Yes it is. The proof is as follows:
We prove the statement by contradiction i.e. Assume that sqrt(6) is a rational number.
Then there exist positive integers p and q with gcd(p,q) = 1 such that p/q = sqrt(6).
Square both sides: p^2 / q^2 = 6,
p^2 = 6q^2.
Now as 2 is a divisor of the right-hand side (RHS), it implies that 2 is also a divisor of the left-hand side (LHS).
This is only possible if 2 is a factor of p.
Let p =2k. Then k is a positive integer as well.
Thus, 4k^2 = 6q^2,
2k^2 = 3q^2.
As 2 is a factor of the LHS, 2 is also a factor of the RHS.
But this is only possible if 2 is a factor of q.
=> gcd(p,q) >= 2. Contradiction!
Thus sqrt(6) is irrational.
yes it is
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The square root of 6 is an irrational number, approximately equal to 2.44948974278. When you take the square root of 6 and continue to do so to infinity, the number will not converge to a specific value but will approach the square root of 6. This means that as you take the square root of the result repeatedly, the number will get closer and closer to approximately 2.44948974278 but will never exactly reach it.
You can find the square root of an irrational number by approximating irrational square roots of them, after you use the calculator. (The calculator gives an approximate root also) For example,1. Approximate the square root of 4.3 to the nearest hundredth.Use the calculator, which shows 2. 0736444135.Since 3 < 5 round down to 2.07 and drop the digits to the right of 7.2. Approximate the negative square root of 10.8 to the nearest hundredth.Use the calculator, which shows -3.286335345Since 6 > 5 round up to -3.29 and drop the digits to the right of 8.
The square root of 36 is 6. 6 cubed is 216.
if this is x squared -6x+6=0 then -6=b, a=1, c=6 6+ square root of -6 squared-4(6x1) - 6+ square root of (36-24) - 6+ square root of 12 - 6+ square root of 4 x square root of 3 - 6 + (2x square root of 3) - that is all divided by 2 multiplied by a meaning it is divided by 2. so x= 6 + or - (2 square root 3) divided by 2 srry steps are jmbled -
6