Help me Please In Proving!

I'm very weak in proving something related to real numbers.I dont face problem in proving geometric proofs but in proving something like this i am just blank-prove that any two consecutive natural nimbers are co-prime .I understand what this sentence means but if someone says me to prove this I become totally blank. So please help me

#NumberTheory

Note by Aman Real
6 years, 1 month ago

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Comments

@calvin sir help me

Aman Real - 6 years, 1 month ago

Co-prime means that the two numbers dont have any common factor other than 1 or their gcd is 1

Aditya Chauhan - 6 years, 1 month ago

Coprime means the gcd of the 2 numbers is one. So the question is asking you to prove that if 2 numbers are consecutive, their only common factor is 1. This is how you do a proof: you write what has to be proved (write RTP- this stands for 'required to prove'), write your working (including dialogue so the reader can understand) and when you have finished your proof, you write Q.E.D. This stands for quad eras demonstratum, which is latin for 'which was to be proved'.

RTP: Any two consecutive natural numbers are co-proime.
Proof: Let the 2 consecutive integers be n and n+1. Frist, find the gcd of n and n+1. Call this gcd(n,n+1) By the ecludian algorithm, gcd(n,n+1) = gcd(n, n+1-n) = gcd(n,1) But the greatest common divisior of any integer and 1, i.e. they are coprime. Thus, any two consecutive natural numbers are co-proime. Q.E.D.

A Former Brilliant Member - 6 years, 1 month ago

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Thanks,but one question i want to ask is that ,why did you did this gcd=(n,n+1-n) ?

Aman Real - 6 years, 1 month ago

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its a theorem called the ecludian algorithm... i will give a wikipedia definition ''The Euclidean algorithm is based on the principle that the greatest common divisor of two numbers does not change if the larger number is replaced by its difference with the smaller number. ''

In this case, the larger number is n+1, and the difference with the smaller number is n+1-n=1, so it is the same as the gcd(n,1). Which is always 1.

A Former Brilliant Member - 6 years, 1 month ago
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