Question

proved prove that 7 under root 3 minus 5 upon 6 is irrational when under root 3 is irrational​

Answer 1

Answer:

suppose 3–√ is rational, then 3–√=ab for some (a,b) suppose we have a/b in simplest form.

3–√a2=ab=3b2

if b is even, then a is also even in which case a/b is not in simplest form.

if b is odd then a is also odd. Therefore:

ab(2n+1)24n2+4n+14n2+4n2n2+2n2(n2+n)=2n+1=2m+1=3(2m+1)2=12m2+12m+3=12m2+12m+2=6m2+6m+1=2(3m2+3m)+1

Since (n^2+n) is an integer, the left hand side is even. Since (3m^2+3m) is an integer, the right hand side is odd and we have found a contradiction, therefore our hypothesis is false.

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