# Lesson Video: Operations on Real Numbers Mathematics

In this video, we will learn how to solve problems involving operations and properties of operations on real numbers.

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### Video Transcript

In this video, we’ll learn how to solve problems involving operations and properties of operations on real numbers. These will include calculations involving surds or radicals, such as the square root of two or the square root of five, and will involve us considering inverse operations. In other words, opposite operations.

We begin by defining some terms. When we talk about the inverse in mathematics, we’re talking about something opposite in effect. An inverse operation is an operation which undoes what was done by the previous operation. We’re going to be looking at the additive inverse and the multiplicative inverse. Now, the additive inverse of a number 𝑎 is the number that when added to 𝑎 gives zero. And a multiplicative inverse of a number 𝑎 is the number that when multiplied by 𝑎 gives one. Another way of considering this is as the reciprocal, one over 𝑎. Let’s have a look at how we can calculate these.

Find the additive inverse of eight minus the square root of 135.

Remember, the additive inverse of a number 𝑎 is the number that when added to 𝑎 gives zero. So, we need to find a number that when we add it to eight minus the square root of 135, we get zero. And one way to answer this is to use algebra. Let’s let 𝑥 be the additive inverse of eight minus the square root of 135. Then, we know that the sum of 𝑥 and eight minus the square root of 135 is zero. And since we’re simply adding here, we don’t actually need these parentheses or brackets.

We want to find the value of 𝑥. Remember, we’re trying to find the additive inverse of our number. And so, we’re going to solve this equation. We have 𝑥 plus eight minus the square root of 135. So, we begin by subtracting eight from both sides of our equation. On the left-hand side, that leaves us with 𝑥 minus the square root of 135. And on the right-hand side, we get negative eight. So, 𝑥 minus the square root of 135 is equal to negative eight. The opposite of subtracting is adding. So, next, we add the square root of 135 to both sides. And so, we see that 𝑥 is equal to negative eight plus the square root of 135. And it’s quite usual to write the positive number first.

And so, we can say that the additive inverse of eight minus the square root of 135 is the square root of 135 minus eight. Now, it follows that since we know that the additive inverse of a number and that number sum to zero. We can check our solution by adding eight minus the square root of 135 and the square root of 135 minus eight. So, that’s eight minus the square root of 135 plus the square root of 135 minus eight. So, we see that eight minus eight is zero and negative root 135 plus root 135 is zero. So, we get zero as required.

Next, we’ll look at an example of how to find the multiplicative inverse.

Find the multiplicative inverse of the square root of six over 30.

Remember, the multiplicative inverse of a number 𝑎 is the number that when we multiply it by 𝑎 gives us one. Another way of considering this is as the reciprocal of that number. So, if we have a number 𝑎, its reciprocal is one over 𝑎. So, here, we need to find the number that when we multiply it by the square root of six over 30, we get one. If we let 𝑥 be the multiplicative inverse of the square root of six over 30, then we could say that 𝑥 times the square root of six over 30 is equal to one.

Now, equivalently, we would achieve this by solving this equation. We said that it is also the reciprocal of the original number. That’s one over the number. So, it’s one over the square root of six over 30. This doesn’t look very nice, though. So, we’re going to recall how we divide fractions. Really, we’re wanting to divide one by the square root of six over 30. So, we write one as one over one. And then recall that to divide by a fraction, we multiply by the reciprocal of that fraction. This is sometimes called keep, change, flip. So, 𝑥 is equal to one over one times 30 over root six.

And if we multiply the numerators and then separately multiply the denominators of our fractions, we get 𝑥 is equal to 30 over root six. Now, in fact, we really didn’t need to perform this step. Given a fraction in the form 𝑎 over 𝑏, its reciprocal is simply 𝑏 over 𝑎. But of course, it’s always good to understand where these things come from. So, we found the multiplicative inverse to be 30 over the square root of six.

But we’re really not finished. We need to rationalize the denominator. In other words, we want the denominator of our fraction to be rational. At the moment, it’s an irrational number. The square root of six cannot be written as a fraction where the numerator and denominator are integers. So, how do we achieve this? Well, we multiply the numerator and denominator of our fraction by the square root of six. That’s the same as multiplying by the square root of six over the square root of six, or just by multiplying by one.

