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Here is the basic outline of number groups: Natural Numbers - Positive whole numbers. 1,2,3,4,... Whole Numbers - The natural numbers and 0. 0,1,2,3,... Integers - Whole numbers and their negatives. -3,-2,-1,0,1,2,... Rational Numbers - Numbers that can be written in the form p/q, where both p and q are integers. 1/2,2/3,56/91,... In the above, all natural numbers are whole numbers, all whole numbers are integers, and all integers are rational numbers. Irrational Numbers - Numbers that can not be expressed in the form p/q, where both p and q are integers. This is typically the most challenging concept for students to understand. Some examples are 'pi' and 'e', as well as the square roots of numbers that are not perfect squares. sqrt(2),'pi'/2,e^'pi',... (Note in the above that although 'pi'/2 is written as a ratio of two numbers, it is not rational because 'pi' is not an integer.) Real Numbers - The combination of rational and irrational numbers. 0.3345,sqrt(e),17,... Imaginary Numbers - Don't be fooled by the name: imaginary numbers are just as 'real' as real numbers. Consider sqrt(4). You know this is 2. Sqrt(9). 3. Sqrt(-1). Did I stump you? Sqrt(-1) is defined as equally 'i'. Thus, when you see 'i', think sqrt(-1). Sqrt(-x) is equal to i*sqrt(x). Simple enough? Any number of the form a+b*i is an imaginary number, where 'a' and 'b' are both real numbers. The value 'a' is often denoted Re(z), and 'b' is often denoted Im(z), where z=a+b*i. Note that imaginary numbers can have a=0, however they can not have b=0. 2+3i,9i,sqrt(7)-'pi'*i,... Pure Imaginary Numbers - Simple enough, imaginary numbers where a=Re(z)=0. Thus they have no 'real' component. 2i,sqrt(19)i,'e'*i,... Complex Numbers - All numbers. As simple as that. The combination of all imaginary and real numbers. 23,sqrt(34320),'pi'^(23)/e-7*i,... Now to directly answer your questions: 1 - Integers, as we have seen above, are not the same as whole numbers. 2 - Sqrt(25) is an integer because of the perfect square exception I noted earlier. This also applies to numbers like cube root of 64 which is 4. 3 - Explained above. 4 - Explained above. 5 - Explained above. Now a simple proof that irrational numbers exist (i.e. there are numbers that can't be written as p/q where both p and q are integers that share no factors): Suppose sqrt(m)=p/q is rational and that p/q is reduced completely (p and q share no integer factors). m is an integer that is not a perfect square. Then (p^2)/(q^2)=m Thus p^2 = m * q^2 Because m and q^2 are both integers, this shows that m must be a factor of p^2 Therefore, sqrt(m) must be a factor of p However, sqrt(m) is not an integer because m is not a perfect square, so if sqrt(m) divides p, m must also divide p Let's then say that p = m * S where S is the integer p/m p^2 = m^2 * S^2 Plugging this in the earlier relation we get m^2 * S^2 = m * q^2 Therefore, m * S^2 = q^2 Using the same argument as before, we see that m is also a factor of q. Because both p and q share the factor m, they could not possibly have been in simplest form, thus proving that sqrt(m) can not equal p/q.