if n is any prime no. and a^square is divisible by n, then n will also divide a. JUSTIFY.
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Answered by
14
Hello aditya
Let the prime factorisation of n be
a=p1*p2*p3........pn
Where p1,p2,.........pm are primes,not neccesarily distinct
So the prime factorisation of square of a is
a²=(p1*p2*p3........pm)*(p1*p2*p3........pn)
It is given that a² is divisible by n
Therefore from the fundamental theorem of arithmetic it states that p is one of the prime factors of a².
The unique factors of a² are p1,p2,....on
So p is one among p1,p2,.....on
we know that a=p1,p2....on
So p divides a
Hope this helps!!!
Let the prime factorisation of n be
a=p1*p2*p3........pn
Where p1,p2,.........pm are primes,not neccesarily distinct
So the prime factorisation of square of a is
a²=(p1*p2*p3........pm)*(p1*p2*p3........pn)
It is given that a² is divisible by n
Therefore from the fundamental theorem of arithmetic it states that p is one of the prime factors of a².
The unique factors of a² are p1,p2,....on
So p is one among p1,p2,.....on
we know that a=p1,p2....on
So p divides a
Hope this helps!!!
adityagupta190:
Hey Wvaish.... thank you for giving me the answer of my question. Would you mind if u give me a brief of this answer or just tell me the answer in simple language.... its little complicated. Thank you in advance!
Answered by
0
Answer:
it was teorem which was already proved
Step-by-step explanation:
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