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1 year ago

3. Angles X and Y have a sum of 159 degrees. Angle X is 43 degrees smaller than angle Y.

Mathematics

1 answer:

Lesechka [4]1 year ago

5 0

$\underline{\underline{\large\bf{Solution:-}}}\\$

$\longrightarrow$ Since angle X and Y have a sum of 159

$\implies \bf Y + X= 159 \_\_\_\_(1)$

$\longrightarrow$ Since X is 43 degrees smaller than Y , so subtracting X from Y would give 43 , therefore equation will be written as -

$\implies \bf Y - X = 43 \_\_\_\_(2)$

<u>Adding Equation (1) and (2) </u>

$\begin{gathered}\\\implies\quad \sf Y+X+Y-X = 159+43 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf Y+Y + X-X = 202 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf 2Y = 202 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf Y = \frac{202}{2} \\\end{gathered}$

$\begin{gathered}\\\implies\quad \boxed{\sf {Y = 101 }} \\\end{gathered}$

<u>Putting Y in eq (1)</u>

$\begin{gathered}\\\implies\quad \sf X + 101 = 153 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \sf X = 153-101 \\\end{gathered}$

$\begin{gathered}\\\implies\quad \boxed{\sf{ X = 52 }}\\\end{gathered}$

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This statement can be proven by contradiction for $n \in \mathbb{N}$ (including the case where $n = 0$ .)

$\text{Let $n \in \mathbb{N}$ be a perfect square}$ .

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$\textbf{Case 2.} ~ \text{$n \in \mathbb{N}$ and $n \ne 0$. Hence $n \ge 1$}$ .

$\text{Assume that $n$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $a \in \mathbb{N}$ s.t. $a^2 = n$}$ .

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$\text{$\iff$ $\exists$ $b \in \mathbb{N}$ s.t. $b^2 = n + 2$}$ .

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$\text{$\iff 1 \ge 2\,a $}$ .

$\text{$\displaystyle \iff a \le \frac{1}{2}$}$ .

$\text{Contradiction (with the assumption that $a \ge 1$)}$ .

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$\text{Hence the claim is true for all $n \in \mathbb{N}$}$ .

Step-by-step explanation:

Assume that the natural number $n \in \mathbb{N}$ is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number $a$ ( $a \in \mathbb{N}$ ) such that $a^2 = n$ .

Assume by contradiction that $n + 2$ is indeed a perfect square. Then there should exist another natural number $b \in \mathbb{N}$ such that $b^2 = (n + 2)$ .

Note, that since $(n + 2) > n \ge 0$ , $\sqrt{n + 2} > \sqrt{n}$ . Since $b = \sqrt{n + 2}$ while $a = \sqrt{n}$ , one can conclude that $b > a$ .

Keep in mind that both $a$ and $b$ are natural numbers. The minimum separation between two natural numbers is $1$ . In other words, if $b > a$ , then it must be true that $b \ge a + 1$ .

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by $(a + 1)$ and use the fact that $b \ge a + 1$ to make the left-hand side $b^2$ .)

$b^2 \ge (a + 1)^2$ .

Expand the right-hand side using the binomial theorem:

$(a + 1)^2 = a^2 + 2\,a + 1$ .

$b^2 \ge a^2 + 2\,a + 1$ .

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$\underbrace{b^2}_{=n + 2)} \ge \underbrace{a^2}_{=n} + 2\,a + 1$ .

$n + 2 \ge n + 2\, a + 1$ .

Subtract $n + 1$ from both sides of the inequality:

$1 \ge 2\, a$ .

$\displaystyle a \le \frac{1}{2} = 0.5$ .

Recall that $a$ was assumed to be a natural number. In other words, $a \ge 0$ and $a$ must be an integer. Hence, the only possible value of $a$ would be $0$ .

Since $a$ could be equal $0$ , there's not yet a valid contradiction. To produce the contradiction and complete the proof, it would be necessary to show that $a = 0$ just won't work as in the assumption.

If indeed $a = 0$ , then $n = a^2 = 0$ . $n + 2 = 2$ , which isn't a perfect square. That contradicts the assumption that if $n = 0$ is a perfect square, $n + 2 = 2$ would be a perfect square. Hence, by contradiction, one can conclude that

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