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Proof by contradiction: suppose that root 7 (I'll write sqrt(7)) is a rational number, then we can write sqrt(7)=a/b where a and b are integers in their lowest form (ie they are fully cancelled). Then square both sides, you get 7=(a^2)/(b^2) rearranging gives (a^2)=7(b^2). Now consider the prime factors of a and b. Their squares have an even number of prime factors (eg. every prime factor of a is there twice in a squared). So a^2 and b^2 have an even number of prime factors. But 7(b^2) then has an odd number of prime factors. But a^2 can't have an odd and an even number of prime factors by unique factorisation. Contradiction X So root 7 is irrational.
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Can multiplying two irrational numbers give you an irrational number?
Yes. For example, the square root of 3 (an irrational number) times the square root of 2(an irrational number) gets you the square root of 6(an irrational number)
Can you give one example for surd which is not irrational?
It is impossible to have a surd that is not irrational. Surds are defined to be an irrational number (square root of a number).
What systems does the square root of 5 belong?
Square root of 5 is a real number, to start with. It is irrational, also. And there are two values which, when squared, give 5 as the answer.Since they are irrational, we can give approximations: +2.236068 and -2.236068
Can an irrational number be a decimal if so give an example?
Irrational numbers are decimal numbers that can't be expressed as fractions. An example is the square root of 2
Give an example of irrational number?
pi square root of 2 e square root of 3