And in doing so, all we’re doing is creating an equivalent fraction. 30 times the square root of six is 30 root six. Then, the square root of six times itself is, of course, simply six. Multiplying a number by itself is squaring it and squaring is the inverse to square rooting. So, we see that our multiplicative inverse is 30 root six over six. Finally, we spot that both 30 and six have a common factor of six. And so, dividing through by six, we get five root six over one, which is simply five root six. The multiplicative inverse of the square root of six over 30 is five root six.

In our next example, we’ll look at how to find the multiplicative inverse of the sum of two radicals.

Find the multiplicative inverse of root six plus root seven, expressing your answer in simplest form.

Remember, the multiplicative inverse of a number 𝑎 is the number that when multiplied by 𝑎 gives one. It’s the reciprocal of that number, one over 𝑎. We’re looking to find the multiplicative inverse of root six plus root seven. So, that’s the reciprocal of root six plus root seven. It’s one over that expression. The problem is, we’re not quite finished. We need to give our answer in simplest form. In other words, we need to rationalize the denominator.

Currently, our denominator is an irrational expression. It cannot be written as a fraction where both the numerator and denominator of that fraction are integers; they’re whole numbers. So, how do we rationalize the denominator of our fraction? We might recall that multiplying a radical expression by its conjugate, that is, multiplying it by an expression where we change the sign between the two terms, gives us a rational result. So, if we change the sign here, we see that the conjugate of root six plus root seven is root six minus root seven.

We can’t just multiply the denominator of our fraction though. We have to do the same to the numerator. This is essentially like multiplying by one. So, we create an equivalent fraction. Let’s begin by evaluating the denominator of our fraction. We’re going to multiply root six plus root seven by root six minus root seven. There are a number of techniques we can use. Let’s use the FOIL method for distributing parentheses or expanding brackets.

We multiply the first term in each expression. The square root of six times the square root of six is simply six. We then multiply the outer terms. That’s the square root of six times negative the square root of seven. By recalling that for real numbers 𝑎 and 𝑏, the square root of 𝑎 times the square root of 𝑏 is the square root of 𝑎𝑏. We see that the square root of six times the square root of seven is root 42. So, our second term is negative root 42. We then multiply the inner terms, and we get positive root 42. Finally, we multiply the last terms in each expression. And since root seven times itself is simply seven, we get negative seven.

We notice now that negative root 42 plus root 42 is zero. And so, we’re left with six minus seven, which is simply negative one. Now, distributing the parentheses on our numerator is a little bit easier. One times root six and one times negative root seven gives us root six minus root seven. So, we have root six minus root seven over negative one.

There’s just one more step. We’re going to divide each term on our numerator by negative one. Root six divided by negative one is negative root six and negative root seven divided by negative one is positive root seven. So, we get negative root six plus root seven, which we can write as root seven minus root six. So, we see that the multiplicative inverse of root six plus root seven is root seven minus root six.

Now, of course, we can always check our result by finding the product by multiplying these two values together. We use the FOIL method again. Root six times root seven is root 42. We then multiply root six by negative root six to get negative six. Root seven times root seven is seven. Then, root seven times negative root six is negative root 42. The root 42s cancel and we’re left with one as required.

We’ll now look at how to apply these processes to a geometric problem.

Given that the dimensions of a rectangle are 57 plus seven root two centimeters and 57 minus seven root two centimeters, find the length of its perimeter.

Remember, the perimeter of a rectangle is the total distance around the outside of the shape. Let’s sketch the rectangle out and label its dimensions. We know that opposite sides in a rectangle are equal in length. This means we can label our opposite sides as shown. The perimeter is then the sum of all of these dimensions. It’s 57 minus seven root two plus 57 plus seven root two plus 57 minus seven root two plus 57 plus seven root two. And in fact, since we’re simply finding the sum, we don’t actually need these brackets or parentheses.

