Find an irrational number that satisfies the given inequality

[chwala]

Homework Statement

Write down an irrational number, $n$ where $31<n< 32$ [1mark]

Relevant Equations

past international exam paper- Numbers

I just came across this question and the ms indicates,

Would $31.5$ be correct? ...i think it is rational as it can be expressed as $31.5 = \dfrac{63}{2}$.

 

[SammyS]

Homework Statement: Write down an irrational number, $n$ where $31<n< 32$ [1mark]
Relevant Equations: past international exam paper- Numbers

I just came across this question and the ms indicates,

View attachment 339123

Would $31.5$ be correct? ...i think it is rational as it can be expressed as $31.5 = \dfrac{63}{2}$.

As you showed, 31.5 is rational, so not correct.

What do you know about rational numbers when expressed in decimal form?

 

[chwala]

As you showed, 31.5 is rational, so not correct.

What do you know about rational numbers when expressed in decimal form?

...they can always be expressed as a fraction...

Let us go with $10π$ cheers @SammyS

It is clear now...

My bad, i misread the ms as ' any number satisfying the inequality'.

 

[Delta2]

@chwala your solution is a clever shortcut where you climb upon the shoulders of Archimedes and $\pi$.

My solution would be to choose the number $x=31.a_1a_2...a_i...$

where I define as $a_i=$ the first digit of the number $i$ in decimal system.

Now who can prove that this number x is indeed irrational? e hehe My intuition tells me that it is irrational cause I cant find an easy way to prove that its digits have a period but i might be wrong.

 

[WWGD]

@chwala your solution is a clever shortcut where you climb upon the shoulders of Archimedes and $\pi$.

My solution would be to choose the number $x=31.a_1a_2...a_i...$

where I define as $a_i=$ the first digit of the number $i$ in decimal system.

Now who can prove that this number x is indeed irrational? e hehe My intuition tells me that it is irrational cause I cant find an easy way to prove that its digits have a period but i might be wrong.

The number looks to be male, so won't likely have a period. ;).

 

[Delta2]

The number looks to be male, so won't likely have a period. ;).

A haha best post of the new year so far....

 

[WWGD]

Maybe you can just use an argument by contradiction. If 31.a1a2... were Rational, then, given Rationals are closed under addition, so would 31.a1a2...-31= 0.a1a2...

Maybe slightly-contrived, but works.

 

[Delta2]

Maybe you can just use an argument by contradiction. If 31.a1a2... were Rational, then, given Rationals are closed under addition, so would 31.a1a2...-31= 0.a1a2...

Maybe slightly-contrived, but works.

ye ok this is correct but the main problem is why 0.a1a2... is irrational, the way I see it is that it seems to (after the initial segment which is 0.123456789) contain subsegments of 11111.... (or 22222.. or 33333..) that seem to have increasing length with no bound on the length , so there is no period, but it is gonna be hard to find a rigorous way .

 

[WWGD]

ye ok this is correct but the main problem is why 0.a1a2... is irrational, the way I see it is that it seems to (after the initial segment which is 0.123456789) contain subsegments of 11111.... (or 22222.. or 33333..) that seem to have increasing length with no bound on the length , so there is no period, but it is gonna be hard to find a rigorous way .

I thought if you just assume there is at least one Irrational in [0,1], that would do it. You could, e.g., use that [0,1] is uncountable, but Rationals; specifically Rationals in [0,1] are countable, then there have to be Irrationals in [0,1] too.

 

[Delta2]

I thought if you just assume there is at least one Irrational in [0,1], that would do it. You could, e.g., use that [0,1] is uncountable, but Rationals; specifically Rationals in [0,1] are countable, then there have to be Irrationals in [0,1] too.

Yes right [0,1] has infinite rationals and infinite irrationals so it is something like a micrography of the set of real numbers. The initial problem can be shifted to find a specific irrational in [0,1].

 

[Orodruin]

@chwala your solution is a clever shortcut where you climb upon the shoulders of Archimedes and $\pi$.

My solution would be to choose the number $x=31.a_1a_2...a_i...$

where I define as $a_i=$ the first digit of the number $i$ in decimal system.

Now who can prove that this number x is indeed irrational? e hehe My intuition tells me that it is irrational cause I cant find an easy way to prove that its digits have a period but i might be wrong.

I would not call that a shortcut - on the contrary, I think it is what is essentially expected. You are making things unnecessarily complicated here for no good reason. It doesn’t have to be $\pi$ (although it fits very well), but any known irrational number will work well multiplied by a suitable rational number.

Another reasonable approach is to take the square root of a number between $31^2 = 961$ and $32^2 = 1024$. Since those numbers are not perfect squares, their prime factorization will contain at least one prime raised to an odd power, resulting in an irrational number. Say $\sqrt{1000} = 10\sqrt{10}$, which is clearly irrational.

