BdMO National Secondary 2011/9
Problem 9:
Prove that \[ \sqrt[100]{\sqrt{3}+\sqrt{2}}+\sqrt[100]{\sqrt{3}-\sqrt{2}}\]
is irrational.
Prove that \[ \sqrt[100]{\sqrt{3}+\sqrt{2}}+\sqrt[100]{\sqrt{3}-\sqrt{2}}\]
is irrational.
"Inspiration is needed in geometry, just as much as in poetry." -- Aleksandr Pushkin
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Re: BdMO National Secondary 2011/9
Suppose,
\[\sqrt{3}=a,
\sqrt{2}=b\]
After point of these numbers,digits aren't same.these are 'going on'.So a+b and a-b are irrational.There are no r and r' where a+b equals to 100th power of r and a-b equals to 100t power of r',while r and r' are rational.
So the number given is irrational.Is mine correct????
\[\sqrt{3}=a,
\sqrt{2}=b\]
After point of these numbers,digits aren't same.these are 'going on'.So a+b and a-b are irrational.There are no r and r' where a+b equals to 100th power of r and a-b equals to 100t power of r',while r and r' are rational.
So the number given is irrational.Is mine correct????
Try not to become a man of success but rather to become a man of value.-Albert Einstein
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Re: BdMO National Secondary 2011/9
Sorry you are incorrect.why the 100th power needed to be equal?Try to prove that if a+1/a is rational then a^n+1/a^n
must be rational.I will post the solution in higher secondary section.
must be rational.I will post the solution in higher secondary section.
Re: BdMO National Secondary 2011/9
but how this works,if $a+\frac 1 a$ irrational,then this does not guarantee that $a^n+\frac 1 {a^n}$ is irrational because then you need to prove the converse or only if part,but you mentioned the only part onlyMehfuj Zahir wrote:Sorry you are incorrect.why the 100th power needed to be equal?Try to prove that if a+1/a is rational then a^n+1/a^n
must be rational.I will post the solution in higher secondary section.
One one thing is neutral in the universe, that is $0$.
Re: BdMO National Secondary 2011/9
brother,are you talking about the bases(r & r')?Mehfuj Zahir wrote:why the 100th power needed to be equal?
Try not to become a man of success but rather to become a man of value.-Albert Einstein
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Re: BdMO National Secondary 2011/9
At first we should try to do this a+1/a is rational then a^n+(1/a)^n must rational.
Now according to the question a^100+(1/a)^100 is irrational.
So a+1/a is is irrational.
Now according to the question a^100+(1/a)^100 is irrational.
So a+1/a is is irrational.
- FahimFerdous
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Re: BdMO National Secondary 2011/9
However, I think I've solved it. But, I'm not sure. Ok, if I'm wrong then correct me. Here's my solution:
Suppose, $(3^{\frac{1}{2}}+2^{\frac{1}{2}})^{\frac{1}{100}}=x$
Then, we can easily show that,
$(3^{\frac{1}{2}}-2^{\frac{1}{2}})^{\frac{1}{100}}=\frac{1}{x}$
We can easily show that $x$ and $\frac{1}{x}$ are individually irrational.
Now, we suppose that $x+\frac{1}{x}$ is not irrational. We suppose it's rational (cause the sum of two irrational number can be rational). So, we can say that,
$x+\frac{1}{x}=\frac{p}{q}$ ($p,q$ is co-prime and $q$ isn't equal to $0$)
Now, if we square both sides, then we get, $x^2+\frac{1}{x^2}=\frac{p^2}{q^2}-2$
Now, if we square both sides again, then we get, $x^4+\frac{1}{x^4}=\{\frac{p^2}{q^2}-2\}^2-2$
We can say from here that the right side of the equation is rational. So, $x^4+\frac{1}{x^4}$ is rational.
Now, we get back at the beginning where we supposed that $x+\frac{1}{x}$ is rational. Now, we must say that, $x-\frac{1}{x}$ is irrational. Because, the sum and the substract of two irrational numbers can't both be rational at the same time. So, $x-\frac{1}{x}=m$, where $m$ is irrational. Now,
$(x+\frac{1}{x})(x-\frac{1}{x})=\frac{p}{qm}$
or, $x^2-\frac{1}{x^2}=n$ [here p/q*m=n, where n is also irrational]
We square both sides and we get,
$x^4+\frac{1}{x^4}=n^2+2$
We know that the square of irrational number is irrational. Now, we can say from here that $x^4+\frac{1}{x^4}$ is irrational. But we previously proved that it's rational. So, it's a contradiction. Which concludes that $x+\frac{1}{x}$ can't be rational, it must be irrational. So, it's proved.
Is my solution correct? :-S
Suppose, $(3^{\frac{1}{2}}+2^{\frac{1}{2}})^{\frac{1}{100}}=x$
Then, we can easily show that,
$(3^{\frac{1}{2}}-2^{\frac{1}{2}})^{\frac{1}{100}}=\frac{1}{x}$
We can easily show that $x$ and $\frac{1}{x}$ are individually irrational.
