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How do you show that 2 plus root2 is an irrational number?
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.
Is the set of irrational numbers closed under division?
No, it is not. Root2 and root 8 are each irrational. Root8 / root2 =2. 2 is not a member of the set.
Write a fortran code to calculate the root of quadratic equation?
implicit double precision(a-h,o-z) write(*,*) "please provide the a,b,c coeff" read(*,*) A,B,C D=B*B-4*A*C if(D.GT.0) then root1=(-B/(2*A))+(SQRT(D))/(2*A) root2=(-B/(2*A))-(SQRT(D))/(2*A) write(*,*) root1,root2 elseif(D.EQ.0) then root1=(-B/(2*A)) root2=root1 write(*,*) root1,root2 else root1=(-B/(2*A))+(SQRT(-D))/(2*A) root2=(-B/(2*A))-(SQRT(-D))/(2*A) a=(root1+root2)/2 b=(root1-root2)/2 write(*,*) 'realpartroot=',a, 'complexpartroot=',b endif stop END
Is the difference of root2 and root3 irrational?
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.
What are the dimensions of largest square that can be inscribed in a circle?
The diagonal of the square is the diameter of the circle (think about it!) so the sides of the square can be pythagorassed ie the square of the diagonal is twice the square of a side. As an example, consider a circle of diameter 50 units. the square of this is 2500 so every side of the square is the square root of 1250 units which equals sqrt 625 x sqrt 2 or 25 root 2 ie 35.355339 units, or 70.71% of the diameter of the circle. More basically, the diagonal is the hypotenuse of a 45/45/90 triangle so the sides are in the ratio 1/1/root2 or if you want the hypotenuse to be 1, the sides would be (root 2)/2. Either way, the sides are root2 times the radius of the circle, which is of course half the diameter. This is in full agreement with the 25 root2 calculated above for a 50 unit diameter circle.
Check Binary tree isomorphism?
//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; }
What is the rational number between root2 and root3?
Write three rational numbers between root2 root3 ?
What is the value of root2?
Root 2 or 2^(1/2) is an irrational number. It is approximately 1.414214
How do you show that 2 plus root2 is an irrational number?
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.
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 set of irrational numbers closed under division?
No, it is not. Root2 and root 8 are each irrational. Root8 / root2 =2. 2 is not a member of the set.
Write a fortran code to calculate the root of quadratic equation?
implicit double precision(a-h,o-z) write(*,*) "please provide the a,b,c coeff" read(*,*) A,B,C D=B*B-4*A*C if(D.GT.0) then root1=(-B/(2*A))+(SQRT(D))/(2*A) root2=(-B/(2*A))-(SQRT(D))/(2*A) write(*,*) root1,root2 elseif(D.EQ.0) then root1=(-B/(2*A)) root2=root1 write(*,*) root1,root2 else root1=(-B/(2*A))+(SQRT(-D))/(2*A) root2=(-B/(2*A))-(SQRT(-D))/(2*A) a=(root1+root2)/2 b=(root1-root2)/2 write(*,*) 'realpartroot=',a, 'complexpartroot=',b endif stop END
What is the square root of 19905.25?
141.08596670115706443962956239886 is the root2 of 19905.25
Do irrational numbers contain the number zero?
Yes indeed. There are infinitely many 0 is Pi and others too root2 etc etc
What is 5m squared root2 +2 in algebraic expression?
6.25
How do you create a program in C that ask the user for two numbers and then displays the square root of both numbers and then multiplies and displays the results of the two square roots?
#include <stdio.h> #include <math.h> /* bear in mind that this has no idiot-proofing whatsoever. For example, it doesn't check to see if the numbers entered are positive (and sqrt doesn't handle imaginary numbers) */ int main(int argc, char *argv[]){ float val1, val2, root1, root2; printf("Gimme a number: "); scanf("%f", &val1); printf("Gimme another number: "); scanf("%f", &val2); root1 = sqrt(val1); root2 = sqrt(val2); printf("The square root of %f is %f.\n", val1, root1); printf("The square root of %f is %f.\n", val2, root2); printf("The product of %f and %f is %f\n", root1, root2, root1 * root2); return 0; }
Why are root2 root3 irrational numbers?
Because they cannot be expressed as ratios of integers.
