Prove that if A - B and B - C then A - CHint - Start with an arbitrary element x - A and show that x - C

Given: A #8834;B and B #8834;C To prove: A #8834;C Proof: Let #160;x #8712;A #8756; #160;x #8712;B #160; #160; #160; #160; #160; #1 ...

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Prove that if...

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Prove that if A $\subset$ B and B $\subset$ C then A $\subset$ C
(Hint : Start with an arbitrary element x $\in$ A and show that x $\in$ C)

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A = x - 2, B = y + 2

C = 2 y - x

A + B + C = 3 y

\frac{a + b x}{a - b x} = \frac{b + c x}{b - c x} = \frac{c + d x}{c - d x}

a = \frac{x}{y + z}

b = \frac{y}{z + x}

c = \frac{z}{x + y}

\frac{a}{1 + a} + \frac{b}{1 + b} + \frac{c}{1 + c} = 1

\frac{y - z}{a (b - c)} = \frac{z - x}{b (c - a)} = \frac{x - y}{c (a - b)}

\frac{x}{3 x - y - z} = \frac{y}{3 y - z - x} = \frac{z}{3 z - x - y}

x + y + z \neq 0

\frac{a x + b y}{x + y} = \frac{b x + a z}{x + z} = \frac{a y + b z}{y + z}

x + y + z \neq 0

\frac{a + b}{2}

\frac{y + z}{a} = \frac{z + x}{b} = \frac{x + y}{c}

\frac{x}{b + c - a} = \frac{y}{c + a - b} = \frac{z}{a + b - c}

\frac{3 x - 5 y}{5 z + 3 y} = \frac{x + 5 z}{y - 5 x} = \frac{y - z}{x - z}

\frac{x}{y}

\int \frac{d x}{\sqrt{{sin}^{3} x {cos}^{5} x}} = a \sqrt{cot x} + b \sqrt{{tan}^{3} x} + c

^{'} a^{'}

^{'} b^{'}

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