# Prove by contradictionWatch

Thread starter 1 week ago
#1
Need help with this question. Thanks
Last edited by dasda; 1 week ago
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1 week ago
#2
(Original post by dasda)
Need help with this question. Thanks
What have you tried? The proof is pretty much the same as for .
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Thread starter 1 week ago
#3
I cubed both sides and ended up with 2=a^3/b^3. I then put 2a^3=b^3. I then said that b=(2n)^3
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1 week ago
#4
(Original post by dasda)
I cubed both sides and ended up with 2=a^3/b^3. I then put 2a^3=b^3. I then said that b=(2n)^3
You are jumping across landscapes here with this argument. Too many missing links to a point where you're confusing yourself.

Firstly, you need to hammer down the fact that you assume cube root 2 is rational, this means expressing it in the irreducible form a/b where a is some integer and b is a natural number. The irreducible part is key here.

Secondly, you cube both sides and get 2 = a^3/b^3 and this is the same as 2b^3 = a^3. You got this the wrong way round in yours. Be careful!
So, we know that a^3 is even. What does this tell us about a itself?
Last edited by RDKGames; 1 week ago
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Thread starter 1 week ago
#5
(Original post by RDKGames)
You are jumping across landscapes here with this argument. Too many missing links to a point where you're confusing yourself.

Firstly, you need to hammer down the fact that you assume cube root 2 is rational, this means expressing it in the form a/b where a and b are irreducible. And that's a key word.

Secondly, you cube both sides and get 2 = a^3/b^3 and this is the same as 2b^3 = a^3. You got this the wrong way round in yours. Be careful!
So, we know that a^3 is even. What does this tell us about a itself?
A is even.
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1 week ago
#6
(Original post by dasda)
A is even.
Therefore you can express it in the form for some integer . What next?
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Thread starter 1 week ago
#7
a=2n^3
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1 week ago
#8
(Original post by dasda)
a=2n^3
Nope. Where did that come from?
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Thread starter 1 week ago
#9
(Original post by RDKGames)
Nope. Where did that come from?
i assumed since a was even and the equation has a^3, we would cube 2n
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1 week ago
#10
(Original post by dasda)
i assumed since a was even and the equation has a^3, we would cube 2n
Yes we do, but you cubed it incorrectly, and didn't write a^3 to indicate it. If you're cubing 2n you need to cube the 2 as well. 0
Thread starter 1 week ago
#11
so 8n^3
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1 week ago
#12
(Original post by dasda)
so 8n^3
Yes, . What next?
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Thread starter 1 week ago
#13
(Original post by RDKGames)
Yes we do, but you cubed it incorrectly, and didn't write a^3 to indicate it. If you're cubing 2n you need to cube the 2 as well. is it 8a^3=b^3
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1 week ago
#14
(Original post by dasda)
is it 8a^3=b^3
Nope.

We have so it becomes .
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Thread starter 1 week ago
#15
this is the step I am unsure about. I would divide both sides by 2
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1 week ago
#16
(Original post by dasda)
this is the step I am unsure about. I would divide both sides by 2
Yep, so . What does this tell you about about ?? Hence ?
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Thread starter 1 week ago
#17
I would say b is even.
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1 week ago
#18
(Original post by dasda)
I would say b is even.
Indeed. Do you see the contradiction now? Since a and b must both be even.

Have a look at my post #4 on this thread and what assumptions we have made if you're not seeing the contradiction immediately.
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Thread starter 1 week ago
#19
(Original post by RDKGames)
Indeed. Do you see the contradiction now? Since a and b must both be even.

Have a look at my post #4 on this thread and what assumptions we have made if you're not seeing the contradiction immediately.
you assumed that the irrational was irreducible. but we got a factor of 2 and 1 so our assumption is incorrect
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Thread starter 1 week ago
#20
is that a correct line of thinking

(Original post by RDKGames)
Indeed. Do you see the contradiction now? Since a and b must both be even.

Have a look at my post #4 on this thread and what assumptions we have made if you're not seeing the contradiction immediately.
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