Question

write the formula for finding irrational number between a and b​

Answer 1

Answer:

We would need to prove that between any two rational numbers, there is an irrational number.

First and foremost, the existence of irrational numbers greater than 1 has already been established (e.g., the positive real value of the square root of 2, the base of the natural logarithm.) the reciprocal of an irrational number is irrational (otherwise it would contradict the closure of the rationals under multiplication). We therefore conclude the existence of irrational numbers between 0 and 1.

Let us set i as an irrational number such that

0<i<1

Let us set p and q as two nonnegative rational numbers,, such that

p<q

. We need a bijection mapping 0 to p and 1 to q

y=(q−p)x+p

is such a bijection.

Let us apply this bijection step by step. Multiply the inequality above by q−p and we get

0<(q−p)i<q−p

Because p<q , q−p is strictly positive, and multiplication preserves the order.

Then we add p (addition by rational numbers preserves order).

p<(q−p)i+p<q

Implicit in the closure of the rationals under multiplication is that the product of an nonzero rational and an irrational is irrational, so (q-p)i is irrational. Implicit in the closure of the rationals under addition is that the sum of a rational and an uirrational is irrational, so (q−p)i+p is irrational. We have thus established that between any two positive rational numbers, there is an irrational number. Specifically, we have proven that (q−p)i+p is between p and q if i is between 0 and 1 .

Let us set m as a rational number. Without loss of generality, we can assume m is positive. then subtract m from the inequality above.

p−m<(q−p)i+p−m<q−m

Order is preserved under subtraction. With large enough m , p−m is a negative rational number. q−m is of course rational. As we have already established that (q−p)i+p is irrational, (q−p)i+p−m must be irrational as well.

QED

Step-by-step explanation:

hope this works

Answer 2

Step-by-step explanation:

ad < bc and now find a non-square integer p such that (ad)^2 < p < (bc)^2. Now sqrt(p)/bd is an irrational number between a/b and c/d.

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