3 quick grammar fix

Consider the hypergeometric function ${}_2F_1(a,b,c;z)$. When $a$, $b$ and $c$ are rational and ${}_2F_1$ is a transcendental function, Siegel sought to prove that apart that--apart from obvious exceptions the exceptions--the function takes transcendental values at algebraic $z$. But it turns out that there are $a$, $b$ and $c$ for which this is false. For example:

$${}_2F_1(1/3,2/3,5/6;27/32)=8/5$$

$${}_2F_1(1/4,1/2,3/4;80/81)=9/5$$

$${}_2F_1(1/12,5/12,1/2;1323/1331)= 11^{1/4}$$

Post Made Community Wiki by S. Carnahan
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Consider the hypergeometric function 2F1(a,b,c:z). ${}_2F_1(a,b,c;z)$. When a,b $a$, $b$ and c $c$ are rational and 2F1 ${}_2F_1$ is a transcendental function, Siegel sought to prove that apart from obvious exceptions the function takes transcendental values at algebraic z. $z$. But it turns out that there are a,b $a$, $b$ and c $c$ for which this is false. For example:

2F1(1/3,2/3,5/6;27/32)=8/5

2F1(1/4,1/2,3/4;80/81)=9/5

2F1(1/12,5/12,1/2;1323/1331)= 11^(1/4)

$${}_2F_1(1/3,2/3,5/6;27/32)=8/5$$

$${}_2F_1(1/4,1/2,3/4;80/81)=9/5$$

$${}_2F_1(1/12,5/12,1/2;1323/1331)= 11^{1/4}$$

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