Proofs: sum, prod. of two rtnl's are rtnl; sqrt[3] is irrtnl

[Guest]

Hi guys, new here, and desperate. It is my first week of calc and everything has made sense up until now. I have the following questions and I'm simply not sure where to get started! I think I might be able to solve it once I'm off and running but I don't know about getting them started, thanks for anyone who can help!

1. Prove that the sum of two rational numbers is a rational number

2. Prove that the product of two rational numbers is a rational number

3. Prove that the square root of 3 is an irrational number

I have the rest of them, but these 3 are giving me a particularly hard time, I'll be back for more help soon I'm sure.

 

[soroban]

Re: Proofs

Hello, Buschmaster!

Unusual problems for a Calculus class.

I must assume you have some basic definitions and theorems to work with.

For example: Integers are closed under addition.
\(\displaystyle \;\;\)That is, the sum of any two integers is an integer.
Similarly, integers are closed under subtraction and multiplication.

Also, we should have a definiton of a rational number.
\(\displaystyle \;\;\)Something like: a ratio of two integers \(\displaystyle \frac{a}{b}\), where \(\displaystyle b\,\neq\,0.\)
In baby-talk: it is "an integer over an integer".

1. Prove that the sum of two rational numbers is a rational number.

A proof might go something like this . . .

Given two rational numbers: \(\displaystyle \frac{a}{b}\) and \(\displaystyle \frac{c}{d}\)

Their sum (by definition of addition) is: \(\displaystyle \L\,\frac{a}{b}\,+\,\frac{c}{d}\:=\:\frac{ad\,+\,bc}{bd}\)

Since integers are closed under addition and multiplication,
\(\displaystyle \;\;ad\,+\,bc\) is an integer and \(\displaystyle bd\) is an integer.

Therefore, \(\displaystyle \L\frac{ad\,+\,bc}{bd}\) is a rational number.

 

[Guest]

Wow that makes sense! Thanks!

This is technically "review" I think but it's for my Calculus class and I guess it has a Calculus spin to it? yeah.. .that's what I hear. I was actually expressing mine in p/q, but this is easily converted! It's great! Also, like I thought, I was not even close to on the right track. I solved the next two, we'll see if tomorrow's assignment conquers me again, I have it in the morning, so you may hear from me again before too long.