Roots of factorial signs.

Number Theory Level 2

Fill in the blanks:

The numbers $\sqrt{2!}, \sqrt{3!} , \sqrt{4!}, \cdots$ are all $\text{\_\_\_\_\_\_\_\_ } .$

Notation: $!$ is the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

rational numbers irrational numbers

2 solutions

Alexander Shannon
Feb 22, 2021

According to Bertrand's postulate, there is always a prime $p$ , such that $\lfloor \frac{x}{2} \rfloor < p < x$ . The prime $p$ has the property $p|x!$ , but $p^2|x!$ . Therefore,

$\sqrt{x!} = \sqrt{R} \times \sqrt{p}$

where $R \times p = x!$ . Also, $R$ and $p$ are coprimes. if $\sqrt{x!}$ is rational, then $\sqrt{x!}=\frac{a}{b}$ , for positive integers $a,b$ ( $a$ and $b$ are coprimes).

$\sqrt{x!} = \sqrt{R} \times \sqrt{p} \implies x!=R \times p \implies x!=\frac{a^2}{b^2}=R\times p \implies$

$a^2= (R \times p) \times b^2 \implies p|a \implies p^2|a^2 \implies p|b \implies gcd(a,b)=p$

which is contradiction. So, $\sqrt{x!}$ is irrational. Note that the reasoning only works for $x>3$ . So for $\sqrt{2!}$ and $\sqrt{3!}$ one should prove the irrationality separately, although the reasoning is similar.

Excluding $0$ and $1$ , then we get the answer $\text { Irrational }$ .

Because $\sqrt { 0! } = \pm 1 = \sqrt { 1! }$ .

. . - 3 months, 2 weeks ago

correct $..$

Alexander Shannon - 3 months, 2 weeks ago

. .
Feb 18, 2021

Substituting $x = 2$ to the expression, $\sqrt { x! }$ , then $\sqrt { 2! } = \sqrt { 2 }$ .

Since $\sqrt { 2 }$ is a irrational number, so the answer is irrational number.

You still have to demonstrate that $\sqrt{3!}, \sqrt{4!} , \sqrt{5!} , \ldots$ are all irrational numbers.

Pi Han Goh - 3 months, 3 weeks ago

I know.

$\sqrt { 6 }, \sqrt { 24 }, \sqrt { 120 }, \sqrt { 720 }, \sqrt { 5040 }, \sqrt { 40320 }, \cdots$ are all irrational because the factorial cannot be a perfect square.

. . - 3 months, 3 weeks ago

But how do you prove a factorial can't be a perfect square? (Hint: think about prime factors; Bertrand's postulate also helps)

Chris Lewis - 3 months, 3 weeks ago

@Chris Lewis – Excepting $0$ and $1$ , then we get the irrational number.

Because $\sqrt { 0! } = \pm 1 = \sqrt { 1! }$ .

Of course, I cannot understand why $0!$ is equal to $1$ .

. . - 3 months, 2 weeks ago

@. . – You still didn't prove that $\sqrt{3!} ,\sqrt{4!} ,\ldots$ are all irrational numbers.

Pi Han Goh - 3 months, 2 weeks ago

@Pi Han Goh – If $\sqrt { 6 }$ is rational, then if $a$ and $b$ are integers, then $\sqrt { 6 } = \frac { b } { a } ( a \ne 0 )$ .

Multiplying $a$ to both sides, then $a \sqrt { 6 } = b$ .

And let $\frac { b } { a }$ to a irraducible fraction.

Then $\sqrt { a ^ { 2 } 6 } = b$ .

So, $a ^ { 2 } \times 6 = b ^ { 2 }$ .

Next, $6 = ( \frac { b } { a } ) ^ { 2 }$ .

And, $6 = \frac { b ^ { 2 } } { a ^ { 2 } }$ .

Then, $6 a ^ { 2 } = b ^ { 2 }$ .

So, $b$ is even because $6 a ^ { 2 }$ is even.

Next, then $a$ is even because $b$ is even.

We let $\frac { b } { a }$ is irreducible, but $a$ and $b$ are all even.

So $\sqrt { 6 }$ is irrational.

Then, we get the answer $\boxed { \text { Irrational Number } }$ .

Much simpler proof exists.

Since $6$ is not a perfect square, so $\sqrt { 6 }$ is irrational.

. . - 3 months, 2 weeks ago

@. . – You have only proved that $\sqrt{3!} = \sqrt{6}$ is irrational number.

What about the other infinitely many numbers, $\sqrt{4!} ,\sqrt{5!} , \sqrt{6!} , \ldots$ ?

Pi Han Goh - 3 months, 2 weeks ago

@Pi Han Goh – $4! = 24, 5! = 120, 6! = 720, 7! = 5040, 8! = 40320, \cdots$ are all not perfect squares, so $\sqrt { 4! }, \sqrt { 5! }, \sqrt { 6! }, \sqrt {7! }, \sqrt { 8! }, \cdots$ are all irrational numbers.

