Suppose sqrt(2) < 1 and assume that only the positive sqrt is being considered.
Then square both sides of the inequality: 2 < 1 which is a contradiction.
So the supposition is wrong ie sqrt(2) > 1
Of course, there is a sqrt of 2 which is not >1, and that is -1.4142..
when the number is greater than 1
Yes, if the number is less than 1
You eventually get the answer 1
By Calculator 65^(1/2) = 8.0622... < 8.3 8.3^(2) = 68.89 And 65 < 68.89
only if x is greater than 1
Yes, if the number is less than '1'.Just the opposite, if the number is greater than '1'.
when the number is greater than 1
False. Only a square number greater than 1 is always bigger than its root. For example, the root of 16 is 4, but the root of 1/16 (0.0625) is 1/4 (0.25) and the square root of 1 is 1.
5 1/4 = 5.25 square root of 26 approximately equal to 5.099 Hence 5.25 > 5.099 or 5 1/4 is greater than square root of 26
Yes, if the number is less than 1
You eventually get the answer 1
only if x is greater than 1
By Calculator 65^(1/2) = 8.0622... < 8.3 8.3^(2) = 68.89 And 65 < 68.89
1 is less than the square root of 2.
The square root of 2 and the square root of 3 both qualify. Both of these are irrational and both are greater than 1 but less than 2. There are, of course, uncountably infinite different irrational numbers in the range between 1 and 2 and countably infinite rational numbers.
No, not always since: if a number is more than 1, then its square root is smaller than the number. if a number is less than 1, then its square root is bigger than the number.
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