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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.

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

If $a=\sqrt{11}+\sqrt{3}, b =\sqrt{12}+\sqrt{2}, c=\sqrt{6}+\sqrt{4}$, then which of the following holds true ?

  1. $c>a>b$
  2. $a>b>c$
  3. $a>c>b$
  4. $b>a>c$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$a=\sqrt{11}+\sqrt{3}$

$a^{2}=11+3+2\sqrt{33}=14+2\sqrt{33}$

$b=\sqrt{12}+\sqrt{2}$

$b^{2}=14+2\sqrt{24}$

As $\sqrt{33} > \sqrt{24}, a^{2} > b^{2}, a>b$

$c=\sqrt{6}+\sqrt{4}$

$c^{2}=10+2\sqrt{24}$

As $14>10, b^{2} > c^{2}, b>c$

Hence, $a>b>c$.
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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 not:
$\left( 3+\sqrt { 5 }  \right) $ is an irrational number. 

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let us suppose $3+\sqrt 5$ is rational.

$=>3+\sqrt 5$ is in the form of $\dfrac pq$ where $p$ and $q$ are integers and $q\neq0$

$=>\sqrt5=\dfrac pq-3$

​$=>\sqrt5=\dfrac{p-3q}{q}$

as $p, q$ and $3$ are integers $\dfrac{p-3q}{q}$ is a rational number.

$=>\sqrt 5$ is a rational number.

But we know that $\sqrt 5$ is an irrational number.

So this is a contradiction.

This contradiction has arisen because of our wrong assumption that $3+\sqrt 5$ is a rational number.

Hence $3+ \sqrt5$  is an irrational number.
Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers

A rational number equivalent to  $ \displaystyle \frac{-5}{-3}  $ is -

  1. $ \displaystyle \frac{25}{15} $
  2. $ \displaystyle \frac{-15}{25} $
  3. $ \displaystyle \frac{-25}{15} $

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

 $ \displaystyle  \because  \frac{-5}{-3} $= $ \displaystyle  \frac{-5}{-3} $X $ \displaystyle  \frac{-5}{-5} $= $ \displaystyle  \frac{25}{15} $

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

Every irrational number is

  1. a surd

  2. a prime number

  3. not a surd

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

An irrational number is a real number that cannot be represented as a ratio or a simple fraction.


By definition, a surd is an irrational root of a rational number. So we know that surds are always irrational and they are always roots.

For eg, $\sqrt2$ is a surd since 2 is rational and $\sqrt 2$ is irrational.

Similarly, the cube root of 9 is also a surd since 9 is rational and the cube root of 9 is irrational.

On the other hand, $\sqrtπ$ is not a surd even though $\sqrtπ$ is irrational because π is not rational.

Thus, to answer the question, every surd is an irrational number, though an irrational number may or may not be a surd


The answer is Option C.