1. Problem Statement: Show that each number given is rational, meaning it can be expressed as a fraction $\frac{p}{q}$ where $p,q$ are integers and $q \neq 0$.\n\n2. Number 0.123123123... : This decimal repeats the block "123" indefinitely. Let $x = 0.123123123...$. Multiply by $10^3=1000$ to shift the decimal:\n$$1000x = 123.123123...$$\nSubtract the original $x$:\n$$1000x - x = 123.123123... - 0.123123... = 123$$\nSo,\n$$999x = 123 \implies x = \frac{123}{999} = \frac{41}{333}$$\nWhich is rational.\n\n3. Number 0.217171717... : This is $0.2$ followed by the repeating block "17". Let $x=0.2171717...$. Multiply by $10$ to move past the non-repeating digit:\n$$10x = 2.171717...$$\nMultiply by $10^2=100$ to move past one repeating block:\n$$1000x = 217.171717...$$\nSubtract $10x$:\n$$1000x - 10x = 217.171717... - 2.171717... = 215$$\nTherefore,\n$$990x = 215 \implies x = \frac{215}{990} = \frac{43}{198}$$\nRational number.\n\n4. Number 2.56565656... : Let $x = 2.565656...$ (repeating '56'). Multiply by $100$:\n$$100x = 256.5656...$$\nSubtract original $x$:\n$$100x - x = 256.5656... - 2.5656... = 254$$\nSo,\n$$99x = 254 \implies x = \frac{254}{99}$$\nRational number.\n\n5. Number 3.929292... : Let $x=3.929292...$ (repeating '92'). Multiply by $100$:\n$$100x = 392.929292...$$\nSubtract $x$:\n$$100x - x = 392.929292... - 3.929292... = 389$$\nThus,\n$$99x = 389 \implies x = \frac{389}{99}$$\nRational number.\n\n6. Number 0.199999... : This is a repeating '9' after '0.1'. Let $x=0.199999...$. Multiply by $10$:\n$$10x = 1.99999...$$\nSubtract $x$:\n$$10x - x = 1.99999... - 0.199999... = 1.8$$\nWhich gives,\n$$9x = 1.8 \implies x = \frac{1.8}{9} = 0.2 = \frac{1}{5}$$\nRational number.\n\n7. Number 0.399999... : Similar process, let $x = 0.399999...$, multiply by $10$:\n$$10x = 3.99999...$$\nSubtract $x$:\n$$10x - x = 3.99999... - 0.39999... = 3.6$$\nSo,\n$$9x = 3.6 \implies x = \frac{3.6}{9} = 0.4 = \frac{2}{5}$$\nRational number.\n\nAll given numbers can be expressed as a fraction of two integers, so all are rational.