prove that each number is irrational?

Question

prove that each number is irrational
a) square root of 3
b) 2 square root 2
c) square root 2 + 1 (note that the 1 is not in square root)

Answer 1

Irrational numbers are those which cant be expressed as terminating decimals

SQRT(3) and SQRT(2) both can't be expressed as terminating decimals.
=> they are irrational.

2 * SQRT(2) = Integer * Irrational Number = Irrational Number

SQRT(2) + 1 = Irrational Number + Integer = Irrational Number

Answer 2

Here is the proof for square root of 3 not being rational. This can also be used to prove the non-rationality of any prime number.

1. Assume that square root of 3 is rational.
2. There exists a fraction of integers that equals the squareroot of 3 (definition of a ration number)
3. Let a and b be integers when if divided equals the squareroot of 3 (From #2)
4. Let a / b be a fraction in simplist form where a / b = sqrt of 3 (From #3)
5. a and b have no common factors (From Algebra & From #4)
6. (a / b)^2 = 3 (From Algebra & #5)
7. a^2 / b^2 = 3 / 1 (From Algebra & #6)
8. If a and b have no common factors then a^2 and b^2 also have no common factors (From Algebra)
9. a^2 / b^2 is in simplist form (From Algebra, #5 & #8)
10. Two equal fractions in simplist form have the same numerator and the same denominator (From Algebra)
11. a^2 = 3 and b^2 = 1 (From #8 & #10)
12. 3 is a prime number and therefore cannot have two of the same integer factors. (From Algebra)
13. a is not an integer (From Algebra, #11 & #12)

Since #2 contridicts #13, and I’ve only made a single assumption (#1), that assumption must be false.

Answer 3

For all instances: Given two rational numbers, to add, subtract, multiply, and divide (except to divide by 0) will only produce a rational number. To prove this, consider two rational numbers, a/b and c/d, where a, b, c, d are integers with b and d not 0, and perform the appropriate operation, showing that the result is rational. Given two irrational numbers, you can find cases in which adding, or subtracting, or multiplying, or dividing will produce either a rational number or an irrational number (but not necessarily the same for every operation). You show this with specific numbers. Given a rational number and an irrational number, adding, subtracting, multiplying (except by 0), and dividing (except by 0) will only produce an irrational number. To prove this, let one number be rational, of the form a/b, where a and b are integers with b not 0, and let the other number be irrational, t, and assume that the result is a rational number c/d, where c and d are integers, with c not 0. Then consider the equation (a/b) & t = c/d or t & (a/b) = c/d, where "&" is your operation (either +, -, *, or /). Then solve for the irrational number t, showing that the result is an operation involving the rational numbers a/b and c/d, which must be rational.

Answer 4

You need to proceed by contradiction. Assume that the numbers are rational (i.e. = q/n for some q, n in Z with gcd(q,n) = 1) then show that this cannot be the case (there is a common factor, therefore gcd(q,n) is not 1).

Answer 5

a)3^1/2 cannot be expressed as a fraction
b)I presume you mean 2(2^1/2)=8^1/2 which also cannot be expressed as a fraction.
c)2^1/2+1 also cannot be expressed as a fraction.

Answer 6

Can't prove it, so they're rational.

Answer 7

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