Next, we notice that negative seven root two and seven root two are additive inverses of one another. An additive inverse of a number 𝑎 is the number that when we add to 𝑎 gives zero. This means the sum of negative seven root two and seven root two is zero. So, we have zero here and zero here. The perimeter is, therefore, simply 57 plus 57 plus 57 plus 57 or 57 times four. That’s 228 or 228 centimeters. And so, the length of the perimeter of our shape is 228 centimeters.

We’ll now consider the effect of squaring on algebraic expressions involving square roots.

Given that 𝑎 is equal to the square root of two and 𝑏 is equal to the square root of six, find the value of 𝑎 squared over 𝑏 squared.

Recall the order of operations. This is sometimes called BIDMAS or PEMDAS. These tell us the order in which we perform a series of operations. And so, looking at our expression 𝑎 squared over 𝑏 squared, we see we have exponents or indices. And this line here means division. In each expression, BIDMAS and PEMDAS, the exponents or indices are calculated before any divisions. So, we’re simply going to begin by calculating the value of 𝑎 squared and 𝑏 squared.

𝑎 is the square root of two and 𝑏 is the square root of six, which means that 𝑎 squared must be root two squared and 𝑏 squared must be root six squared. We could rewrite each of these as the square root of two times the square root of two and root six times root six, respectively. Alternatively, we recall that square rooting and squaring are inverse operations. They are the opposite of one another, and they undo what the other one does.

This means the square root of two squared is simply two, whilst the square root of six squared is six. So, we’ve calculated 𝑎 squared and 𝑏 squared. We replace 𝑎 squared with two and 𝑏 squared with six in our original expression. And we see that 𝑎 squared over 𝑏 squared is two over six. And since both the numerator and denominator of our fraction share a common factor of two, we divide them both by two. Two divided by two is one and six divided by two is three. So, the value of 𝑎 squared over 𝑏 squared is one-third.

In our very final example, we’ll consider one more real-life application of these processes.

A square has a side length of 𝑙 centimeters and an area of 63 square centimeters. Find the area of a square whose side length is six 𝑙 centimeters.

Remember, the area of a rectangle is calculated by multiplying its width by its height. A square is simply a rectangle whose sides are the same length. And so, we can say that the area of a square is its side length multiplied by itself or its side length squared. Now, we’re told that our square has an area of 63 square centimeters. We’re also told that its side length is 𝑙. So, replacing area with 63 and side length with 𝑙, we form an equation. We get 63 is equal to 𝑙 squared.

Now, the question wants us to find the area of a square whose side length is six 𝑙 centimeters. So, what we’re going to do is begin by calculating the value of 𝑙. In other words, we’re going to solve this equation for 𝑙. To do so, we perform an inverse operation. Currently, 𝑙 is being squared. The opposite of squaring is square rooting a number. This essentially undoes the previous operation.

And so, if we square both sides of our equation, we get simply 𝑙 on the right-hand side. Then, the left-hand side is equal to the square root of 63. Now, it’s worth recalling that when we find the square root in an equation, we look to find the positive and negative square roots of the number. But this is a side length, so we can’t have a negative value. And 𝑙 is equal to the square root of 63. We might be interested in simplifying this radical or surd. But in fact, we’re not quite done with it. So, we’ll leave it as it is for now.

Our new square has a side length of six 𝑙 centimeters. We calculated 𝑙 to be equal to the square root of 63. So, six 𝑙 must be six times this. It’s six root 63. Since this is the new side length of our square, the area of this square is this value squared. It’s six root 63 times six root 63. We distribute the two over both parts of this value. So, we get six squared times the square root of 63 squared. Six squared is 36.

And of course, squaring and finding the square root are inverse operations of one another. They undo the other operation. So, the square root of 63 squared is just 63. This means the area of our square is 36 times 63, which is 2,268. And since we’re working in centimeters, the units here are square centimeters.

In this video, we’ve learned that when we talk about the inverse in mathematics, we’re talking about something opposite in effect. An inverse operation is one which undoes what was done by the previous operation. We also learned about additive inverses. We said the additive inverse of a number 𝑎 is the number that when we add it to the original number 𝑎 gives zero. We also saw that multiplicative inverses of a number 𝑎 are the numbers that when multiplied by 𝑎 give one. And another way of considering this is as the reciprocal, one over 𝑎.