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Cube root of 6 is irrational. (please check if my proof is correct)

  • Thread starter pankaz712
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  • #1
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Prove that cube root of 6 irrational.

Solution: I am trying to prove by contradiction.

Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0)
Cubing both sides : 6=a^3/b^3
a^3 = 6b^3
a^3 = 2(3b^3)
Therefore, 2 divides a^3 or a^2 * a . By Euclid's Lemma if a prime number divides the product of two integers then it must divide one of the two integers. Since all the terms here are the same we conclude that 2 divides a.
Now there exists an integer k such that a=2k
Substituting 2k in the above equation
8k^3 = 6b^3
b^3 = 2{(2k^3) / 3)}
Therefore, 2 divides b^3. Using the same logic as above. 2 divides b.
Hence 2 is common factor of both a & b. But this is a contradiction of the fact that a & b are co-prime.
Therefore, the initial assumption is wrong. cube root 6 is irrational
 

Answers and Replies

  • #2
eumyang
Homework Helper
1,347
10
Prove that cube root of 6 irrational.

Solution: I am trying to prove by contradiction.

Assume cube root 6 is rational. Then let cube root 6 = a/b ( a & b are co-prime and b not = 0)
Cubing both sides : 6=a^3/b^3
a^3 = 6b^3
a^3 = 2(3b^3)
I don't think you need the step which I bolded. You can just say from the previous line,
a^3 = 6b^3
that 6 divides a^3, and consequently, 6 divides a. So if 6 divides a, there exists an integer k such that a = 6k. Replace in
a^3 = 6b^3
and see what happens.
 
  • #3
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I don't think you need the step which I bolded. You can just say from the previous line,
a^3 = 6b^3
that 6 divides a^3, and consequently, 6 divides a. So if 6 divides a, there exists an integer k such that a = 6k. Replace in
a^3 = 6b^3
and see what happens.
i did that because i thought Euclid's Lemma was true only for prime numbers.
 
  • #4
1,006
105
How do you know that (2k^3) / 3 is an integer?
 
  • #5
eumyang
Homework Helper
1,347
10
I remembered seeing this proof before, so I guess I am not remembering it correctly.

Let's try this:
Now there exists an integer k such that a=2k
Substituting 2k in the above equation
8k^3 = 6b^3
b^3 = 2{(2k^3) / 3)}
Something bothers me about the bolded. Do we know for sure that "(2k^3) / 3)" is an integer? Instead, I would say this:
8k^3 = 6b^3
4k^3 = 3b^3

The LHS contains at least one factor that is even, so the entire LHS is even. Therefore the RHS is even and b^3 is even. Etc., etc.

EDIT: Beaten to it. :wink:
 
  • #6
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The LHS contains at least one factor that is even, so the entire LHS is even. Therefore the RHS is even and b^3 is even. Etc., etc.
thnx i understood ur method. I have just one more question:

How to prove that root n is irrational, if n is not a perfect square. Also, if n is a perfect square then how does it affect the proof.
 

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