rewriting rules of exponent allows this
x=(a/b) This can be because a and b are any arbitrary integer not
equal to zero. Just substitution.
thus a contradiction since we claim x is irrational, which by
definition are numbers that cannot be written as fraction of two
integers.
QED
or by Proof by Contrapositive
1.-x-->-x^3 premise (x is a rational implies x^3 is rational)
conclude: -x^3
2.-x=a/b definition a of -x, a rational number
3.-x^3=a/b definition for rational number
4.-x= (a/b)^3 arithmetic simplification line 3
5.-x= (a^3/b^3) arithmetic rules of exponent line 4
6.-x = (a/b) arithmetic substitution of line 5 since a and b are any
two integers
7.-x^3 6,1 modus ponens
QED
correct?
David C. Ullrich |
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