 

[Delta2]

I would not call that a shortcut - on the contrary, I think it is what is essentially expected. You are making things unnecessarily complicated here for no good reason. It doesn’t have to be $\pi$ (although it fits very well), but any known irrational number will work well multiplied by a suitable rational number.

Yes ok true I just thought that this $a_i$ might intrigue the minds of the readers here :D
On your path of thought, we can take as number $12e-1$ just to use $e$ the other big star of irrationals in mathematics.

Another reasonable approach is to take the square root of a number between $31^2 = 961$ and $32^2 = 1024$. Since those numbers are not perfect squares, their prime factorization will contain at least one prime raised to an odd power, resulting in an irrational number. Say $\sqrt{1000} = 10\sqrt{10}$, which is clearly irrational.

Is that a well known result that the square root of a prime raised to an odd power is an irrational?

 

[Orodruin]

Is that a well known result that the square root of a prime raised to an odd power is an irrational?

Yes. You can prove it along the same lines as proving that $\sqrt{2}$ is irrational.

Not only that, but a product of any number of distinct primes is irrational.

 

[Delta2]

Not only that, but a product of any number of distinct primes is irrational.

Slight correction, you mean the square root (or even possible the n-th root for any n>1) of such a product is irrational.

 

[PeroK]

Is that a well known result that the square root of a prime raised to an odd power is an irrational?

If the square root of a positive integer is not an integer, then it is irrational. It is never a proper rational number. This follows directly from unique prime factorisation.

 

[Delta2]

If the square root of a positive integer is not an integer, then it is irrational. It is never a proper rational number. This follows directly from unique prime factorisation.

Not so directly but yeah I think you mean the proof is kind of similar as to that the square root of 2 is irrational.

Ye kinda sketch the proof in my mind, it ends up that an integer would have two different prime factorizations unless the square root is irrational or the square root is integer.

 

[PeroK]

Not so directly but yeah I think you mean the proof is kind of similar as to that the square root of 2 is irrational.

I didn't mean that. The usual proof of the irrationality of the square root of 2 avoids using prime factorisation.

 

[Delta2]

I didn't mean that. The usual proof of the irrationality of the square root of 2 avoids using prime factorisation.

Yeah but it can be done with prime factorization and I actually meant that it is a proof by contradiction where you assume that the square root is $$\frac{m}{n}$$ where m,n coprimes and then you end up with a contradiction. So it can be either n=1 or the square root is irrational.

 

[PeroK]

Yeah but it can be done with prime factorization and I actually meant that it is a proof by contradiction where you assume that the square root is $$\frac{m}{n}$$ where m,n coprimes and then you end up with a contradiction. So it can be either n=1 or the square root is irrational.

The proof does not involve contradiction; and does not start by assuming that the square root of some integer is a proper rational.

 

[PeroK]

Let $\frac m n$ be a proper rational in its lowest form (i.e. $n \ne 1$ and $m, n$ have no common factors (except 1), hence no primes factors in common). Then $\frac {m^2}{n^2}$ is a proper rational ($n^2 \ne 1$ and $m^2, n^2$ have no common factors, as they have no primes factors in common.)

Therefore: The square of a proper rational is itself a proper rational.

Hence: the square root of an integer is never a proper rational.

Hence: any integer is either a perfect square or its square root is irrational.

 

[Delta2]

The proof does not involve contradiction; and does not start by assuming that the square root of some integer is a proper rational.

Yes ok you can see it that way too, the way I had in mind it doesn't involve exactly a contradiction but a false conclusion, that $m^2$ will have two different prime factorizations.

 

[Orodruin]

Let $\frac m n$ be a proper rational in its lowest form (i.e. $n \ne 1$ and $m, n$ have no common factors (except 1), hence no primes factors in common). Then $\frac {m^2}{n^2}$ is a proper rational ($n^2 \ne 1$ and $m^2, n^2$ have no common factors, as they have no primes factors in common.)

Therefore: The square of a proper rational is itself a proper rational.

Hence: the square root of an integer is never a proper rational.

Hence: any integer is either a perfect square or its square root is irrational.

Euclid’s proof (which is quite common) is to assume that $\sqrt{p} = a/b$ with $a$ and $b$ being coprime. However, this leads to $p$ being a factor in both $a$ and $b$, which is a contraction.

 

[PeroK]

Euclid’s proof (which is quite common) is to assume that $\sqrt{p} = a/b$ with $a$ and $b$ being coprime. However, this leads to $p$ being a factor in both $a$ and $b$, which is a contraction.

The heart of the Euclid proof is that if $p = \dfrac{a^2}{b^2}$, then $p$ is a common factor of $a$ and $b$. This uses the primeness of $p$, which is ironically irrelevant, as $b \ne 1$ must have at least one prime factor.

If $n = \dfrac{a^2}{b^2}$, then every prime that divides $b$ also divides $a$. There is no need for $n$ to be prime. [There's only the trivial case where $b = 1$ and $n$ is a perfect square.]