2 proof questions help

[D1Kuta]

Hi, would greatly appreciate some help or even just some tips in answering these two questions. Not sure which proof is required to complete each of these questions and how to answer it. Very confused?

1) Suppose that m and n are positive integers m>n. Prove that the square root of m^2 - n^2 cannot be the same as m-n.

2) Suppose that b is a rational number and x^2-2bx+c=0 has two distinct real solutions. Prove that one solution is irrational if and only if the other solution is irrational.

 

[MarkFL]

Hello, and welcome to FMH!

1.) Let's state:

[MATH]\sqrt{m^2-n^2}\ne m-n[/MATH]
Since \(m\) and \(n\) are positive integers where \(m>n\), we know both sides of the inequation are positive, and so we may square both sides...what do you get when you do?

An alternate approach would be to factor the radicand on the LHS as the difference of squares, and then divide through by a common factor.

 

[Steven G]

Given m>n.
The only expression you can take the square root to get m-n is (m-n)²
So you want (m-n)² = m²-n². Under what conditions will this be true?

 

[pka]

Hi, would greatly appreciate some help or even just some tips in answering these two questions. Not sure which proof is required to complete each of these questions and how to answer it. Very confused.

1) Suppose that m and n are positive integers m>n. Prove that the square root of m^2 - n^2 cannot be the same as m-n.

2) Suppose that b is a rational number and x^2-2bx+c=0 has two distinct real solutions. Prove that one solution is irrational if and only if the other solution is irrational.

1) Suppose that \(\displaystyle \begin{align*}\sqrt{m^2-n^2}&=m-n \\m^2-n^2&=m^2-2mn+n^2\\2mn&=0 \end{align*}\)
What does that tell you about \(\displaystyle m\text{ and }n~?\) Is that a contradiction?

2) The roots of \(\displaystyle x^2-2bx+c=0\) are \(\displaystyle x=\dfrac{2b\pm\sqrt{4b^2-4c}}{2}\text{ implies }x=b\pm\sqrt{b^2-c}.\)
To have two real roots we must have \(\displaystyle b^2-c>0\). Can you explain why?
Do you know that \(\displaystyle \sqrt{~x~}\) is irrational if \(\displaystyle x\) is not a square?

 

[D1Kuta]

Greatly appreciate your help. For 1) It is a contradiction because m and n are supposed to be positive integers but they cannot be positive integers if 2mn=0 so they are neutral integers which contradicts m>n.

However I am still very confused about number 2. We must have two real roots because the discriminant will intersect the x-axis twice. However I am still confused regarding the irrational number and how I would be able to prove this.

 

[MarkFL]

Suppose there is a rational root \(r_r\) and an irrational root \(r_i\)...then we may state:

[MATH]x^2-2bx+c=\left(x-r_r\right)\left(x-r_i\right)[/MATH]
Expand the RHS, and compare the coefficients of the linear terms...what conclusion do you draw?

 

[Steven G]

Greatly appreciate your help. For 1) It is a contradiction because m and n are supposed to be positive integers but they cannot be positive integers if 2mn=0 so they are neutral integers which contradicts m>n.

No, you have the wrong contradiction. That is your proof is not valid by any means. I assume by a neutral integer you mean 0, am I correct?
If 2mn=0 it does NOT imply m and n are both 0. After all if m= 5 and n=0, then 2mn = 0. What you can conclude from 2mn=0 is that m or n is 0. Since m and n are both positive integers neither m nor n can be 0. That is your contradiction!

 

[JeffM]

Greatly appreciate your help. For 1) It is a contradiction because m and n are supposed to be positive integers but they cannot be positive integers if 2mn=0 so they are neutral integers which contradicts m>n.

However I am still very confused about number 2. We must have two real roots because the discriminant will intersect the x-axis twice. However I am still confused regarding the irrational number and how I would be able to prove this.

Mark's answer is certainly the most direct, but it may not be quite so intuitive as this:

[MATH]d = b^2 - 4c.[/MATH]
[MATH]\text {By hypothesis, } b \in \mathbb Q,\ \dfrac{-\ b + \sqrt{d}}{2} \in \mathbb R,\ \text { and } \dfrac{-\ b - \sqrt{d}}{2} \in \mathbb R.[/MATH]
[MATH]\sqrt{d} \in \mathbb Q \implies \dfrac{-\ b + \sqrt{d}}{2} \in \mathbb Q \text { and } \dfrac{-\ b - \sqrt{d}}{2} \in \mathbb Q[/MATH]
[MATH]\text {because the rational numbers are closed under addition and subtraction.}[/MATH]
[MATH]\text {HOWEVER, a rational plus or minus an irrational is necessarily irrational.} [/MATH]
[MATH]\therefore \sqrt{d} \not \in \mathbb Q \implies \dfrac{-\ b + \sqrt{d}}{2} \not \in \mathbb Q \text { and } \dfrac{-\ b - \sqrt{d}}{2} \not \in \mathbb Q[/MATH]
[MATH]\text {In short, }\dfrac{-\ b + \sqrt{d}}{2} \in \mathbb Q \iff \dfrac{-\ b - \sqrt{d}}{2} \in \mathbb Q.[/MATH]