A multiplicative inverse of 5 mod7 would be a number n ( not a unique one) such that 5n=1Let's look at the possible numbers5x1=5mode 75x2=10=3 mod 75x3=15=1 mod 7 THAT WILL DO IT3 is the multiplicative inverse of 5 mod 7.What about the others? 5x4=20, that is -1 mod 7 or 65x5=25 which is 4 mod 75x6=30 which is -5 or 2 mod 7How did we know it existed? Because 7 is a prime. For every prime number p and positive integer n, there exists a finite field with pn elements. This is an important theorem in abstract algebra. Since it is a field, it must have a multiplicative inverse. So the numbers mod 7 make up a field and hence have a multiplicative inverse.
The modulus of a number relative to a base is the positive remainder when itis divided by that base. So -24 mod 7 = (-28 + 4) mod 7 = -28 mod 7 + 4 mod 7 = 4 mod 7 = 4while 24 mod 7 = (21 + 3) mod 7 = 21 mod 7 + 3 mod 7 = 3 mod 7 = 3.
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
Multiplicative inverses are two numbers whose product is one.Another name for multiplicative inverse is reciprocal. The reciprocal of 2/3 is 3/2. 2/3 X 3/2 = 6/6 = 1. The multiplicative inverse of 7 is 1/7. 7 X 1/7 = 7/7 = 1.
To find the units digit of 399 the question being asked is: What is (399) MOD 10? This does not necessitate evaluation of 399 before the modulus is done, as it can be done whenever it is possible during the multiplication as any multiple of 10 multiplied by 3 is still a multiple of 10. The first few powers of 3 modulus 10 are: 31 MOD 10 = 3 32 MOD 10 = (3 x 31) MOD 10 = (3 x 3) MOD 10 = 9 33 MOD 10 = (3 x 32) MOD 10 = (3 x 9) MOD 10 = 27 MOD 10 = 7 34 MOD 10 = (3 x 33) MOD 10 = (3 x 7) MOD 10 = 81 MOD 10 = 1 35 MOD 10 = (3 x 34) MOD 10 = (3 x 1) MOD 10 = 3 36 MOD 10 = (3 x 35) MOD 10 = (3 x 3) MOD 10 = 9 At this point, it can be seen that the answer is a repeating pattern of 3, 9, 7, 1, 3, 9, ... So we need the 99th element of this pattern. The pattern is a repeat of 4 digits, so we calculate 99 MOD 4 = 3. So the 3rd element of the repeating part is the answer: 7. (If the power MOD 4 had been 0, it would have been the 4th element of the pattern: 1)
A multiplicative inverse of 5 mod7 would be a number n ( not a unique one) such that 5n=1Let's look at the possible numbers5x1=5mode 75x2=10=3 mod 75x3=15=1 mod 7 THAT WILL DO IT3 is the multiplicative inverse of 5 mod 7.What about the others? 5x4=20, that is -1 mod 7 or 65x5=25 which is 4 mod 75x6=30 which is -5 or 2 mod 7How did we know it existed? Because 7 is a prime. For every prime number p and positive integer n, there exists a finite field with pn elements. This is an important theorem in abstract algebra. Since it is a field, it must have a multiplicative inverse. So the numbers mod 7 make up a field and hence have a multiplicative inverse.
The modulus of a number relative to a base is the positive remainder when itis divided by that base. So -24 mod 7 = (-28 + 4) mod 7 = -28 mod 7 + 4 mod 7 = 4 mod 7 = 4while 24 mod 7 = (21 + 3) mod 7 = 21 mod 7 + 3 mod 7 = 3 mod 7 = 3.
2 1/3 = 7/3 inverse of 7/3 is 3/7
The multiplicative inverse of -24/7 is -7/24
To find the inverse of a function, simply switch the variables x and y. So for the function y=7x+3, the inverse would be x=7y+3, or y=(x-3)/7.
3 and a half = 7/2 so its multiplicative inverse is 2/7.
Look at the powers of 5 mod 7: 5¹ mod 7 = 5 5² mod 7 = 5 × (5¹ mod 7) mod 7 = (5 × 5) mode 7 = 25 mod 7 = 4 5³ mod 7 = 5 × (5² mod 7) mod 7 = (5 × 4) mod 7 = 20 mod 7 = 6 5⁴ mod 7 = 5 × (5³ mod 7) mod 7 = (5 × 6) mod 7 = 30 mod 7 = 2 5⁵ mod 7 = 5 × (5⁴ mod 7) mod 7 = (5 × 2) mod 7 = 10 mod 7 = 3 5⁶ mod 7 = 5 × (5⁵ mod 7) mod 7 = (5 × 3) mod 7 = 15 mod 7 = 1 5⁷ mod 7 = 5 × (5⁶ mod 7) mod 7 = (5 × 1) mod 7 = 5 mod 7 = 5 At this point, it is obvious that the remainders will repeat the cycle {5, 4, 6, 2, 3, 1} There are 6 remainders in the cycle, so the remainder of 30 divided by 6 will tell you which remainder to use; if the remainder is 0, use the 6th element. 30 ÷ 6 = 5 r 0 →use the 6th element which is 1, so 5³⁰ ÷ 7 will have a remainder of 1. 1 ≡ 5³⁰ mod 7.
3 1/2 = 7/2 in improper fractionmultiplicative inverse is the reciprocal of 7/2 = 2/7
Additive inverse just means that when you add them, it equals 0. So, 1 and -1; 12 and -12; -74 and 74; 0 and -0. It can also be applied to things other than integers: x+12 and -x-12; 2x-3 and -2x+3 (which is the same as 3-2x) Or, if you know anything about modular arithmetic, then: 7+x≡0 (mod 9) and 2; 6-x≡4 (mod 7) and -2
38/7 x 7/38 = 1 = multiplicative inverse 38/7 + -38/7 = 0 = additive inverse
Divide 1 by the number. The multiplicative inverse of 7 is 1/7, for example.
Swap the numerator and denominator. For example, the multiplicative inverse of 5/7 is 7/5