prove that √5, √7, √11 are irrational numbers.
Answers
Answer:
root 5 , root 7 , root 11 all of them give irrational number when solved
Answer:
Step 1: First we write 5 as 5 00 00 00 and pair digits starting from one's place.
Step 2: Now find a number whose square results in a number less than 5.
Step 3: The number obtained is 2. The quotient becomes 2 and the remainder obtained is 1.
Step 4: Now the 00 is brought down and the quotient obtained in step 3 is doubled, i.e; (2×2 = 4).
Step 5: The unit digit with 4 should be 2 as 42×2 = 84 as it is less than 100. On subtracting 84 from 100 we get 16.
Step 6: 00 is brought down and the quotient obtained in step 5 is doubled. Thus, 22 is doubled and 44 is obtained as the starting of a new divisor.
Step 7: The unit digit with 44 should be 3 as 443×3 = 1329 as it is less than 1600. On subtracting 1329 from 1600 we get 271.
Step 8: Now bring the 00 down and the quotient obtained in step 7 is doubled. Thus, 223 is doubled and 446 is obtained as the starting of a new divisor.
Step 9: The unit digit with 446 should be 6 as 4466×6 = 26796 as it is less than 27100. On subtracting 26796 from 27100 we get 304.
Step 10: The long division is continued further in order to obtain the required number of digits after the decimal.
√7 = p/q
On squaring both the side we get,
=> 7 = (p/q)2
=> 7q2 = p2……………………………..(1)
p2/7 = q2
So 7 divides p and p and p and q are multiple of 7.
⇒ p = 7m
⇒ p² = 49m² ………………………………..(2)
From equations (1) and (2), we get,
7q² = 49m²
⇒ q² = 7m²
⇒ q² is a multiple of 7
⇒ q is a multiple of 7
Hence, p,q have a common factor 7. This contradicts our assumption that they are co-primes. Therefore, p/q is not a rational number
√7 is an irrational number.
Step 1: Add pairs of 0 after 11 as 11.00 00 00 and pair the digits starting from the right and find a number whose square is less than or equal to the number 11, it will be our first divisor and quotient. We have 3, square the number and subtract the result from 11, 2 is the remainder.
Step 2: Take the next pair of 0 down after the remainder, as 00 is brought down, we get 200 as the next dividend, and double the first quotient to get the partial divisor of this step. The unit digit the divisor will be the number which on multiplying with the complete divisor thus formed, gives a number equal or less than the new dividend. Here, we get 3 at the units place, and 63 is our divisor and 3 is our quotient. Subtract the result after multiplying 63 with 3 from 200, and note down the remainder.
Step 3: Take the next pair of 0 down after the remainder of the previous step to get the dividend, 1100 is the new dividend. Add the units place of the divisor obtained in the previous step to the divisor itself and get the partial divisor of this step. Here, we get 66. The unit digit the divisor will be the number which on multiplying with the complete divisor thus formed, gives a number equal or less than the new dividend. Here, we get 1 at the units place, and 661 is our divisor and 1 is our new quotient. Subtract the result after multiplying 661 with 1 from 1100, and note down the remainder.
Step 4: Repeat the process until the required number of digits after the decimal is obtained.
See the following image to see a few of the steps in the long division of the 11.