# 2 Proof Questions

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Dec 30th 2006, 08:49 PM
ruprotein
2 Proof Questions
Anyone got a step by step approach to solve this, it aint that hard just checkin my work

Prove (There exists a real number x)(For all real numbers y)(4+6y-y^2<x)

and the 2nd one is

Prove sqrt(2) +sqrt(5) is irrational

the way i did this was by contradiction and i proved that sqrt(10) is irrational because via contradiction i said sqrt(10) i srational then for integers m, n swrt(10) = m/n and m and n are both not even integers, but as i worked it i showed that m and n were both even therefore it cantbe rational, was my approach correct?
• Dec 30th 2006, 09:19 PM
malaygoel
Quote:

Originally Posted by ruprotein
Anyone got a step by step approach to solve this, it aint that hard just checkin my work

Prove (There exists a real number x)(For all real numbers y)(4+6y-y^2<x)

and the 2nd one is

Prove sqrt(2) +sqrt(5) is irrational

the way i did this was by contradiction and i proved that sqrt(10) is irrational because via contradiction i said sqrt(10) i srational then for integers m, n swrt(10) = m/n and m and n are both not even integers, but as i worked it i showed that m and n were both even therefore it cantbe rational, was my approach correct?

for the first, $\displaystyle 4+6y-y^2$ is a real number for every real number y and you can always a real number greater than a given real number.
hence there exists infinite x for every y

for the second, your approach is absolutely correct.

Keep Smiling
Malay
• Dec 30th 2006, 09:38 PM
ruprotein
thanks

Can you just say that for the 1st part though im sure you have to say it in more detail, i said somethinlike what u said i received 4 out of 12 pts out of it and i dunno why

Step 1: Assume y is an arbirary real number
Step 2: Then 4+6y^2-y^2 is a real number [Axiom (For all x)(x+y element of real #'s)
Step 3: Assume x is some real number... dunno how to finish it off
• Dec 30th 2006, 10:11 PM
malaygoel
Quote:

Originally Posted by ruprotein
thanks

Can you just say that for the 1st part though im sure you have to say it in more detail, i said somethinlike what u said i received 4 out of 12 pts out of it and i dunno why

Step 1: Assume y is an arbirary real number
Step 2: Then 4+6y^2-y^2 is a real number [Axiom (For all x)(x+y element of real #'s)
Step 3: Assume x is some real number... dunno how to finish it off

It may be like this
let a is a positive real number
let x= 4+6y^2-y^2 + a=sum of two real numbers.
now x is a real numbers and for any postive real value of a, it is greater than 4+6y^2-y^2

Keep Smiling
• Dec 30th 2006, 10:45 PM
CaptainBlack
Quote:

Originally Posted by malaygoel
for the first, $\displaystyle 4+6y-y^2$ is a real number for every real number y and you can always a real number greater than a given real number.
hence there exists infinite x for every y

You are supposed to show that there is a $\displaystyle x \in \mathbb{R}$ for which $\displaystyle 4+6y-y^2 < x$ for all $\displaystyle y \in \mathbb{R}$.

RonL
• Dec 30th 2006, 10:48 PM
CaptainBlack
Quote:

Originally Posted by ruprotein
Prove sqrt(2) +sqrt(5) is irrational

the way i did this was by contradiction and i proved that sqrt(10) is irrational because via contradiction i said sqrt(10) i srational then for integers m, n swrt(10) = m/n and m and n are both not even integers, but as i worked it i showed that m and n were both even therefore it cantbe rational, was my approach correct?

Why are you working with $\displaystyle \sqrt{10}$?

$\displaystyle \sqrt{2}+ \sqrt{5} \ne \sqrt{10}$

so if this approach is to work you need to show how $\displaystyle \sqrt{10}$ is related to $\displaystyle \sqrt{2}+ \sqrt{5}$.

RonL
• Dec 30th 2006, 11:02 PM
ruprotein
(sqrt(2)+sqrt(5))^.5 = 7 + 2*sqrt(2)*sqrt(5) = 7 + 2*sqrt(10)
• Dec 30th 2006, 11:36 PM
earboth
Quote:

Originally Posted by ruprotein
(sqrt(2)+sqrt(5))^.5 = 7 + 2*sqrt(2)*sqrt(5) = 7 + 2*sqrt(10)

Hello,

this transformation isn't correct:

$\displaystyle \sqrt{\left( \sqrt{2}+\sqrt{5} \right)^2}=\sqrt{ 7+2 \cdot \sqrt{10} }$

But does this really help?
• Dec 31st 2006, 12:20 AM
CaptainBlack
Quote:

Originally Posted by ruprotein
Anyone got a step by step approach to solve this, it aint that hard just checkin my work

Prove (There exists a real number x)(For all real numbers y)(4+6y-y^2<x)

Show that there exists an $\displaystyle x \in \mathbb{R}$ such that for all $\displaystyle y \in \mathbb{R}:\ \ 4+6y-y^2 < x$.

Suppose otherwise, then for any $\displaystyle X \in \mathbb{R}$ there exists a $\displaystyle k \in \mathbb{R},\ k>X$ such that:

$\displaystyle 4+6y-y^2=k$,

or:

$\displaystyle y^2-6y+(k-4)=0$

The quadratic formula can be used to find those $\displaystyle y$'s that satisfy this, and these are:

$\displaystyle y=\frac{6\pm\sqrt{36-4(k-4)}}{2}$

This has real roots only if the discriminant $\displaystyle 36-4(k-4) \ge 0$ or $\displaystyle k \le 13$.

