prove that rootl2 and root 5 is an a irrational number
Answers
Step-by-step explanation:
√5
- contradiction-
Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.
⇒√5=p/q
⇒5=p²/q² {Squaring both the sides}
⇒5q²=p² (1)
⇒p² is a multiple of 5. {Euclid's Division Lemma}
⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}
⇒p=5m
⇒p²=25m² (2)
From equations (1) and (2), we get,
5q²=25m²
⇒q²=5m²
⇒q² is a multiple of 5. {Euclid's Division Lemma}
⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}
Hence, p,q have a common factor 5. this contradicts that they are co-primes. Therefore, p/q is not a rational number. This proves that √5 is an irrational number.
Now,√2
Let's suppose √2 is a rational number. Then we can write it √2 = a/b where a, b are whole numbers, b not zero.
We additionally assume that this a/b is simplified to lowest terms, since that can obviously be done with any fraction. Notice that in order for a/b to be in simplest terms, both of a and b cannot be even. One or both must be odd. Otherwise, we could simplify a/b further.
From the equality √2 = a/b it follows that 2 = a2/b2, or a2 = 2 · b2. So the square of a is an even number since it is two times something.
From this we know that a itself is also an even number. Why? Because it can't be odd; if a itself was odd, then a · a would be odd too. Odd number times odd number is always odd. Check it if you don't believe me!
Okay, if a itself is an even number, then a is 2 times some other whole number. In symbols, a = 2k where k is this other number. We don't need to know what k is; it won't matter. Soon comes the contradiction.
If we substitute a = 2k into the original equation 2 = a2/b2, this is what we get:
2 = (2k)2/b2
2 = 4k2/b2
2*b2 = 4k2
b2 = 2k2
This means that b2 is even, from which follows again that b itself is even. And that is a contradiction!!!
WHY is that a contradiction? Because we started the whole process assuming that a/b was simplified to lowest terms, and now it turns out that a and b both would be even. We ended at a contradiction; thus our original assumption (that √2 is rational) is not correct. Therefore √2 cannot be rational
Answer:
Proving that √5 is irrational?
Let's prove this by the method of contradiction-
Let's prove this by the method of contradiction-Say, √5 is a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m⇒p²=25m² (2)
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m⇒p²=25m² (2)From equations (1) and (2), we get,
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m⇒p²=25m² (2)From equations (1) and (2), we get,5q²=25m²
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m⇒p²=25m² (2)From equations (1) and (2), we get,5q²=25m²⇒q²=5m²
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m⇒p²=25m² (2)From equations (1) and (2), we get,5q²=25m²⇒q²=5m²⇒q² is a multiple of 5. {Euclid's Division Lemma}
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m⇒p²=25m² (2)From equations (1) and (2), we get,5q²=25m²⇒q²=5m²⇒q² is a multiple of 5. {Euclid's Division Lemma}⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}
a rational number. ∴ It can be expressed in the form p/q where p,q are co-prime integers.⇒√5=p/q⇒5=p²/q² {Squaring both the sides}⇒5q²=p² (1)⇒p² is a multiple of 5. {Euclid's Division Lemma}⇒p is also a multiple of 5. {Fundamental Theorm of arithmetic}⇒p=5m⇒p²=25m² (2)From equations (1) and (2), we get,5q²=25m²⇒q²=5m²⇒q² is a multiple of 5. {Euclid's Division Lemma}⇒q is a multiple of 5.{Fundamental Theorm of Arithmetic}Hence, p,q have a common factor 5. this contradicts that they are co-primes. Therefore, p/q is not a rational number. This proves that √5 is an irrational number.