Proof that if x^3 is divisible by 13 then so is x? Watch

lightningdoritos
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  1. No idea where to start with this
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Louisb19
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(Original post by lightningdoritos)
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For  x \in \mathbb{R} ?
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TeeEm
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(Original post by Louisb19)
For  x \in \mathbb{R} ?
divisibility has to be integers or better natural
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DylanJ42
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(Original post by lightningdoritos)
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Posting so I can follow this, very interested to see the solution. I have zero idea also
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Pronged Lily
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Think about the prime factorisation of x^3...
(compared to the prime factorisation of x)
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Louisb19
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(Original post by TeeEm)
divisibility has to be integers or better natural
My bad, it makes sense that the question would be referring to natural numbers. I couldn't think of any cube numbers which are divisible by 13 off the top of my head.

I was thinking of using an inductive method however I cannot think of a base case. Any thoughts?
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Renzhi10122
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(Original post by lightningdoritos)
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Use Euclid's lemma, (though I'm not sure if this is allowed).
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TeeEm
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(Original post by Louisb19)
My bad, it makes sense that the question would be referring to natural numbers. I couldn't think of any cube numbers which are divisible by 13 off the top of my head.

I was thinking of using an inductive method however I cannot think of a base case. Any thoughts?
I am not a purist but my thought is that

x3 it is of the form Πp3n, where p is a prime factor


since 13 is primeif x3 is divisible by 133n and therefore x must be divisible too
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Pronged Lily
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(Original post by Louisb19)
My bad, it makes sense that the question would be referring to natural numbers. I couldn't think of any cube numbers which are divisible by 13 off the top of my head.

I was thinking of using an inductive method however I cannot think of a base case. Any thoughts?
Use contradiction?
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16Characters....
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(Original post by Pronged Lily)
Use contradiction?
Or possibly contraposition?
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Pronged Lily
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(Original post by 16Characters....)
Or possibly contraposition?
Yeah that would probably be quickest
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Louisb19
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So you have the base case of x = 13, clearly x^3 and x are both divisible by 13.

You could extend this to any value of x = 13n where n is a natural number > 0.

I have a hunch that because 13 is a prime number and can therefore not be further broken down, the only values of x which will work are as I described. I doubt this is thorough enough, I don't know how to prove it. I would appreciate it if someone could offer me a hint.

Time for bed.
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Number Nine
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(Original post by Louisb19)
So you have the base case of x = 13, clearly x^3 and x are both divisible by 13.

You could extend this to any value of x = 13n where n is a natural number > 0.

I have a hunch that because 13 is a prime number and can therefore not be further broken down, the only values of x which will work are as I described. I doubt this is thorough enough, I don't know how to prove it. I would appreciate it if someone could offer me a hint.

Time for bed.
Yeah there needs to be a factor of 13 in there and if it's not coming from x then where is it coming from? This cannot be through some magic multiplication as 13 is a prime number

QED
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math42
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If you can use the fact that every natural number (greater than or equal to two, not that that matters) can be expressed as a product of prime factors then this seems quite trivial...for instance write out x in prime factorized form (allowing for primes to the power of zero so that in the multiplication you can just write primes to arbitrary powers in order up to some arbitrary largest prime divisor) Say the power of 13 in the prime factorization of x is "i" for instance and then note what the power of 13 is in x^3
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lightningdoritos
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Thanks everyone, the contrapositive worked!
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