So if $\displaystyle X=14$, there is no $\displaystyle k$ having the properties assumed, which is a contradiction, hence there exists an $\displaystyle x$ such that:

$\displaystyle 4+6y-y^2 < x$

for all $\displaystyle y$, and $\displaystyle x=14$ is one such.

RonL
• Dec 31st 2006, 06:56 AM
Soroban
Hello, ruprotein!

In #2, $\displaystyle m$ and $\displaystyle n$ are relative prime.
. . It does not involve even/odd properties.

Quote:

Prove $\displaystyle \sqrt{2} + \sqrt{5}$ is irrational.

First, we will prove that $\displaystyle \sqrt{10}$ is irrational.

Assume $\displaystyle \sqrt{10}$ is rational.
Then: $\displaystyle \sqrt{10} \:=\:\frac{a}{b}$ for relatively prime integers $\displaystyle a$ and $\displaystyle b.$
. . That is, $\displaystyle \frac{a}{b}$ is already reduced to lowest terms.

Square both sides: .$\displaystyle 10 \:=\:\frac{a^2}{b^2}\;\;\Rightarrow\;\;10b^2\:=\:a ^2$ [1]
Since $\displaystyle a^2$ is a multiple of 10, then $\displaystyle a$ is a multiple of 10: .$\displaystyle a \,=\,10p$

Substitute into [1]: .$\displaystyle 10b^2\:=\:(10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p ^2\;\;\Rightarrow\;\;b^2\:=\:10p^2$
Since $\displaystyle b^2$ is a multiple of 10, then $\displaystyle b$ is a multiple of 10: .$\displaystyle b\,=\,10q$

But this contradicts the fact that $\displaystyle a$ and $\displaystyle b$ are relatively prime.
. . Therefore, $\displaystyle \sqrt{10}$ is irrational.

Assume $\displaystyle \sqrt{2} + \sqrt{5}$ is rational.

Then: .$\displaystyle \sqrt{2} + \sqrt{5} \:=\:\frac{m}{n}$ for relatively prime integers $\displaystyle m$ and $\displaystyle n.$
. . That is, $\displaystyle \frac{m}{n}$ is already reduced to lowest terms.

Square both sides: .$\displaystyle \left(\sqrt{2} + \sqrt{5}\right)^2 \:=\:\left(\frac{m}{n}\right)^2$
. . and we have: .$\displaystyle 2 + 2\sqrt{10} + 5 \:=\:\frac{m^2}{n^2}\;\;\Rightarrow\;\;\sqrt{10} \:=\:\frac{\frac{m^2}{n^2} - 7}{2}$
Hence: .$\displaystyle \sqrt{10}\:= \:\frac{m^2-7n^2}{2n^2}$

The right side is rational. .It is the product, difference and quotient of integers.
. . But we have shown that the left side is irrational.

Hence, our assumption was incorrect.
. . Therefore: .$\displaystyle \sqrt{2} + \sqrt{5}$ is irrational.

• Dec 31st 2006, 07:31 AM
ThePerfectHacker
The set of point,
$\displaystyle 4+6y-y^2$
Has a maximum value.
(An upside down parabola).
Thus, if you choose a number above the maximum point then you are safe.
• Dec 31st 2006, 07:38 AM
Jenny20
[QUOTE=Soroban;33016]Hello, ruprotein!

In #2, $\displaystyle m$ and $\displaystyle n$ are relative prime.
. . It does not involve even/odd properties.

[size=3]
First, we will prove that $\displaystyle \sqrt{10}$ is irrational.

Assume $\displaystyle \sqrt{10}$ is rational.
Then: $\displaystyle \sqrt{10} \:=\:\frac{a}{b}$ for relatively prime integers $\displaystyle a$ and $\displaystyle b.$
. . That is, $\displaystyle \frac{a}{b}$ is already reduced to lowest terms.

Square both sides: .$\displaystyle 10 \:=\:\frac{a^2}{b^2}\;\;\Rightarrow\;\;10b^2\:=\:a ^2$ [1]
Since $\displaystyle a^2$ is a multiple of 10, then $\displaystyle a$ is a multiple of 10: .$\displaystyle a \,=\,10p$

Substitute into [1]: .$\displaystyle 10b^2\:=\:(10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p ^2\;\;\Rightarrow\;\;b^2\:=\:10p^2$
Since $\displaystyle b^2$ is a multiple of 10, then $\displaystyle b$ is a multiple of 10: .$\displaystyle b\,=\,10q$

But this contradicts the fact that $\displaystyle a$ and $\displaystyle b$ are relatively prime.
. . Therefore, $\displaystyle \sqrt{10}$ is irrational.

=========================

Here is the other proof of sqrt(10) is irrational:

Theorem. If nth root of a is not an integer , then in fact nth root of a is irrational.

sqrt(9) < sqrt(10)< sqrt(16)
3 < sqrt(10) < 4

So sqrt(10) is not an integer, hence it is irrational.
• Dec 31st 2006, 07:42 AM
ThePerfectHacker
Quote:

Originally Posted by Jenny20

Here is the other proof of sqrt(10) is irrational:

Theorem. If nth root of a is not an integer , then in fact nth root of a is irrational.

sqrt(9) < sqrt(10)< sqrt(16)
3 < sqrt(10) < 4

So sqrt(10) is not an integer, hence it is irrational.

I think the user is asked not to use that method. Otherwise it is too easy.
• Dec 31st 2006, 07:44 AM
Jenny20
i see. :)
• Dec 31st 2006, 09:39 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
The set of point,
$\displaystyle 4+6y-y^2$
Has a maximum value.
(An upside down parabola).
Thus, if you choose a number above the maximum point then you are safe.

Pointing at a diagram does not constitute a proof. You will have the
Bourbaki police after you at this rate.

RonL
Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last