Prove that root 2 is irrational. Some simple method will be appreceated
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let√2 be rational
∴√2 = p/q for co - prime integers p and q where q ≠0
2 = p²/q²
2q² = p²
therefore p² is divisible by 2
∴p is divisible by 2
∴ p = 2c for some integer c
substituting it in 2q² = p²
2q² = 4c²
2c² = q²
∴q² is divisible by 2
∴q is divisible by 2
∴both p and q are divisible by 2
but this is not possible as p and q are co-prime
this contradiction has arisen from our assumption that √2 is rational
∴√2 is irrational
pls mark as brainliest
∴√2 = p/q for co - prime integers p and q where q ≠0
2 = p²/q²
2q² = p²
therefore p² is divisible by 2
∴p is divisible by 2
∴ p = 2c for some integer c
substituting it in 2q² = p²
2q² = 4c²
2c² = q²
∴q² is divisible by 2
∴q is divisible by 2
∴both p and q are divisible by 2
but this is not possible as p and q are co-prime
this contradiction has arisen from our assumption that √2 is rational
∴√2 is irrational
pls mark as brainliest
GovindKrishnan:
Plz mark this answer as brainliest …
u really dont have to do to that
Answered by
4
√2 = 2½
( ½ is a fraction…)
So √2 is irrational …
To be frank, I'm not sure with my answer …
( ½ is a fraction…)
So √2 is irrational …
To be frank, I'm not sure with my answer …
also if this is for10 grade cbse, this answer wiil not be concidered
Is it correct now? I made some changes…
Simple method was asked… So I tried to make it simple …
u typed exponent of a rational number should be a fraction when u should have typed exponent of a rational number should not be a fraction
also 4 ^1/2 is rational ...........so i am not sure if ur method can be recomended for this sum
K… I just gave it a try… Ur answer will be brainliest … :-)
bro i am not correcting u so that i can get the brainliest answer , i am just doing it so u know its not as right as it seems to be
it was quite an innovative idea though...............i'll give u that
No probs… :-)
You did the right thing… :-)
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