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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

Can anyone solve it explicitly out?

find positive irrantional a, who satisfies a²=10+2√5

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
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it has no solution, my software told me

**X'(y-Xβ)=0**

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**MathsIsFun****Administrator**- Registered: 2005-01-21
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Software?

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 13,517

a²=10+2√5

There are two possible roots, a and -a. Therefore, a is the positive irrational root.

Character is who you are when no one is looking.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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...≈3.80423

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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But there's another thing:

You have to prove that a is irrational

IPBLE: Increasing Performance By Lowering Expectations.

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**George,Y****Member**- Registered: 2006-03-12
- Posts: 1,306

eazy, if a is a rational, it can be presented as N/M , where N and M are both integers.

N²/M²=a², thus a is a rational. but a is not.

to be strict 10+2√5 is rational <=>√5 is rational (proved by simple fraction algebra)

The last part is very difficult, it usually lies on a Math Analysis book's page.

Proporsition: √5 cannot be expressed as N/M, where N and M are both integers.

Proof:

suppose √5 can be expressed as N/M, thus its simpified form would be p or p/q, where p and q are both integers. it cannot be p alone, since no integer p satisfy pp=25

p²/q²=5, p²= p p =5, thus p|5 , then p²|25 and q|5 is invalid(don't know the english words)

∴when p²/q² is an integer L, L|25 but L cannot.

Hence the assumption is false.

*Last edited by George,Y (2006-04-05 23:14:54)*

**X'(y-Xβ)=0**

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**George,Y****Member**- Registered: 2006-03-12
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you can simplify

as 2+√5**X'(y-Xβ)=0**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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If a = p/q, then:

where p,q elem N.

But √5 is irrational, and the right side of the equation is rational, so there don't exist (p,q) elem N: p/q=a, so a is irrational.

IPBLE: Increasing Performance By Lowering Expectations.

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**krassi_holmz****Real Member**- Registered: 2005-12-02
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Sorry for my late post, George.

For the other thing:

There' s more generalized formula:

Let Q be the set of all retional numbers:

Q={p/q|p,q ∈ N}, where N = {1,2,3...}

Let Ir is the set of all numbers of the kind x^(1/y):

Ir={x^(1/y)|x,y ∈ N}.

Then:

Q || Ir = N.

Proof:

Let a,b,c,d ∈ N and

But d ∈ N => exists k ∈ N : k^c=d.

But then

.

Particularry, this means that if an integer n is not a square, then √n is irrational.

*Last edited by krassi_holmz (2006-04-06 00:31:35)*

IPBLE: Increasing Performance By Lowering Expectations.

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**George,Y****Member**- Registered: 2006-03-12
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But d ∈ N => exists k ∈ N : k^c=d.

for more illustrative

<=> ⇒ a/b=q∈N ⇒ d=q[sup]c[/sup]*Last edited by George,Y (2006-04-06 13:09:51)*

**X'(y-Xβ)=0**

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**krassi_holmz****Real Member**- Registered: 2005-12-02
- Posts: 1,908

agree.

IPBLE: Increasing Performance By Lowering Expectations.

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