Question

(√2 + √3) (√2 + √3) is a​

Answer 1

Answer: (√2 + √3) (√2 + √3) is a​ n Irrational number,,

Answer 2

Answer:

HII!

Step-by-step explanation:

The answer is 1.

1. Explanation:

Using the rule: (a+b) (a-b) = a^2 - b^2

Our equation: (2+sqrt3) (2-sqrt3) = 2^2 - sqrt3^2

= 4 - 3 = 1

Notes:

1.1 The definition of square root:

sqrt(n) * sqrt(n) = n.

This is by the definition of square root:

…. If r = sqrt(n)

…. then r X r = n.

1.2 Any n which is a result of squaring - is necessarily a positive number.

1.3 The square root of three

Unimportant information follows:

The ancient Greeks proved that the square root of 3 is an irrational number,

that cannot be obtained by dividing one number by another.

In decimal notation you will always need to “add numbers after the dot” to get more and more precise but that process will never end.

Fortunately, in our case there’s no problem because the squaring of the square root simply eliminates the term and we get the positive number of 3.

… sqrt(3) x sqrt(3) = 3.

2. Explanation of the explanation

2.1 Brute force (a+b)(c+d)

Any equation (a + b) (c +d) is solved as follows:

Example:

(2 + 3)(4 + 5) = Lets calculate each term in the parenthesis:

First term: (2+3) = 5

Second term: (4+5) = 9

So (2+3) (4+5) = 5 x 9 = 45. Clear?

2.2 Rule: a(b+c) = ab + ac

But we can use a rule: because a(b+c) = ab + ac

Example:

2 (3+4) = (2x3) + (2x4). Lets check that to see why its true

Take a chockolate bar with rows of 3. Then add to the row another 4 squares.

You now have rows of 7 (that’s 3 + 4).

When you count squares in two rows of 7 ,

its the same as counting two rows of 3 then counting two rows of 4.

[] [] [] + [] [] [] [] = [] [] [] [] [] [] []

[] [] [] _ [] [] [] [] _ [] [] [] [] [] [] []

Lets just check:

…2 x 7 = 2 x 3 + 2 x 4 ??

…14 = 6 + 8. yup!

3. Rule: (a+b) (c+d) = ac + ad + bc + bd

This is simply a breakdown of rule 2.1

3.1 Definition: (a+b) is a number. Lets call it n. n = (a+b)

We can write (a+b) x (c + d) = n (c + d)

which according to 2.1 (remember the chockolate bar?) is :

… n(c+d) = nc + nd.

But lets open up the n in nc + nd

Remember n = (a+b) ? Take a look back. That was our definition in 3.1.

nc = (a+b)c

Lets start with the nc: which is actually (a+b)c

According to rule 2.2 we can open the parenthesis (a+b)c

(a+b)c = c(a+b) = ca + cb

Next we’ll do the same with nd: which is actually (a+b)d

(a+b)d = d(a+b) = da + db

Wrapping up:

(a+b)(c+d) = ab+ac + bc+bd

ok?

4. The (a+b)(a-b) rule

4.1 Some notes about negative numbers:

4.1.1 Adding -n instead of subtracting n

The subtraction of a-b is the same as adding -b to a.

…a-b = a + (-b) = a + -b

…a + (-b) = a-b

4.1.2 Negative numbers are number multiplied by -1

-b = b(-1) = (-1)b

4.1.3 Multiplication with negative numbers

Multiplying by a negative number is multiplying by the number and then negating.

a (-b) = -1 ab = -ab

(-a)(-b) = (-1)a(-1)b = (-1)(-1)ab = ab

4.1.4 Negation by subtraction

ab + -ab = ab - ab = 0

ab + -ab = -ab + ab = 0

4.2 Solution:

Lets open our equation up according to rule 3:

….(a+b)(a-b) = (a+b)(a + -b)

….= aa + a(-b) + ba + b(-b)

….= aa + -ab + ab + -bb

…. = aa - bb