Which is the irrational number?[tex]-\frac{16}{4}[/tex], [tex]\sqrt[3]{64}[/tex], [tex]\frac{15}{8}[/tex], or [tex]\sqrt{26}[/tex]

Accepted Solution

a rational value is a value that can be expressed as a ration or fraction, so -16/4 and 15/8 are pretty much there already so we can skip those two.now let's take a peek at the roots, βˆ›(64), well 4Β³ = 64, thus βˆ›(64) = βˆ›(4Β³) = 4, and we can write any integer is over 1, namely 4/1, so that's rational.[tex]\bf \sqrt{26}\implies \sqrt{2\cdot 13}\implies \sqrt{2}\cdot \sqrt{13}[/tex]well, 2 and 13 are both prime numbers, so they can't be factored into two values that will ever give either one, so we have one irrational value, √2 times another irrational value √13 in this case, so, no way we can express those in a rational fashion.factoid:Ancient Greeks were aware of √2 as irrational, and kinda never liked it very much, thus they didn't deal with it much.