Is there an easy way to raise the imaginary number i to the power of 37? (Written on a computer as i^37)
There is! It is as easy as dividing the power by 4 and looking at the remainder for the answer. Let me explain: First, think about the first four powers of i, because we will use these answers: i^1 = the square root of negative 1 or just i 1^2 = i times itself or (i )(i) = (the square root of negative 1)(the square root of negative 1) = -1 i^3 = i times i squared or (i^1)(i^2). We just did those. Let us substitute from above so (i^1)(i^2) is i (-1) = -i i^4 = (i^2) (i^2) and substituting from above (-1)(-1) = 1 Let us think about the power of i to the 4th power equaling 1, or i^4 = 1 This means i^8 = (i^4)(i^4) = (1)(1) = 1 and i^12 = (i^4)(i^4) (i^4) = (1)(1)(1) = 1 and i to any power that is divisible by 4 will also equal 1. This is fantastic! What is i^37? It is (i^36)(i^1) and now we know that i^36 = 1 and (from above) (i^1) =i. Thus, (i^36)(i^1) = 1i or just i. In short, look at the power i is raised to and divided it by four. Whatever the remainder is will be the answer. 37 divided by 4 = 9 remainder 1. We only care about the remainder of 1, with the answer being i raised to the power of 1. More examples: i^18 Easy way: 18 divided by 4 = 4 remainder 2. Then, i^18 = i^2 or (from above) = -1 Proof: i^18 = (i^16)(i^2) = 1 (-1) = -1
Why doesn't -x squared equal (-x) squared?
Remember that the squaring function--unless it is outside of a parentheses--only applies to what it is right next to. So, negative x raised to the power of two means that you square the x and then make it negative. Effectively - (x)(x) equals negative x squared. On the other hand, when the negative is next to the x on the inside of the parentheses, you multiply the negative x by itself twice. In other words (-x) times (-x) is equal to a positive x squared.
The sum of three consecutive integers is 36. Find the integers.
First of all, integers include the number 0, the counting numbers like 1, 2, 3 and so on, and their opposites such as -1, -2, -3 and so on. Next, we consider what it means to have consecutive numbers. Well, 1, 2, and 3 are consecutive numbers. Notice that 2 can be thought of as 1+1 and 3 can be thought of as 1+2. After this we consider that sum means the answer to an addition problem. Finally, we bring all of this together. We know nothing about the first integer, so we can call this x. The next consecutive integer will be one more than x or simply, x+1. The third consecutive integer will be 2 more than x or simply, x+2. Now we know the sum of these numbers is 36, so we are going to add these expressions that represent the three integers to create an equation: x + x+1 + x+2 = 36 --> Gathering like terms this becomes 3x + 3 = 36 --> Subtracting 3 from both sides we have 3x = 33 --> Dividing both sides of the equation by 3, we arrive at x = 11. Finally, we find the first integer is 11, the next consecutive integer is x + 1 or 12, and the last consecutive integer is 13. Checking our work, 11+12+13=36. Excellent!