show that √3 is a irrational number
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Let √3 be a rational no. which can be written in the simplest form of a/b. And HCF of a & b is 1.
√3 =a/b
√3b = a
by squaring both sides
3b² = a² ......(i)
3 is the factor of a² , in this way we can also say that 3 is factor of a.
Now, put a = 3c
3b² = 9c²
b² = 3c² .......(ii)
3 is the factor of b². Also, 3 is the factor of b.
From (i) & (ii) 3 is also the factor of a and b which contradict our the fact.
Hence,
√3 is irrational
√3 =a/b
√3b = a
by squaring both sides
3b² = a² ......(i)
3 is the factor of a² , in this way we can also say that 3 is factor of a.
Now, put a = 3c
3b² = 9c²
b² = 3c² .......(ii)
3 is the factor of b². Also, 3 is the factor of b.
From (i) & (ii) 3 is also the factor of a and b which contradict our the fact.
Hence,
√3 is irrational
Answered by
2
Answer:
Let us assume to the contrary that √3 is a rational number.
It can be expressed in the form of p/q
where p and q are co-primes and q≠ 0.
⇒ √3 = p/q
⇒ 3 = p2/q2 (Squaring on both the sides)
⇒ 3q2 = p2………………………………..(1)
It means that 3 divides p2 and also 3 divides p because each factor should appear two times for the square to exist.
So we have p = 3r
where r is some integer.
⇒ p^2 = 9r^2………………………………..(2)
from equation (1) and (2)
⇒ 3q^2 = 9r^2
⇒ q^2 = 3r^2
Where q2 is multiply of 3 and also q is multiple of 3.
Then p, q have a common factor of 3. This runs contrary to their being co-primes. Consequently, p / q is not a rational number. This demonstrates that √3 is an irrational number.
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