A rational number is defined as a number that can be expressed in the form of a division of two integers, i.e. p/q, where q is not equal to 0. √3 = 1.7320508075688772... and it keeps extending. Since it does not terminate or repeat after the decimal point, √3 is an irrational number.
Let us assume on the contrary that 3 is a rational number.
Then, there exist positive integers a and b such that
3=ba where, a and b, are co-prime i.e. their HCF is 1
Now,
3=ba
⇒3=b2a2
⇒3b2=a2
⇒3 divides a2[∵3 divides 3b2]
⇒3 divides a...(i)
⇒a=3c for some integer c
⇒a2=9c2
⇒3b2=9c2[∵a2=3b2]
⇒b2=3c2
⇒3 divides b2[∵3 divides 3c2]
⇒3 divides b...(ii)
From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.
Hence, 3 is an irrational number.
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Answer:
A rational number is defined as a number that can be expressed in the form of a division of two integers, i.e. p/q, where q is not equal to 0. √3 = 1.7320508075688772... and it keeps extending. Since it does not terminate or repeat after the decimal point, √3 is an irrational number.
Explanation:
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Verified answer
Answer:
Let us assume on the contrary that 3 is a rational number.
Then, there exist positive integers a and b such that
3=ba where, a and b, are co-prime i.e. their HCF is 1
Now,
3=ba
⇒3=b2a2
⇒3b2=a2
⇒3 divides a2[∵3 divides 3b2]
⇒3 divides a...(i)
⇒a=3c for some integer c
⇒a2=9c2
⇒3b2=9c2[∵a2=3b2]
⇒b2=3c2
⇒3 divides b2[∵3 divides 3c2]
⇒3 divides b...(ii)
From (i) and (ii), we observe that a and b have at least 3 as a common factor. But, this contradicts the fact that a and b are co-prime. This means that our assumption is not correct.
Hence, 3 is an irrational number.
hi if it was helpful than please mark me brainlist
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