![\sqrt{2} \sqrt{2}](https://tex.z-dn.net/?f=+%5Csqrt%7B2%7D+)
is irrational
Answers
Answered by
0
Answer:
Yes, under root 2 is irrational number because it's decimal expansion is non-terminating and non-repeating.
Answered by
1
Answer:
yes
Step-by-step explanation:
Because √2 is not an integer (2 is not a perfect square), √2 must therefore be irrational. This proof can be generalized to show that any square root of any natural number that is not the square of a natural number is irrational.
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