1. ## 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?

2. 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, $4+6y-y^2$ 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

3. 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

4. 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

5. Originally Posted by malaygoel
for the first, $4+6y-y^2$ 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 $x \in \mathbb{R}$ for which $4+6y-y^2 < x$ for all $y \in \mathbb{R}$.

RonL

6. 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 $\sqrt{10}$?

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

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

RonL

7. (sqrt(2)+sqrt(5))^.5 = 7 + 2*sqrt(2)*sqrt(5) = 7 + 2*sqrt(10)

8. 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:

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

But does this really help?

9. 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 $x \in \mathbb{R}$ such that for all $y \in \mathbb{R}:\ \ 4+6y-y^2 < x$.

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

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

or:

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

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

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

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

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

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

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

RonL

10. Hello, ruprotein!

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

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

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

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

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

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

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

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

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

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

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: . $\sqrt{2} + \sqrt{5}$ is irrational.

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

12. [QUOTE=Soroban;33016]Hello, ruprotein!

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

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

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

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

Substitute into [1]: . $10b^2\:=\10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p^2\;\;\Righ tarrow\;\;b^2\:=\:10p^2" alt="10b^2\:=\10p)^2\;\;\Rightarrow\;\;10b^2\:=\:100p^2\;\;\Righ tarrow\;\;b^2\:=\:10p^2" />
Since $b^2$ is a multiple of 10, then $b$ is a multiple of 10: . $b\,=\,10q$

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

13. 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.

14. i see.

15. Originally Posted by ThePerfectHacker
The set of point,
$4+6y-y^2$
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

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