
Between any two rational numbers there is no rational number.
[a] True
[b] False
Answer
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Hint: In order to prove that a statement is incorrect we have to come up with a counterexample and in order to prove it correct we have to come up with a formal proof. Try finding a counterexample in the above case, i.e. find two rational numbers which contain another rational number in between them. Use the fact that integers are also rational numbers.
Complete step-by-step answer:
Rational Numbers: Number which can be expressed in the form of $\dfrac{p}{q}$ where p and q are integers and $q\ne 0$ are called rational numbers. Since every integer n can be expressed in the form of $\dfrac{n}{1}$ and 1,n both are integers and $1\ne 0$, every integer is a rational number.
We will prove that the above statement is incorrect by using the method of contradiction.
Let us assume that the above statement is correct.
Hence \[\forall n,m\in \mathbb{Z},m>n, {\exists }p\in \mathbb{Z}\]such that $n < p < m$.
So we have $0 < m-n \le 1$ because if $m-n > 1$ then let p = 1+n, we have $m > p > n$ which is not possible
Hence $\forall m,n\in \mathbb{Z},m>n$ we have $0 < m-n \le 1$
But since m and n are integers, we have $m-n\ge 1$
Hence, we have m = 1+ n
Hence we have $\left| \mathbb{Z} \right|=2$, because if $\left| \mathbb{Z} \right|\ge 3$ then choose p,q,r such that p>q>r.
Since p>q, p = 1+q
Since p>r, p = 1+r
Since q>r, q = 1+r
Hence p = q
But p = 1+q
So, we have 1+q = q
i.e. 1 = 0
which is incorrect.
Hence, we have $\left| \mathbb{Z} \right|=2$
But we know that $\left| \mathbb{Z} \right|=\infty $, so we have
$2=\infty $which is incorrect.
Hence the assumption that there exists no rational number between two rational numbers is incorrect. Hence option [b] is correct.
Note: [1] Alternatively, we know that 1<2<3 and $1\in Q,2\in Q,3\in Q$. Hence, we found an example at which the above statement fails to hold true.
Hence the above statement is incorrect. Hence option [b] is correct.
[2] We have a stronger relation over the field of rational numbers known as the denseness property of rational numbers. It states rational numbers are dense, i.e. between any two rational numbers there exists another rational number. From the denseness property of rational numbers, the above statement is incorrect. Hence option [b] is correct.
Complete step-by-step answer:
Rational Numbers: Number which can be expressed in the form of $\dfrac{p}{q}$ where p and q are integers and $q\ne 0$ are called rational numbers. Since every integer n can be expressed in the form of $\dfrac{n}{1}$ and 1,n both are integers and $1\ne 0$, every integer is a rational number.
We will prove that the above statement is incorrect by using the method of contradiction.
Let us assume that the above statement is correct.
Hence \[\forall n,m\in \mathbb{Z},m>n, {\exists }p\in \mathbb{Z}\]such that $n < p < m$.
So we have $0 < m-n \le 1$ because if $m-n > 1$ then let p = 1+n, we have $m > p > n$ which is not possible
Hence $\forall m,n\in \mathbb{Z},m>n$ we have $0 < m-n \le 1$
But since m and n are integers, we have $m-n\ge 1$
Hence, we have m = 1+ n
Hence we have $\left| \mathbb{Z} \right|=2$, because if $\left| \mathbb{Z} \right|\ge 3$ then choose p,q,r such that p>q>r.
Since p>q, p = 1+q
Since p>r, p = 1+r
Since q>r, q = 1+r
Hence p = q
But p = 1+q
So, we have 1+q = q
i.e. 1 = 0
which is incorrect.
Hence, we have $\left| \mathbb{Z} \right|=2$
But we know that $\left| \mathbb{Z} \right|=\infty $, so we have
$2=\infty $which is incorrect.
Hence the assumption that there exists no rational number between two rational numbers is incorrect. Hence option [b] is correct.
Note: [1] Alternatively, we know that 1<2<3 and $1\in Q,2\in Q,3\in Q$. Hence, we found an example at which the above statement fails to hold true.
Hence the above statement is incorrect. Hence option [b] is correct.
[2] We have a stronger relation over the field of rational numbers known as the denseness property of rational numbers. It states rational numbers are dense, i.e. between any two rational numbers there exists another rational number. From the denseness property of rational numbers, the above statement is incorrect. Hence option [b] is correct.
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