Mystery Math: Hard Made Easy In Math

HARD MADE EASY

Lets consider a simple case which shows how our understanding of Rational Numbers is based (mainly) on our knowledge of the integers.

(Generalized) Question 1: If $x$ and $y$ are rational numbers that are not squares of other rational numbers, show that $\sqrt{x} + \sqrt{y}$ is not rational.

This seems hard to approach, because there seem to be a lot of unknowns, and we could be unfamiliar with what rational numbers are. Let’s try the following:

Question 2: If $x$ and $y$ are integers that are not squares of other integers, show that $\sqrt{x} + \sqrt{y}$ is not an integer.

Now, this is more friendly, but still a little tricky. In the spirit of this post, let’s make it even easier.

Question 3: If $x$ is an integer that is not the square of another integer, show that $\sqrt{x}$ is not an integer.

Now this is almost silly, and seems to be just a play on words. The proper way to prove this statement that doesn’t make my head run around in circles, is to show the contrapositive – To show that $P \Rightarrow Q$ , it is equivalent to show the contrapositive which is $\lnot Q \Rightarrow \lnot P$ .

Proof: If $\sqrt{x}$ is an integer $n$ , then $x=n^2$ , which is the square of the integer $n$ . $_\square$

Now, what is the generalized version?

(Generalized) Question 4: If $x$ is a rational number that is not the square of another rational number, show that $\sqrt{x}$ is not a rational number.

Proof: Test Yourself 1. $_\square$

Corollary: If $y$ is an integer such that $\sqrt{y}$ is rational, then $y$ must be the square of an integer.

Proof: From Question 4, we know that y must be the square of a rational $\frac {p}{q}$ with $\gcd(p,q)=1$ . Hence $y = \frac {p^2} {q^2}$ , and the only way for this to be an integer is $q=1$ , so $y=p^2$ . $_\square$

Now, back to question 2.

Proof: Suppose $\sqrt{x} + \sqrt{y} = n$ is an integer. Consider $\sqrt{x} = n - \sqrt{y}$ . Squaring both sides, we obtain that $x = n^2 - 2n\sqrt{y} + y$ , so this means that $\sqrt{y} = \frac {n^2 - x - y}{2n}$ is rational. By the corollary above, $y$ must be an square. Similarly, $x$ must be an square. Hence we are done. $_\square$

And finally, back to question 1.

Proof: Proof by contradiction. Suppose $x = \frac {p_x} {q_x}, y = \frac {p_y} {q_y}$ such that $\sqrt{x} + \sqrt{y} = \frac {p_z} {q_z}$ , where $p_x, p_y, p_z, q_x, q_y, q_z$ are all integers. Then, $\sqrt{ q_y ^2 q_z ^2 p_x q_x} + \sqrt{q_x ^2 q_z ^2 p_y q_y} = q_x q_y p_z$ . This contradicts question 2! $_\square$

Corollary: If $x$ and $y$ are rational numbers such that $\sqrt{x} + \sqrt{y}$ is rational, then $\sqrt{x}$ and $\sqrt{y}$ are both rational.

The slightly surprising result, is that in simplifying Question 1 to Question 2, we ended up using Question 2 to prove Question 1. In fact, as is often the case with rational numbers, it is sufficient to consider the integer case, and then clear out denominators by multiplying throughout.

Get A Try & Test Yourself

1. Complete the proof of Question 4.

2. Prove that if $x$ , $y$ and $z$ are rational numbers such that $\sqrt{x} + \sqrt{y} + \sqrt{z}$ is rational, then $\sqrt{x}$ , $\sqrt{y}$ and $\sqrt{z}$ are all rational. How many terms can you show this for?

3. How many ordered triples of integers $(x, y, z)$ are there such that $\sqrt{x} + \sqrt{y} + \sqrt{z} = \sqrt{2000}$ ?

4. (**) Prove that if $a, b, c, x, y, z$ are rational numbers such that $\sqrt{a} + \sqrt{b} + \sqrt{c} + \sqrt{x} + \sqrt{y} + \sqrt{z}$ is rational, then $\sqrt{a}, \sqrt{b}, \sqrt{c}, \sqrt{x}, \sqrt{y}$ and $\sqrt{z}$ are all rational. Note: This is extremely hard, and not approachable by the methods discussed in this post.

Sunday 24 February 2013

Hard Made Easy In Math

Get A Try & Test Yourself

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