It's a little tricky one🤔, Let √5 is a rational number. Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime. a = √5b ⇒ a² = 5b² Therefore, a² is divisibleRead more
It’s a little tricky one🤔,
Let √5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.
a = √5b
⇒ a² = 5b²
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer
(5k)² = 5b²
⇒ 5k² = b²
This means that b² is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
Hence, √5 cannot be expressed as p/q or it can be said that √5 is irrational.
Let 3+2√5 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 3+2√5 = a/b 2√5 = (a/b)-3 √5 = 1/2((a/b) - 3) Since a andbare integers, 1/2((a/b) - 3) will also be rational and therefore, √5 is rational. This contradicts the fact that √5 is irrational. Hence, our assumptioRead more
Let 3+2√5 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
3+2√5 = a/b
2√5 = (a/b)-3
√5 = 1/2((a/b) – 3)
Since a andbare integers, 1/2((a/b) – 3) will also be rational and therefore, √5 is rational.
This contradicts the fact that √5 is irrational. Hence, our assumption that 3 + 2√5 is rational is false. Therefore, 3 +2√5 is irrational.
Let 1/√2 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 1/√2 = a/b Or √2 = b/a b/a is as a and b are integers. Therefore, √2 is rational which contradicts to the fact that √2 is irrational. Hence, our assumption is false and 1/√2 is irrational. Video Explanation😀
Let 1/√2 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
1/√2 = a/b
Or
√2 = b/a
b/a is as a and b are integers.
Therefore, √2 is rational which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and 1/√2 is irrational.
Let 7√5 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 7√5 = a/b ⇒ √5 = a/7b a/7b is rational as a and b are integers. Therefore, √5 should be rational. This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false. Hence,Read more
Let 7√5 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
7√5 = a/b
⇒ √5 = a/7b
a/7b is rational as a and b are integers.
Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.
I can Solve this question😁, If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5. Prime factorisation of 6ⁿ = (2 × 3)ⁿ According to Fundamental Theorem of Arithmetic, the factorisation is uniqe. It can be observed thaRead more
I can Solve this question😁,
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5.
Prime factorisation of 6ⁿ = (2 × 3)ⁿ
According to Fundamental Theorem of Arithmetic, the factorisation is uniqe. It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.
Prove that √5 is irrational.
It's a little tricky one🤔, Let √5 is a rational number. Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime. a = √5b ⇒ a² = 5b² Therefore, a² is divisibleRead more
It’s a little tricky one🤔,
Let √5 is a rational number.
Therefore, we can find two integers a, b (b ≠ 0) such that √5 = a/b Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume that a and b are co-prime.
a = √5b
⇒ a² = 5b²
Therefore, a² is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer
(5k)² = 5b²
⇒ 5k² = b²
This means that b² is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
Hence, √5 cannot be expressed as p/q or it can be said that √5 is irrational.
See this 👇
See lessProve that 3 + 2√5 is irrational.
Let 3+2√5 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 3+2√5 = a/b 2√5 = (a/b)-3 √5 = 1/2((a/b) - 3) Since a andbare integers, 1/2((a/b) - 3) will also be rational and therefore, √5 is rational. This contradicts the fact that √5 is irrational. Hence, our assumptioRead more
Let 3+2√5 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
3+2√5 = a/b
2√5 = (a/b)-3
√5 = 1/2((a/b) – 3)
Since a andbare integers, 1/2((a/b) – 3) will also be rational and therefore, √5 is rational.
This contradicts the fact that √5 is irrational. Hence, our assumption that 3 + 2√5 is rational is false. Therefore, 3 +2√5 is irrational.
See here 😇
See lessProve that the following is irrational: 1/√2
Let 1/√2 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 1/√2 = a/b Or √2 = b/a b/a is as a and b are integers. Therefore, √2 is rational which contradicts to the fact that √2 is irrational. Hence, our assumption is false and 1/√2 is irrational. Video Explanation😀
Let 1/√2 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
1/√2 = a/b
Or
√2 = b/a
b/a is as a and b are integers.
Therefore, √2 is rational which contradicts to the fact that √2 is irrational.
Hence, our assumption is false and 1/√2 is irrational.
Video Explanation😀
See lessProve that the following are irrational: 7√5
Let 7√5 is rational. Therefore, we can find two co-prime integers a, b (b ≠ 0) such that 7√5 = a/b ⇒ √5 = a/7b a/7b is rational as a and b are integers. Therefore, √5 should be rational. This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false. Hence,Read more
Let 7√5 is rational.
Therefore, we can find two co-prime integers a, b (b ≠ 0) such that
7√5 = a/b
⇒ √5 = a/7b
a/7b is rational as a and b are integers.
Therefore, √5 should be rational.
This contradicts the fact that √5 is irrational. Therefore, our assumption that 7√5 is rational is false. Hence, 7√5 is irrational.
See this 👇
See lessCheck whether 6ⁿ can end with the digit 0 for any natural number n.
I can Solve this question😁, If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5. Prime factorisation of 6ⁿ = (2 × 3)ⁿ According to Fundamental Theorem of Arithmetic, the factorisation is uniqe. It can be observed thaRead more
I can Solve this question😁,
If any number ends with the digit 0, it should be divisible by 10 or in other words, it will also be divisible by 2 and 5 as 10 = 2 × 5.
Prime factorisation of 6ⁿ = (2 × 3)ⁿ
According to Fundamental Theorem of Arithmetic, the factorisation is uniqe. It can be observed that 5 is not in the prime factorisation of 6ⁿ.
Hence, for any value of n, 6n will not be divisible by 5. Therefore, 6n cannot end with the digit 0 for any natural number n.
Here is the Video explanation 👇
See less