Grade 8 - Number Theory #1

Andrew, Bob, Catherine, David and Emma are discussing about rational numbers and recurring decimals.


Andrew: Infinite decimals cannot be rational numbers.

Bob: Finite decimals and recurring decimals are all rational numbers.

Catherine: Natural numbers can always be expressed in the form of a recurring decimal, while there exists an integer that cannot be expressed in that form.

David: Some recurring decimals cannot be expressed in the form of a fraction having both of its numerator and denominator as rational numbers.

Emma: Fractions that have their denominator as a power of 10 must be rational numbers.


Andrew has number 1 on his shirt, Bob has number 2, Catherine has number 4, David has number 8 and Emma has number 16.

Add up all of the numbers on the shirts of the people whose statement is correct.


Detail:

  • We don't consider expressions such as 3.00000 3.00000\cdots and 67.00000 67.00000\cdots as recurring decimals.

This problem is a part of <Grade 8 - Number Theory> series .


The answer is 6.

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1 solution

Boi (보이)
Jul 4, 2017

Know that the definition of rational number is:

"A number that can be expressed as a b , where a and b are both integers. ( b 0 )" \text{"A number that can be expressed as }\frac{a}{b}\text{, where }a\text{ and }b\text{ are both integers. (}b\neq0\text{)"}


Andrew:

Counterexample: 0.33333 = 1 3 0.33333\cdots=\dfrac{1}{3} .

F A L S E \therefore\boxed{FALSE}


Bob:

Let a n {a_n} satisfy 0 < a i < 10 0<a_i<10 and a i N a_i\in \mathbb{N} for all natural numbers i i .

Finite decimal: 0. a 1 a 2 a 3 a 4 a n = a 1 a 2 a 3 a 4 a n 1 0 n 0.\overline{a_1a_2a_3a_4\cdots a_n}=\dfrac{\overline{a_1a_2a_3a_4\cdots a_n}}{10^n}

Recurring decimal: 0. a 1 a 2 a 3 a 4 a n = a 1 a 2 a 3 a 4 a n 1 0 n 1 0.\overline{a_1a_2a_3a_4\cdots a_n}=\dfrac{\overline{a_1a_2a_3a_4\cdots a_n}}{10^n-1}

T R U E \therefore\boxed{TRUE}


Catherine:

Natural number n n can always be expressed as ( n 1 ) . 99999 \overline{(n-1).99999\cdots} .

Also, negative integer m m can always be expressed as ( m + 1 ) . 99999 \overline{(m+1).99999\cdots} .

However, the integer 0 0 cannot be expressed as such form.

T R U E \therefore\boxed{TRUE}


David:

Already proven wrong while solving Bob .

F A L S E \therefore\boxed{FALSE}


Emma:

Counterexample: 2 1 0 2 = 0.014142135 \dfrac{\sqrt{2}}{10^2}=0.014142135\cdots .

F A L S E \therefore\boxed{FALSE}


From Andrew~Emma, we now know that only Bob and Catherine said a correct statement.

2 + 4 = 6 \therefore~2+4=\boxed{6} .

For your counterexample to Emma's statement, 0.03333...... is a rational number. However, I do agree that Emma's statement is false, but please change your example to √2/10 which is indeed irrational.

Siva Budaraju - 3 years, 11 months ago

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Ahh yes I've made a mistake. Sorry, and thank you!

Boi (보이) - 3 years, 11 months ago

Can you please explain me why is David wrong?..I see no fault in his statement.

Every recurring decimal is a rational number and hence it can be represented in that form wherein the numerator and denominator are rational.

Ankit Kumar Jain - 3 years, 11 months ago

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Yeah... But mm, I think you misread the question.

David said:

"Some recurring decimals cannot be expressed in the form of a fraction having both of its numerator and denominator as rational numbers."

So what he said is false.

Boi (보이) - 3 years, 11 months ago

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Oh..Sorry ..! Thanks for the consideration.

Ankit Kumar Jain - 3 years, 11 months ago

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@Ankit Kumar Jain De nada!

Haha it's nothing. 👍

Boi (보이) - 3 years, 11 months ago

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