I am preparing for pre-medical Exam. It requires basic mathematics to higher mathematics for Physics and Chemistry.
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There is a whole mathematical field called
https://en.m.wikipedia.org/wiki/Order_theory
I shall let you make what you can from that introductory article. Not attempting to be rigorous, we might say
\(\displaystyle a > b \iff (a - b) > 0.\)
That is, we say 3 is greater than 2 because we can take 2 objects from 3 objects and still have 1 or more objects left. That is a primitive intuition about what we mean by "greater than." We extend the concept from non-negative integers to real numbers a and b.
\(\displaystyle a > b \iff (a - b) > 0.\)
Now from this we can devise certain additional criteria, such as
\(\displaystyle a > b \text { and } a > 0 \implies 1 > \dfrac{a}{b}.\)
If we divide 3 into 1, we get a result less than 1.
\(\displaystyle c > 0 \text { and } a > b \implies ac > bc.\)
The product of 5 and 3 is 15, which is greater than 10, the product of 5 and 2.
Good to here?
At this point, say that we do not know what is the order relationship between
\(\displaystyle 30\) and \(\displaystyle 30\sqrt{2}.\)
Now a square root is non-negative by definition, but it can be zero. Let's assume
\(\displaystyle \sqrt{2} = 0 \implies \sqrt{2} * \sqrt{2} = \sqrt{2} * 0 \implies 2 = 0.\)
Well, 2 is not zero so that assumption was absurd.
\(\displaystyle \therefore \sqrt{2} > 0.\)
Actually this is a general theorem: \(\displaystyle x > 0 \iff \sqrt{x} > 0.\)
Maybe \(\displaystyle 1 > \sqrt{2} > 0 \implies \sqrt{2} * 1 > \sqrt{2} * \sqrt{2} \implies\)
\(\displaystyle \sqrt{2} > 2 \implies 1 > 2 \ \because \ 1 > \sqrt{2} \text { by assumption.}\)
And 1 is not greater than 2 so that assumption also was absurd.
But maybe the square root of 2 equals 1 so let's assume it to be so.
\(\displaystyle 1 = \sqrt{2} \implies 1 * 1 = \sqrt{2} * \sqrt{2} \implies 1 = 2.\)
More nonsense. So 1 is neither greater than nor equal to the square root of 2. Therefore, 1 is less than the square root of 2.
In fact, this is another general theorem \(\displaystyle x > 1 \iff \sqrt{x}> 1.\)
But 2 is greater than 1 so our conclusion is absurd, which means our assumption was false.
\(\displaystyle \therefore \sqrt{2} > 1 \implies 30\sqrt{2} > 30 * 1 = 30.\)
Most of the ideas about the real numbers are simple extensions of ideas that you already know about the natural numbers. There are rigorous developments of them studied in fields like foundations of mathematics, but that requires a course, not a tutoring site.