prove that √10 is an irrational number.
please!! please!! give this answer by step by step explanation.
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prove that √10 is an irrational number.
please!! please!! give this answer by step by step explanation.
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Answer:
Step-by-step explanation:
Therefore √10 = a/b where a and b are coprime integers. Then: √10 = a/b 10 = a^2/b^2 10b^2 = a^2 2*(5b^2) = a^2 Since a^2 is a multiple of 2, a must also be a multiple of 2 (if you square an even number, you get an even number, but if you square an odd number, you get an odd number).
Answer:
We assume that √10 is a rational number
=> √10 = p/q ( where, p & q belong to the set of integers, q is not equal to 0)
=> Now, let's cancel out all common factors of p/q & represent it in its least form a/b
So, √10 = a/b, here a & b will be co primes. ie, a & b have no common factor other than one.
=> 10 = a² / b²
=> 10 b² = a² ………….(1)
This concludes that, 10 divides a² exactly.
=> 10 divides ‘a’ exactly. ( as a is an integer)
Or, we can say that 10 is a factor of 'a'
which can be stated as…
=> a = 10 * c
=> a² = 100 c² …………(2)
By (1) & (2)
10b² = 100c²
=> b² = 10 c²
This concludes that , 10 divides b² exactly
=> 10 divides b ( as b is an integer)
Or, 10 is a factor of 'b'
This way, we concluded that 10 is a factor of ‘a’ & ‘b’ both. That shows that a & b are not coprimes…
where as we supposed initially, a & b are coprimes.
This contradiction concludes that our initial assumption is wrong.
Hence, √10 should be an irrational number.
[ Hence Proved]