1. True or False? Give your reason. There is a largest rational number less than 1.
Solution FALSE. For any rational number x less than 1, (1+x)/2 is another rational number less than 1 which is greater than x.
2. True or False? Give your reason. (-\sqrt(3)+1)/(\sqrt(3)-1) is a rational number.
Solution This number is in fact -1, which is rational.
3. Verify that x=1+i\sqrt(3) satisfies the equation x^2-2x+4=0.
Solution x^2=1+2i\sqrt(3)+3i^2=-2+2i\sqrt(3), -2x=-2-2i\sqrt(3), so x^2-2x+4=0.
4. Prove that the cube root of 5 is irrational.
Solution Suppose that the cube root of 5 does equal p/q, for integers p and q. Cubing this equation we have 5=p^3/q^3 or 5*q^3=p^3. Now consider how many 5's are in the prime factorization of the right side. We must a multiple of 3 number of 5's. The left side must have one more than a multiple of three number of fives, and this contradicts the fundamental theorem of arithmetic.
5. Suppose that w is complex number such that w^6=-1. Can you find w^15?
Solution w^15=w^6*w^6*w^3=(-1)*(-1)*w^3=w^3. Also w^3 squared is -1, so w^3 is either i or -i.
6. Let p(x)= x^3+10*x^2+x. Show that there are three distinct real values values of x such that p(x)=0.
Solution p(x)=x(x^2+10x+1), so x=0 solves p(x)=0. There are two other real solutions. To see this either write them out using the quadratic formula, (-10\pm \sqrt(100-4))/2, or use the intermediate value theorem x^2+10x+1 is 1 at x=-10, -8 at x=-1 (so there is zero between -10 and -1), and is 1 at x=0 (so there is zero between -1 and 0).
1. Let x=(1/2+i)/(2/3-i). Find a rational number r such that x-ir is a real number.
Solution x=(1/2+i)*(2/3+i)/(13/9)=(-2/3+7/6i)/(13/9) so r=7/6*9/13.
2. True or False? Give your reason. The cube root of a positive integer m is irrational unless m is perfect cube.
Solution TRUE. Could m=p^3/q^3? Then we would have m*q^3=p^3. If m is not a perfect cube, then some prime r in the prime factorization of m occurs either 1 mod 3 times or 2 mod 3 times. This contradicts m*q^3=p^3. (see prob 1-4).
3. Prove that the number x=(\sqrt(3)-\sqrt(5))/4 is irrational.
Solution 4x=(\sqrt(3)-\sqrt(5)) so 16x^2=3+5-2*\sqrt(15), or \sqrt(15)=-8x^2+4. If x were rational then the right side of this equation would be rational, so \sqrt(15) would be rational which we know is wrong. So x cannot be rational.
4. Suppose that x^6=1. Find rational numbers a,b,c,d,e,f such that x^1000=a+b*x+c*x^2+d*x^3+e*x^4+f*x^5.
Solution 1000=6*166+4, so x^1000=x^4 and we can take a=b=c=d=f=0, d=1.
5. Show that there is some number x between 1 and 3 such that x^7+x-3=0.
SolutionIf x=0 the polynomial is -3, if x=2 the polynomial is 128+2-3>0, so the IVT applies.
1. True or False? Give your reason. There is positive real number x such that x^3+3x^2-100*x-4=0
Solution Let p(x)=x^3+3x^2-100*x-4. Since p(0)=-4 and p(1000)>0, the IVT implies there is some x between 0 and 1000 such that p(x)=0. TRUE.
2. Give an example of two irrational numbers x,y such that x/y is rational.
Solution x=\sqrt(3)=y, x/y=1.
3. True or false: If x and y are any two real numbers, then there is some rational number r between x and y.
Solution TRUE. We did this in class. Find the first place where the decimal representations of x and y disagree, and choose a digit for r in between. Then add all zeros.
4. The infinite decimal which consists of the consecutive integers 0.1234567891011121314... is an irrational number.
Solution TRUE. This infinite decimal does not terminate. Could it be repeating? Suppose the repeating part has some length, say 17. Eventually we come to the integer 1111111111111111 which appears in our given number, so our repeating part would have to be 1111111111111111. But the next number in our expansion would be 11111111111111112, so it does not repeat!
5. Let z=(2-i)/i. Find the a+bi form, a and b rational, for z^2.
Solution z^2=(2-i)^2/i^2=(4-4i+i^2)/(-1)=-3+4i, so a=-3, b=4.
6. Let z=1-2*\sqrt(5). Find the a+b\sqrt(5) form, a and b rational, for z^2.
Solution z^2=1-4*\sqrt(5)+20=21-4*\sqrt(5), so a=21, b=-4.
1. Prove that the real number (2-\sqrt(2))/\sqrt(2) is irrational.
Solution x=(2-\sqrt(2))/\sqrt(2)=\sqrt(2)-1, so x+1=\sqrt(2). If x were rational, the left side would be rational, and then the right would be rational. But we know that \sqrt(2) is irrational.
2. Let z be the complex number z=(1+i)/\sqrt(2). Prove that z^2=i.
3.Suppose that x is a complex number such that x^2+x+1=0. Prove that x^3=1.
4. True or False: The average of two irrational numbers is always irrational.
Solution FALSE. Try \sqrt(2) and -\sqrt(2) whose average is 0.
5. True or False: The square of an irrational number is rational.
Solution FALSE The square of the cube root of 2, is the cube root of 4 which is irrational. Anotehr exmaple is the square of (\sqrt(2)+\sqrt(3)) is 5+2*\sqrt(6) which is irrational.
6. Prove that the set of algebraic numbers is countable.
Solution First we list all of the polynomials with rational coefficients: degree one first row, degree two second row etc, then use our diagonalization method to make a countable list of all of these polynomials. Each polynomial of degree n has n zeros, again list them, and use another diagonal argument.