prove that is a irrational number 5+7root3
![Let us assume, 5−3 is a rational number \ \textless \ br /\ \textgreater \ ⇒5−3=qp, where p,q∈z,q=0\ \textless \ br /\ \textgreater \ 5−qp=3\ \textless \ br /\ \textgreater \ ⇒q5q−p=3\ \textless \ br /\ \textgreater \ ⇒3 is a rational number ∵q5q−p is rational\ \textless \ br /\ \textgreater \ but 3 is not a rational number.\ \textless \ br /\ \textgreater \ This gives us a contradiction. \ \textless \ br /\ \textgreater \ ∴ our assumption that 5−3 is a rational number is wrong \ \textless \ br /\ \textgreater \ ⇒5−3 is an irrational number. \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \](https://tex.z-dn.net/?f=Let+us+assume%2C%C2%A05%E2%88%923%E2%80%8B%C2%A0is+a+rational+number%C2%A0%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+%E2%87%925%E2%88%923%E2%80%8B%3Dqp%E2%80%8B%2C+where%C2%A0p%2Cq%E2%88%88z%2Cq%EE%80%A0%3D0%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+5%E2%88%92qp%E2%80%8B%3D3%E2%80%8B%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+%E2%87%92q5q%E2%88%92p%E2%80%8B%3D3%E2%80%8B%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+%E2%87%923%E2%80%8B%C2%A0is+a+rational+number%C2%A0%E2%88%B5q5q%E2%88%92p%E2%80%8B%C2%A0is+rational%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+but%C2%A03%E2%80%8B%C2%A0is+not+a+rational+number.%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+This+gives+us+a+contradiction.%C2%A0%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+%E2%88%B4%C2%A0our+assumption+that%C2%A05%E2%88%923%E2%80%8B%C2%A0is+a%C2%A0rational+number+is+wrong%C2%A0%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+%E2%87%925%E2%88%923%E2%80%8B%C2%A0is+an+irrational+number.%C2%A0%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+%5C++%5Ctextless+%5C+br+%2F%5C++%5Ctextgreater+%5C+ "Let us assume, 5−3 is a rational number \ \textless \ br /\ \textgreater \ ⇒5−3=qp, where p,q∈z,q=0\ \textless \ br /\ \textgreater \ 5−qp=3\ \textless \ br /\ \textgreater \ ⇒q5q−p=3\ \textless \ br /\ \textgreater \ ⇒3 is a rational number ∵q5q−p is rational\ \textless \ br /\ \textgreater \ but 3 is not a rational number.\ \textless \ br /\ \textgreater \ This gives us a contradiction. \ \textless \ br /\ \textgreater \ ∴ our assumption that 5−3 is a rational number is wrong \ \textless \ br /\ \textgreater \ ⇒5−3 is an irrational number. \ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \")
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