You are watching: Is the square root of 8 irrational

This is the proof ns used. Walk anyone recognize if that is correct?

**Assume the the sqrt(8) is a reasonable number. Sqrt(8) = n/m i think no typical factors in between n and m 8 = n^2/m^2 8m^2 = n^2 therefore n^2 should be an even number since anything muliplied by 8 is even.n should be even as well because even numbers squared space even, odd number squared space odd. Represent n together 4kn^2 = (4k)^2 --> 4^2k^2 --> 16k^2 8m^2 = 16k^2m^2 = 2k^2So, m^2 is evenSo, m is evenTherefore m and n have a usual factor, that disproves my assumption**

This is the proof i used. Does anyone know if the is correct? Assume that the sqrt(8) is a reasonable number. Sqrt(8) = n/m i think no usual factors in between n and m 8 = n^2/m^2 8m^2 = n^2 so n^2 should be an also number because anything muliplied by 8 is even.n have to be even too because also numbers squared room even, odd number squared are odd. Represent n together 4kn^2 = (4k)^2 --> 4^2k^2 --> 16k^2 8m^2 = 16k^2m^2 = 2k^2So, m^2 is evenSo, m is evenTherefore m and also n have actually a usual factor, that disproves my assumption

This is the proof i used. Go anyone know if that is correct? Assume the the sqrt(8) is a rational number. Sqrt(8) = n/m assume no typical factors in between n and m 8 = n^2/m^2 8m^2 = n^2 therefore n^2 should be an even number since anything muliplied by 8 is even.n should be even also because also numbers squared space even, odd number squared space odd. Represent n as 4kn^2 = (4k)^2 --> 4^2k^2 --> 16k^2 8m^2 = 16k^2m^2 = 2k^2So, m^2 is evenSo, m is evenTherefore m and also n have a typical factor, it disproves my assumption

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This is the proof ns used. Walk anyone know if it is correct? Assume the the sqrt(8) is a rational number. Sqrt(8) = n/m assume no typical factors in between n and m 8 = n^2/m^2 8m^2 = n^2 therefore n^2 have to be an also number since anything muliplied by 8 is even. //true n^2 will certainly be evenn need to be even too because also numbers squared room even, odd number squared room odd. // however if n^2 is also its not correct to say the n will be also as the square root might or may not exist..its choose saying 8 is even as such sqrt(8) is even which is no true. Represent n as 4k //even if n was also it will be represented as 2k not 4k choose 6 is even but we cant write it as 4kn^2 = (4k)^2 --> 4^2k^2 --> 16k^2 8m^2 = 16k^2m^2 = 2k^2So, m^2 is evenSo, m is evenTherefore m and also n have actually a common factor, it disproves my assumption

This is the proof i used. Does anyone know if the is correct? Assume that the sqrt(8) is a reasonable number. Sqrt(8) = n/m i think no usual factors in between n and m 8 = n^2/m^2 8m^2 = n^2 so n^2 should be an also number because anything muliplied by 8 is even.n have to be even too because also numbers squared room even, odd number squared are odd. Represent n together 4kn^2 = (4k)^2 --> 4^2k^2 --> 16k^2 8m^2 = 16k^2m^2 = 2k^2So, m^2 is evenSo, m is evenTherefore m and also n have actually a usual factor, that disproves my assumption

This is the proof i used. Go anyone know if that is correct? Assume the the sqrt(8) is a rational number. Sqrt(8) = n/m assume no typical factors in between n and m 8 = n^2/m^2 8m^2 = n^2 therefore n^2 should be an even number since anything muliplied by 8 is even.n should be even also because also numbers squared space even, odd number squared space odd. Represent n as 4kn^2 = (4k)^2 --> 4^2k^2 --> 16k^2 8m^2 = 16k^2m^2 = 2k^2So, m^2 is evenSo, m is evenTherefore m and also n have a typical factor, it disproves my assumption

See more: Black Stool After Drinking Red Wine, Is This Normal Or Am I Going To Die

This is the proof ns used. Walk anyone know if it is correct? Assume the the sqrt(8) is a rational number. Sqrt(8) = n/m assume no typical factors in between n and m 8 = n^2/m^2 8m^2 = n^2 therefore n^2 have to be an also number since anything muliplied by 8 is even. //true n^2 will certainly be evenn need to be even too because also numbers squared room even, odd number squared room odd. // however if n^2 is also its not correct to say the n will be also as the square root might or may not exist..its choose saying 8 is even as such sqrt(8) is even which is no true. Represent n as 4k //even if n was also it will be represented as 2k not 4k choose 6 is even but we cant write it as 4kn^2 = (4k)^2 --> 4^2k^2 --> 16k^2 8m^2 = 16k^2m^2 = 2k^2So, m^2 is evenSo, m is evenTherefore m and also n have actually a common factor, it disproves my assumption