Tag: proofs of irrationality

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Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which of the following is an irrational number? 

  1. $0.14$
  2. $0.14 \overline{16}$
  3. $1.1 {416}$
  4. $0.4014001400014....$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The decimal expansion of a rational number is either terminating or non-terminating repeating.

(A) $0.14$ is terminating, so it is a rational number

(B) $0.14\bar{16}=0.141616....$ 

is also rational ( non-terminating repeating ), where digits $16$ are repeating.

(C) $0.1416$ is terminating, so it is a rational number

(D) $0.4014001400014.....$

is an irrational number because it is neither terminating nor repeating.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

State whether the following statement is true or false.

The following number is irrational
$7\sqrt {5}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume $7\sqrt{5}$ is rational.
Hence, $7\sqrt{5}$ can be written in form $\dfrac{a}{b}$

Where, $a$ and $b$ $(n\ne 0)$ are co-prime.
Hence, $7\sqrt{5}=\dfrac{a}{b}$
$\Rightarrow$  $\sqrt{5}=\dfrac{1}{7}\times\dfrac{a}{b}$

Here, $\dfrac{a}{7b}$ is a rational number, but $\sqrt{5}$ is irrational.

Since, Rational $\ne$ Irrational
This is a contriadition
$\therefore$  Our assumption is incorrect.
$\therefore$  $7\sqrt{5}$ is irrational number. 
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

State whether the following statement is true or false.

The following number is irrational
$6+\sqrt {2}$

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
An Irrational Number is a real number that cannot be written as a simple fraction.

Let us assume $6+\sqrt{2}$ is rational.
Hence, $6+\sqrt{2}$ can be written in form $\dfrac{a}{b}$

Where, $a$ and $b$ $(n\ne 0)$ are co-prime.
Hence, $6+\sqrt{2}=\dfrac{a}{b}$
$\Rightarrow$  $\sqrt{2}=\dfrac{a}{b}-6$

$\Rightarrow$  $\sqrt{2}=\dfrac{a-6b}{b}$

Here, $\dfrac{a-6b}{b}$ is a rational number, but $\sqrt{2}$ is irrational.

Since, Rational $\ne$ Irrational
This is a contriadition
$\therefore$  Our assumption is incorrect.
$\therefore$  $6+\sqrt{2}$ is irrational number. 
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

Which of the following is always true 

  1. $irrational + irrational =irrational $
  2. $\dfrac{rational }{rational }=rational $
  3. $\dfrac{integer }{integer}=integer$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Counter-example for A: $(\sqrt{2}) + (4-\sqrt{2}) = 4$

Counter-example for C: $\dfrac{1}{2}=0.5$

Proof for B:

Let $q _1, q _2$ be two rational numbers such that $q _2\neq0$.


As they are rational, they can be written as $a/b, c/d$ respectively for some integers $a, b, c, d$. $(b,c,d\neq0)$

$\dfrac{q _1}{q _2}=\dfrac{a/b}{c/d}=\dfrac{ad}{bc}$

Since, $a,b,c,d$ were integers, even $ad$ and $bc(\neq0)$ are integers and therefore, the above expression is rational.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

If the product of two irrational numbers is rational, then which of the following can be concluded?

  1. The ratio of the greater and the smaller numbers is an integer.

  2. The sum of the numbers must be rational.

  3. The excess of the greater irrational number over the irrational number must be rational.

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a and b are irrational and ab is rational, let a = x*sqrt(k) and b = y*sqrt(k). The ratio a/b = x/y, which is rational. The options provided are limited, but A is the most mathematically sound conclusion.