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The proof is by the method of reductio ad absurdum. We start by assuming that sqrt(30) is rational. That means that it can be expressed in the form p/q where p and q are co-prime integers. Thus sqrt(30) = p/q. This can be simplified to 30*q^2 = p^2 Now 2 divides the left hand side (LHS) so it must divide the right hand side (RHS). That is, 2 must divide p^2 and since 2 is a prime, 2 must divide p. That is p = 2*r for some integer r. Then substituting for p gives, 30*q^2 = (2*r)^2 = 4*r^2 Dividing both sides by 2 gives 15*q^2 = 2*r^2. But now 2 divides the RHS so it must divide the LHS. That is, 2 must divide 15*q^2. Since 2 does not divide 15, 2 must divide q^2 and then, since 2 is a prime, 2 must divide q. But then we have 2 dividing p as well as q which contradicts the requirement that p and q are co-prime. The contradiction implies that sqrt(30) cannot be rational.
Wiki User
∙ 9y ago
Wiki User
∙ 7y agoThe square root of 30 is an irrational number because it can't be expressed as a fraction.
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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.
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What set does the square root of thirty belong to?
Irrational
How do you prove that root 2 root 3 is an irrational number?
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.
Is the square root of 6 rational?
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.
Is the square root of a 143 a irrational number?
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