So except 0 and 1, then all squares of factorial are all $\text { Irrational number }$ . And except negative numbers also.

. . - 3 months, 2 weeks ago

@. . – And why are they not perfect squares?

Pi Han Goh - 3 months, 2 weeks ago

@Pi Han Goh – It is natural that they are not perfect squares.

For $x > 1$ , $x = \text { Natural Numbers }$ , then we get $x! = a ^ { 2 }$ .

$a = \text { Integers }$ .

Then you will understand.

If $a, b, c, \cdots$ are natural numbers, then $( a ^ { b } \times b ^ { c } \times c ^ { a } ) ^ { abc }$ must be a natural number, and between $a$ , $b$ , and $c$ , there exists a even number.

Next, if $\sqrt { x! } = a$ , then we get $x! = a ^ { 2 }$ .

We cannot find the number $a$ that satisfies the expression above.

Because a factorial means $x! = x \times ( x - 1 ) \times ( x - 2 ) \times \cdots 4 \times 3 \times 2 ( \times 1 ).$ .

We must find the number $x$ .

But we cannot find it.

So the value of $\sqrt { x! }, x > 1, x = \text { Natural number }$ is always $\boxed { \text { Irrational Number } }$ .

. . - 3 months, 2 weeks ago

@. . –

We cannot find the number $a$ that satisfies the expression above.

This is not a valid proof. You have only tested a finite number of cases and found no evidence. When in fact, you should be proving that all such numbers don't satisfy this constraint.

You can't do a proof by exhaustion when there are infinitely many cases.

It's like saying

There are infinitely many prime numbers. Here's my proof:

11 is the biggest prime number? Nope, 13 is also a prime number.

No matter what prime number if you give me, I will always find a larger prime.

Pi Han Goh - 3 months, 2 weeks ago

@Pi Han Goh – Then can you tell me the largest prime number?

. . - 3 months, 2 weeks ago

@. . – You're completely missing the point.

You have only proven that the square root of some factorials are irrational, when in fact, you should be proving that the square roots of ALL factorials are irrational.

Pi Han Goh - 3 months, 2 weeks ago

@Pi Han Goh – Ok.

But you are making me really tired.

First, if $\sqrt { 24 }$ is rational number, then there is a number which satisfies $\sqrt { 24 } = \frac { a } { b } ( b \ne 0 )$ .

And $a$ and $b$ are integers.

Then $24 = \frac { a ^ { 2 } } { b ^ { 2 } }$ .

So, $24 b ^ { 2 } = a ^ { 2 }$ .

Then I proved it as $\sqrt { 3! }$ , $a$ , and $b$ are all even.

And $\sqrt { 5! }, \sqrt { 6! }, \sqrt { 7! }, \sqrt { 8! }, \cdots$ are all irrational number.

Prooves are the same.

I am so tired, so pls don't reply to me anymore.

. . - 3 months, 2 weeks ago

@. . – Yikes, this got heated. The point is that you can't prove something holds for an infinite set by going case by case.

Classic examples of this are Euclid's proof that there are infinitely many primes, or any proof of Pythagoras' theorem.

As @Pi Han Goh says, it's not enough to say that because $x!$ is not a perfect square for $x=2,3,4,5\ldots$ it follows that $x!$ is never a perfect square; how can you be sure (like this) that there isn't some enormous factorial that also happens to be a square?

The only way to approach this kind of problem is to think about something fundamental and common to all factorials, and show that that is incompatible with being a perfect square.

I'll show you a proof below; to make it easier to read, I'll avoid a very formal structure, and I'll include examples (though these aren't necessary).

Fact $1$ : the exponent of every prime factor of a square number is even. [Proof: every natural number $n$ has a prime factorisation. Write $n$ in this form; square it; the exponents of each prime in the factorisation of $n^2$ are twice those in the factorisation of $n$ . For example, $6=2^1\cdot 3^1$ and $6^2=36=2^2 \cdot 3^2$ .]

Fact $2$ : (Bertrand's postulate) for $x>1$ , the largest prime not exceeding $x$ is greater than $\frac12 x$ . [Example: the largest prime not exceeding $10$ is $7$ and $7>\frac{10}{2}$ ]

Now we can prove the result. Assume that for some $x>1$ , $x!$ is a perfect square. Let $p$ be the largest prime not exceeding $x$ . From fact $2$ , the list $1,2,3,\cdots ,x$ contains exactly one multiple of $p$ (ie $p$ itself).

But then the exponent of $p$ in the prime factorisation of $x!$ is one, which is odd; and this contradicts fact $1$ .

Therefore no such $x$ exists.

I hope that makes sense and you can see how it works for all $x$ .

Chris Lewis - 3 months, 2 weeks ago

Roots of factorial signs.

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