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Prove that $\sqrt{5}$ is irrational.
Prove that $\sqrt{5}$ is irrational.!
This Question has 1 answers.
Assume the contrary: $ \sqrt{5} $ is rational.
Then, $ \sqrt{5} = \frac{a}{b} $, where $ a $ and $ b $ are coprime integers.
Squaring both sides:
$ 5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2 $
This implies that $ a^2 $ is divisible by 5. Since 5 is prime, $ a $ must also be divisible by 5.
Therefore, we can write:
$ a = 5k $ for some integer $ k $
Substituting $ a = 5k $ into the equation $ a^2 = 5b^2 $:
$ (5k)^2 = 5b^2 $
$ \Rightarrow 25k^2 = 5b^2 $
$ \Rightarrow 5k^2 = b^2 $
This implies that $ b^2 $ is also divisible by 5, and so $ b $ must be divisible by 5.
Now, both $ a $ and $ b $ are divisible by 5, which contradicts the assumption that $ a $ and $ b $ are coprime.
Therefore, $ \sqrt{5} $ is irrational.
Then, $ \sqrt{5} = \frac{a}{b} $, where $ a $ and $ b $ are coprime integers.
Squaring both sides:
$ 5 = \frac{a^2}{b^2} \Rightarrow a^2 = 5b^2 $
This implies that $ a^2 $ is divisible by 5. Since 5 is prime, $ a $ must also be divisible by 5.
Therefore, we can write:
$ a = 5k $ for some integer $ k $
Substituting $ a = 5k $ into the equation $ a^2 = 5b^2 $:
$ (5k)^2 = 5b^2 $
$ \Rightarrow 25k^2 = 5b^2 $
$ \Rightarrow 5k^2 = b^2 $
This implies that $ b^2 $ is also divisible by 5, and so $ b $ must be divisible by 5.
Now, both $ a $ and $ b $ are divisible by 5, which contradicts the assumption that $ a $ and $ b $ are coprime.
Therefore, $ \sqrt{5} $ is irrational.
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