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$10$ a Use a calculator to find $\newline$ $1$ . $1$ Irrational numbers $\newline$ i $\newline$ $(\sqrt{2}+1)\times(\sqrt{2}-1)$ $\newline$ ii $\newline$ $(\sqrt{3}+1)\times(\sqrt{3}-1)$ $\newline$ b Continue the pattern of the multiplications in part a. $\newline$ iii $\newline$ $(\sqrt{4}+1)\times(\sqrt{4}-1)$ $\newline$ c Generalise the results to find $\newline$ $(\sqrt{N}+1)\times(\sqrt{N}-1)$ where $\newline$ $N$ is a positive integer. $\newline$ d Check your generalisation with further examples. $\newline$ $11$ Here is a decimal: $5.020020002000020000$ examples.

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Negative exponents

Full solution

Q.

10

a Use a calculator to find

\newline

1

.

1

Irrational numbers

\newline

i

\newline

(\sqrt{2}+1)\times(\sqrt{2}-1)

\newline

ii

\newline

(\sqrt{3}+1)\times(\sqrt{3}-1)

\newline

b Continue the pattern of the multiplications in part a.

\newline

iii

\newline

(\sqrt{4}+1)\times(\sqrt{4}-1)

\newline

c Generalise the results to find

\newline

(\sqrt{N}+1)\times(\sqrt{N}-1)

where

\newline

N

is a positive integer.

\newline

d Check your generalisation with further examples.

\newline

11

Here is a decimal:

5.020020002000020000

examples.

Calculate Formula: Calculate $(\sqrt{2}+1)\times(\sqrt{2}-1)$ Use the difference of squares formula: $(a+b)(a-b) = a^2 - b^2$ Here, $a = \sqrt{2}$ and $b = 1$ . $(\sqrt{2}+1)\times(\sqrt{2}-1) = (\sqrt{2})^2 - (1)^2 = 2 - 1 = 1$
Use Difference of Squares: Calculate $(\sqrt{3}+1)\times(\sqrt{3}-1)$ $\newline$ Again, use the difference of squares formula: $(a+b)(a-b) = a^2 - b^2$ $\newline$ Here, $a = \sqrt{3}$ and $b = 1$ . $\newline$ $(\sqrt{3}+1)\times(\sqrt{3}-1) = (\sqrt{3})^2 - (1)^2$ $\newline$ $= 3 - 1$ $\newline$ $= 2$
Calculate Result: Calculate $(\sqrt{4}+1)\times(\sqrt{4}-1)$ Use the difference of squares formula: $(a+b)(a-b) = a^2 - b^2$ Here, $a = \sqrt{4}$ and $b = 1$ . $(\sqrt{4}+1)\times(\sqrt{4}-1) = (\sqrt{4})^2 - (1)^2 = 4 - 1 = 3$
Generalize for $\sqrt{N}$ : Generalize the result for $(\sqrt{N}+1)\times(\sqrt{N}-1)$ Using the pattern from the previous steps, we can generalize: (\sqrt{N}+\(1)\times(\sqrt{N} $-1$ ) = (\sqrt{N})^ $2$ - ( $1$ )^ $2$ = N - $1$
Check with Examples: Check the generalization with further examples $\newline$ Let's check with $N = 5$ and $N = 6$ : $\newline$ For $N = 5$ : $\newline$ $(\sqrt{5}+1)\times(\sqrt{5}-1) = (\sqrt{5})^2 - (1)^2$ $\newline$ $= 5 - 1$ $\newline$ $= 4$ $\newline$ For $N = 6$ : $\newline$ $(\sqrt{6}+1)\times(\sqrt{6}-1) = (\sqrt{6})^2 - (1)^2$ $\newline$ $= 6 - 1$ $\newline$ $= 5$ $\newline$ Both results match the generalization.

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