Suppose we start with the opposite view. That is, the square root of 5 is rational. This means that it can be expressed as a ratio of two integers. Let the square root of 5 in its simplest terms be a/b so that a and b are integers with no common factors and b > 0.
sqrt(5) = a/b so 5 = a2/b2
that is, 5b2= a2
5 is a factor of the left hand side (LHS) so 5 must be a factor of the right hand side (RHS).
Since 5 is a prime and a is an integer, 5 must be a factor of a. That is, a = 5c for some integer c and since a and b are coprime b and c must also be coprime.
But now,
5 = (5c)2/b2so that 5 = 25c2/b2and so b2= 5c2. As before, this implies that 5 must be a factor of b. But that contradicts the supposition that a and b are coprime.
The contradiction implies that the original assumption was incorrect. That is, sqrt(5) cannot be rational.
This is impossible to prove, as the square root of 2 is irrational.
No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.
The square root of 5 is an irrational number
I linked a good resource that explains what you asked below.
yes
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This is impossible to prove, as the square root of 2 is irrational.
The square root of 2 is 1.141..... is an irrational number
Because 3 is a prime number and as such its square root is irrational
It is known that the square root of an integer is either an integer or irrational. If we square root2 root3 we get 6. The square root of 6 is irrational. Therefore, root2 root3 is irrational.
They are +5 and -5, which are both rational.
7 plus the square root of 5 is an irrational number because the square root of 5 is a never ending decimal number that can't be expressed as a fraction.
No; you can prove the square root of any positive number that's not a perfect square is irrational, using a similar method to showing the square root of 2 is irrational.
You cannot. The square root of 5 is irrational.
This is an easy question : root 2, root 3, root 5. Any square root of an integer which is not integral itself is irrational.
5
Yes.