13. Let S, be the area of the right triangle that is formed by the straight line tx + (t+1)y-2=0 with the two coordinate axes. Suppose S1+S2 + S3+ +

1. Assume the contrary, namely that √2 + √3 + √5 = r, where r is a rational number.

Square the equality √2 + √3 = r − √5 to obtain 5 + 2

√6 = r2 + 5 − 2r

√5. It follows

that 2√6 + 2r

√5 is itself rational. Squaring again, we find that 24 + 20r2 + 8r

√30

is rational, and hence √30 is rational, too. Pythagoras’ method for proving that √2 is

irrational can now be applied to show that this is not true. Write √30 = m

n in lowest

terms; then transform this into m2 = 30n2. It follows that m is divisible by 2 and because

2( m

2 )2 = 15n2 it follows that n is divisible by 2 as well. So the fraction was not in lowest

terms, a contradiction. We conclude that the initial assumption was false, and therefore

√2 + √3 + √5 is irratio

what r u talking about dude?

Step-by-step explanation: