# What Are X And Y In Math latest 2023

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## Pythagoras Theorem and Irrational Numbers

It is interesting how the Pythagorean theorem helps to identify the location of an irrational number on the number line. Consider a number x which is a rational number but not a perfect square. It follows that the square root of x must be irrational, that is, a nonterminal, nonrecurring decimal number. Now our interest is in determining where this is on the number line. To do this, consider a right-angled triangle whose base is (x-1)/2 and the hypotenuse is (x+1)/2. What would be the height of this right triangle, i.e. the other arm of the right triangle? Let’s assume it’s there. The Pythagorean theorem tells us that the sum of the squares of the arms of a right triangle is equal to the square of the hypotenuse. Thus, in the triangle that we have just considered, we can write:

[(x – 1)/2]^2 + y^2 = [(x + 1)/2]^2

Where y^2 = [(x + 1)/2]^2 – [(x – 1)/2]^2

= [(x^2 + 2x + 1) / 4] – [(x^2 – 2x + 1) / 4]

= [(x^2 + 2x + 1) – (x^2 – 2x + 1)] / 4

= [x^2 + 2x + 1 – x^2 + 2x – 1)] / 4

= 4x ​​/ 4

=x

i.e. y^2 = x âŸ¹ y = √x

This is exactly what we were looking for, the square root of x which is irrational. Now the length of the third arm of the triangle we have constructed can be marked on the number line using a compass.

So if you’re looking for the size of √x, here’s how we do it. Mark a point A. Mark B such that AB = x units. Label C such that BC = 1 unit. In other words, AC = x+1. Bisector AC. If D is the bisection point of AC, AD = DC = (x+1)/2. Now what would be the length of DB? Since DC = (x+1)/2 and BC = 1, DB = DC – BC = [(x+1)/2] – 1. Either (x+1-2)/2 or (x-1)/2.

Now let’s construct the triangle. Draw a line perpendicular to AC in B. From D, intersect the vertical line in E such that DE = AD. We now have a right triangle whose base is (x-1)/2 and the hypotenuse is (x+1)/2. Can you see what the measure of BE will be? Of course, as we showed above, it will be √x. You can now transfer this length of BE to the number line, using the compass.

Try to represent √5, √7, √11, √6.8 and √9.5 on the number line. Each of them should barely take a few minutes or less.

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