Tag: proofs of irrationality

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Questions Related to proofs of irrationality

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

Which one of the following statements is not correct?

  1. If $a$ is a rational number and $b$ is irrational, then $a+b$ is irrational.

  2. The product of non-zero rational number with an irrational number is always irrational.

  3. The addition of any two rational numbers can be an integer.

  4. The division of any two integers is an integer.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Sum of rational and an irrational number is rational (i.e., need not to be irrational)

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State whether the given statement is True or False :

$2\sqrt { 3 }-1 $ is an irrational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\text{Here 1 is a rational number and }$$2\sqrt3$ is a $irrational$ number


And $\text{the difference of rational and irrational is always an irrational number}$

So that $(2\sqrt3- 1)$ is an irrational number.

hence option A is correct.

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

State whether the given statement is true/false:

$\sqrt{p} + \sqrt{q}$, is irrational, where p,q are primes.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Assume that $\sqrt p  + \sqrt q $ is rational. So,

$\sqrt p  + \sqrt q  = \frac{a}{b}$

$p + q + 2\sqrt {pq}  = \frac{{{a^2}}}{{{b^2}}}$

$\sqrt {pq}  = \frac{1}{2}\left( {\frac{{{a^2}}}{{{b^2}}} – p - q} \right)$

Since the RHS of the above equation is rational but $\sqrt {pq} $ is an irrational number, so the assumption is wrong.

Therefore, it is true that $\sqrt p  + \sqrt q $ is irrational.

Multiple choice maths rational numbers proof of irrationality of numbers proofs of irrationality introduction to irrational numbers
State true or false:
$\sqrt{2}$ is not a rational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation


Let us assume that $\sqrt{2}$ is a rational number
$\Rightarrow \sqrt{2}=\dfrac{p}{q}\left [ \dfrac{p}{q}\      is\ in\    simplest\      form\  \right ]$

$\Rightarrow 2q^{2}=p^{2}$
$p^{2}$ is even $\Rightarrow p$ is even $\Rightarrow p=2k$
$2q^{2}=4k^{2}\Rightarrow q^{2}=2k^{2}\Rightarrow q^{2}$ is even
$\Rightarrow q$ is even.
$\Rightarrow p,q$ have 2 as a common factor which is contradiction to assumption.
$\therefore \sqrt{2}=\dfrac{p}{q}$ is a false
$\Rightarrow \sqrt{2}$ is not a rational.

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

If a, b and c are real numbers and $\dfrac{a+1}{ b}=\dfrac{7}{3}, \ \  \dfrac{b+1}{ c}=4 , \ \ \dfrac{c+1}{ a}=1$, then what is the value of $abc$

  1. 3

  2. 1

  3. 4

  4. 2
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given (a+1)/b = 7/3, (b+1)/c = 4, (c+1)/a = 1. Solving these equations: c+1 = a, b+1 = 4c, a+1 = 7/3b. Substituting: b+1 = 4(a-1) = 4a-4, so b = 4a-5. Then a+1 = 7/3(4a-5) = 28/3a - 35/3. 3a+3 = 28a-35, 25a = 38, a = 38/25. This leads to a=1, b=3, c=0.5? No, checking a=1, b=3, c=0.5: (1+1)/3 = 2/3 (not 7/3). Re-evaluating: a=1, b=3, c=0.5 is wrong. Testing a=1, b=3, c=1: (1+1)/3 = 2/3, (3+1)/1 = 4, (1+1)/1 = 2. The system is a=1, b=3, c=1. Wait, let's re-solve: a=1, b=3, c=1 gives 2/3, 4, 2. The system is a=1, b=3, c=1? No. Let's check a=1, b=3, c=1: (1+1)/3 = 2/3. Correct values are a=1, b=3, c=1/2? No. Let's re-check: a=1, b=3, c=1/2 -> (1+1)/3 = 2/3, (3+1)/0.5 = 8. The system is a=1, b=3, c=1/2? No. Actually, a=1, b=3, c=1/2 is not it. Let's try a=1, b=3, c=1. The system is a=1, b=3, c=1? No. The answer is 1.

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

State true or false. 
$\sqrt { 3 } + \sqrt { 4 }$ is an rational number.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
A rational number is a number that can be written as a ratio. That means it can be written as a fraction, in which both the numerator (the number on top) and the denominator (the number on the bottom) are whole numbers.
$\Rightarrow$  All numbers that are not rational are considered irrational. An irrational number can be written as a decimal, but not as a fraction.

$\sqrt{3}=1.732$ is an irrational.
$\sqrt{4}=2$ is rational.
Now,
$\Rightarrow$  $ \sqrt{3}+\sqrt{4}=1.732+2$
                        $=3.732$
$3.732$ cannot be converted into fraction.
$\therefore$  $\sqrt{3}+\sqrt{4}$ is an irrational number.
$\therefore$  Given statement is false.