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Yes, irrational.

Let p = root 2 and q = root 3. Then (q - p)2 = 5 - 2root6, which is irrational because it is the sum of an integer (5) and an irrational (2root6), and so q - p (which is root3 - root2) is irrational.

Q: Is the difference of root2 and root3 irrational?

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No, it is not. Root2 and root 8 are each irrational. Root8 / root2 =2. 2 is not a member of the set.

Assume it's rational. Then 2 + root2 = some rational number q. Then root2 = q - 2. However, the rational numbers are well-defined under addition by (a,b) + (c,d) = (ad + bc, bd) (in other words, you can add two fractions a/b and c/d and always get another fraction of the form (ad + bc)/bd.) Therefore, q - 2 = q + (-2) is rational, since both q and -2 are rational. This implies root2 must be rational, which is a contradiction. Therefore the assumption that 2 + root2 is rational must be false.

Yes.

It could be either.

No, it is always irrational.

Related questions

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.

Because they cannot be expressed as ratios of integers.

Write three rational numbers between root2 root3 ?

No, it is not. Root2 and root 8 are each irrational. Root8 / root2 =2. 2 is not a member of the set.

Root 2 or 2^(1/2) is an irrational number. It is approximately 1.414214

2-root3 = -1

Yes indeed. There are infinitely many 0 is Pi and others too root2 etc etc

The value of root3 in math is 1.732

It is simple. Take conjugate 2 times. first treat root 2 and root 3 as a single term and do calculations. answer is (6*root2+4*root3-2*root30)/24

//not sure if it is correct bool isomorphic(struct Node* root1,struct Node* root2) { if(root1 root2->value) return ( isomorphic(root1->left,root2->left) && isomorphic(root1->right,root2->right) isomorphic(root1->right,root2->left) && isomorphic(root1->left,root2->right) ); else return false; }

Assume it's rational. Then 2 + root2 = some rational number q. Then root2 = q - 2. However, the rational numbers are well-defined under addition by (a,b) + (c,d) = (ad + bc, bd) (in other words, you can add two fractions a/b and c/d and always get another fraction of the form (ad + bc)/bd.) Therefore, q - 2 = q + (-2) is rational, since both q and -2 are rational. This implies root2 must be rational, which is a contradiction. Therefore the assumption that 2 + root2 is rational must be false.

rational and irrational