It can be shown by … It can be shown by identifying that it can’t be described as the quotient of two rational numbers. The mathematician who discovered this was thrown overboard from a ship for that discovery. It was mathematical heresy at that time.
We can say that m … We can say that m is even, because since 4 times an integer will always be even, and we have 4k=m(squared), m(squared) must be even. By definition the square root of an even integer is always even. Therefore m is even. Now we have m, and n both being even, which cannot be the case, as m and n could be simplified further, therefore we have a contradiction on the proof that radical 2 is a rational number, so radical 2 must be irrational. (if anyone actually understood that kudos to them)
I will continue … I will continue here. We have 2n(squared)= m(squared). By definition, we could say that n is even, because when you trake n(squared) and multiply it by two, you will come uip with an even number. Then we will take another integer k, and say that 2k=n. So we will plug 2k into the equationfor n, to come up with the equation 2(2k)=m(squared). 2 times 2k equals 4k, so we come up with the equation 4k=m(squared). I will finish my proof on the next post.
the classical proof … the classical proof that radical 2 is irrational is a proof by contradiction. You start by assuming that radical 2 is rational, and since it is you can then say that radical 2 becomes a fraction m/n. You can then place it into an equation radical 2=m/n, with n not equaling zero, and both m and n not being even. Then you would square both sides to get 2=m(squared)/n(squared) you would then multiply n(squared) on both sides of the equation to come up with 2n(squared)=m(squared). I’m out of room
@cjunk351 because …
@cjunk351 because you can go on forever and it will never be exact.
It can be shown by …
It can be shown by identifying that it can’t be described as the quotient of two rational numbers. The mathematician who discovered this was thrown overboard from a ship for that discovery. It was mathematical heresy at that time.
Nice job…what …
Nice job…what level was this being taught to?
Kind of got it.
…
Nice one
Kind of got it.
We can say that m …
We can say that m is even, because since 4 times an integer will always be even, and we have 4k=m(squared), m(squared) must be even. By definition the square root of an even integer is always even. Therefore m is even. Now we have m, and n both being even, which cannot be the case, as m and n could be simplified further, therefore we have a contradiction on the proof that radical 2 is a rational number, so radical 2 must be irrational. (if anyone actually understood that kudos to them)
I will continue …
I will continue here. We have 2n(squared)= m(squared). By definition, we could say that n is even, because when you trake n(squared) and multiply it by two, you will come uip with an even number. Then we will take another integer k, and say that 2k=n. So we will plug 2k into the equationfor n, to come up with the equation 2(2k)=m(squared). 2 times 2k equals 4k, so we come up with the equation 4k=m(squared). I will finish my proof on the next post.
the classical proof …
the classical proof that radical 2 is irrational is a proof by contradiction. You start by assuming that radical 2 is rational, and since it is you can then say that radical 2 becomes a fraction m/n. You can then place it into an equation radical 2=m/n, with n not equaling zero, and both m and n not being even. Then you would square both sides to get 2=m(squared)/n(squared) you would then multiply n(squared) on both sides of the equation to come up with 2n(squared)=m(squared). I’m out of room
How can you prove …
How can you prove that the square root of 2 is irrational??
Fabulous. I can …
Fabulous. I can tell this is a great teacher that these students have.