Now, we suppose that $x+\frac{1}{x}$ is not irrational. We suppose it's rational (cause the sum of two irrational number can be rational). So, we can say that,
$x+\frac{1}{x}=\frac{p}{q}$ ($p,q$ is co-prime and $q$ isn't equal to $0$)
Now, if we square both sides, then we get, $x^2+\frac{1}{x^2}=\frac{p^2}{q^2}-2$
Now, if we square both sides again, then we get, $x^4+\frac{1}{x^4}=\{\frac{p^2}{q^2}-2\}^2-2$
We can say from here that the right side of the equation is rational. So, $x^4+\frac{1}{x^4}$ is rational.
Now, we get back at the beginning where we supposed that $x+\frac{1}{x}$ is rational. Now, we must say that, $x-\frac{1}{x}$ is irrational. Because, the sum and the substract of two irrational numbers can't both be rational at the same time. So, $x-\frac{1}{x}=m$, where $m$ is irrational. Now,
$(x+\frac{1}{x})(x-\frac{1}{x})=\frac{p}{qm}$
or, $x^2-\frac{1}{x^2}=n$ [here p/q*m=n, where n is also irrational]
We square both sides and we get,
$x^4+\frac{1}{x^4}=n^2+2$
We know that the square of irrational number is irrational. Now, we can say from here that $x^4+\frac{1}{x^4}$ is irrational. But we previously proved that it's rational. So, it's a contradiction. Which concludes that $x+\frac{1}{x}$ can't be rational, it must be irrational. So, it's proved.
Is my solution correct? :-S
Last edited by Zzzz on Sun Feb 20, 2011 8:23 pm, edited 4 times in total.
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Your hot head might dominate your good heart!
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Re: BdMO National Secondary 2011/9
Fahim if x+1/x is rational then you cannot say that x-1/x is irrational.And the square of irrational number is not always irrational.ExampleFahimFerdous wrote:However, I think I've solved it. But, I'm not sure. Ok, if I'm wrong then correct me. Here's my solution:
Suppose, (3^1/2+2^1/2)^1/100=x
Then, we can easily show that,
(3^1/2-2^1/2)^1/100=1/x
We can easily show that x and 1/x are individually irrational.
Now, we suppose that (x+1/x) is not irrational. We suppose it's rational (cause the sum of two irrational number can be rational). So, we can say that,
x+1/x=p/q (p,q is co-prime and q isn't equal to 0)
Now, if we square both sides, then we get, x^2+1/x^2=(p^2/q^2)-2
Now, if we square both sides again, then we get, x^4+1/x^4={(p^2-q^2)-2}^2-2
We can say from here that the right side of the equation is rational. So, x^4+1/x^4 is rational.
Now, we get back at the beginning where we supposed that x+1/x is rational. Now, we must say that, x-1/x is irrational. Because, the sum and the substract of two irrational numbers can't both be rational at the same time. So, x-1/x=m, where m is irrational. Now,
(x+1/x)(x-1/x)=p/q*m
or, x^2-1/x^2=n [here p/q*m=n, where n is also irrational]
We square both sides and we get,
x^4+1/x^4=n^2+2
We know that the square of irrational number is irrational. Now, we can say from here that x^4+1/x^4 is irrational. But we previously proved that it's rational. So, it's a contradiction. Which concludes that x+1/x can't be rational, it must be irrational. So, it's proved.
Is my solution correct? :-S
\[\sqrt{3^{2}}\]
- FahimFerdous
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Re: BdMO National Secondary 2011/9
@Mehfuz vaia: I'm really sorry that I didn't notice that the square of irrational number can be rational. It was foolish of me. That was a great mistake.
But, if x is an irrational number and x+1/x is rational, then x-1/x must be irrational.
If both can be rational then we get,
x+1/x=p/q and x-1/x=m/n
If we add these two then we get,
x=(p/q)(m/n)/2
Here, the left side is rational while the right side is irrational. So, it's a contradiction. In this case, I think I'm right.
But, if x is an irrational number and x+1/x is rational, then x-1/x must be irrational.
If both can be rational then we get,
x+1/x=p/q and x-1/x=m/n
If we add these two then we get,
x=(p/q)(m/n)/2
Here, the left side is rational while the right side is irrational. So, it's a contradiction. In this case, I think I'm right.
Your hot head might dominate your good heart!
Re: BdMO National Secondary 2011/9
yes,otherwise if $x-\frac 1 x$ is rational too,then $x+\frac 1 x+x-\frac 1 x=2x$ is rational,contradictionFahimFerdous wrote:@Mehfuz vaia: I'm really sorry that I didn't notice that the square of irrational number can be rational. It was foolish of me. That was a great mistake.
But, if x is an irrational number and x+1/x is rational, then x-1/x must be irrational.
If both can be rational then we get,
x+1/x=p/q and x-1/x=m/n
If we add these two then we get,
x=(p/q)(m/n)/2
Here, the left side is rational while the right side is irrational. So, it's a contradiction. In this case, I think I'm right.
One one thing is neutral in the universe, that is $0$.