How do i prove that and irrational number / by an irrational number gives an irrational answer when not 1?

Question

Proof of irrational and rational numbers. Are answers rational or irrational when have two numbers and can only divide, subtract, add or multiply them? How do i prove all instances?

Answer 1

You can't prove it, because it's not true. If the source numbers are both rational (and disallowing division by 0), then the answer will always be rational under these operations. If one is rational and the other irrational, the result will always be irrational except in the cases of multiplying 0 by an irrational or dividing 0 by an irrational. If both are irrational, the result may be rational or irrational.

It's a worthwhile exercise to try and prove the first two parts of this. For the third part (both numbers irrational), just find examples of each.

For the first part (both numbers rational), just write the numbers as a/b and c/d where a, b, c, d are integers and b, d are not 0. Get expressions for a/b + c/d, a/b - c/d, a/b * c/d and a/b / c/d in terms of a, b, c, d with only integer expressions on top and bottom (and bottom is not zero); then you know the answer is rational. For instance, a/b + c/d = (ad + bc) / bd, and ad + bc and bd are both integers; also bd is not zero since b and d are not 0. So a/b + c/d is rational.

For the second part, use reverse operations and the result of the first part. For instance, if x is rational and y is irrational, x + y = z must be irrational, since if it was rational we would have z - x = y, so two rational numbers would subtract to give an irrational number, which you should have proved in the first part cannot happen.

Answer 2

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 3

This statement isn't always true. (2*pi) / pi = 2.

sqrt(8) / sqrt(2) = 2.

I don't think there's a general principle to prove here.

For any irrational number, you can find another irrational number such that the two will add to 7.

An irrational number plus a rational number might be something to work with.

Answer 4

Emily, As I comprehend it, listed here are some definitions: Integers: ....., -3, -2, -a million, 0, a million, 2, 3, 4, ..... total Numbers: 0, a million, 2, 3, 4, ..... organic Numbers: a million, 2, 3, 4, 5, ..... So, "2" is an integer, an entire selection, and a organic selection. 0 is an integer, an entire selection, yet not a organic selection "-3," or "unfavourable 3" is an integer, and neither an entire selection nor a organic selection. solutions on your questions: a million-Integers comprise total numbers, yet additionally comprise unfavourable numbers 2-The sq. root of 25 is an integer; your reasoning is actual 3- A rational selection is any selection which would be written as a fragment in lowest words. Examples: a) 3/4, -2/3, b) terminating decimals, e.g., 0.25, which could be written as 25/a hundred or a million/4 c) integers, e.g., 572, which could be written as 572/a million, or -6, which could be written as -6/a million d) 0.11111111111111.....(or the different repeating decimal), which could be rewritten as a million/9 e) sq. roots of appropriate squares, ?sixty 4, ?9, and so on. -An irrational selection is any selection that may not rational. some examples are "pi," and sq. roots of numbers that are actually not appropriate squares. The sq. root of any best selection, as an occasion, is irrational. 4)a real selection is any rational or irrational selection. Examples are integers, organic numbers, total numbers, any fraction, any decimal, any sq. root of a favorable selection, "pi," best numbers, and so on. Any selection which would be represented on a selection line is a real selection. Examples of numbers that are actually not actual are ?(-a million), and log(-a million). 5)See above good success!

Answer 5

You only need to disprove one. Take the irrational numbers e (the base of natural logarithms) and 2*e. One is certainly derived from the other, but 2*e/e = 2, not 1.

Answer 6

take examples.
for example
(2^1/2)/2^1/2 = 1
but try dividing by any no
it is irrational.
so subtract the same thing you get 0 so you can prove like that

Answer 7

do a google search for quantifiers....that is the way you have to solve this....i just dont have the time right now to answer fully, sorry