Hard Made Easy II

by Calvin on February 6, 2013

This is a continuation of the Hard Made Easy series, in which we feature certain questions where simplifying the problem actually helps us approach it.

Below, is 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.

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.

From → General, Level 2

7 Comments

Trever permalink

Many mathematicians do the same thing. If you can’t solve a hard problem, work up by solving easier problems. Solving similar problems works the same way.

I am currently reading, “The Man Who Loved Only Numbers” tells a great story of shared cooperation in mathematics and solving a problem may just be contributing something small and maybe someone else can break the problem down further.

Reply
- Calvin permalink
  
  Indeed. One important strategy of problem solving is to break down a problem into simpler versions and understand them.
  
  This series of posts is choosing to focus on particular problems which are made easier when we make simplifying assumptions, and are later used to solve the problem again. Common examples include translating the domain (see previous post), dealing with integers as opposed to rationals (this post). If you have suggestions about other procedures, let me know
  
  Reply
rishabh gupta permalink

sir it is very good to all of us thank you sir

Reply
hemang sarkar permalink

awesome stuff sir. i love the thought process.

Reply
Carlos M. Gozun permalink

I really appreciate your enthusiasm in teaching others. I am a math teacher as MathEd. I do not understand pure math, please send me some notes which I can be used so I may understand puremath later. tnx

Reply
Ed Joshua Mañalac permalink

I have a questions sir, Right if the statement says, if….isn’t it to be prove two ways? 1 forward the other backward?

Reply
- Calvin permalink
  
  No. You are thinking of the phrase “If and only if”.
  
  Reply

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