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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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Q: Give a proof that the square root of 7 is an irrational number?

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It is a prime number that has only factors of itself and one therefore it is an irrational number like all prime numbers are.

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)

It is impossible to have a surd that is not irrational. Surds are defined to be an irrational number (square root of a number).

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No, but you can add an irrational number and a rational number to give an irrational.For example, 1 + pi is irrational.

sqrt(8) = sqrt(4*2) = 2*sqrt(2).Even without given that sqrt(2) is a rational, you can give that the square root of 2 starts converging onto the "Pythagoras Constant" eventually, as it takes an infinite amount of digits to square root an integer that is not perfectly squared.Thus, an rational x irrational = irrational, thus the sqrt(8) is irrational (an approximation is 2.8284271247...).

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

Any irrational number, when multiplied by 0.5 will give an irrational number.

Any irrational number, added to 0.4 will give an irrational number.

Yes normally it does

Irrational numbers are decimal numbers that can't be expressed as fractions. An example is the square root of 2

The square root of any positive square number is always rational as for example the square root of 36 is 6 which is a rational number.

Any irrational number will do.

It might seems like it, but actually no. Proof: sqrt(0) = 0 (0 is an integer, not a irrational number) sqrt(1) = 1 (1 is an integer, not irrational) sqrt(2) = irrational sqrt(3) = irrational sqrt(4) = 2 (integer) As you can see, there are more than 1 square root of a positive integer that yields an integer, not a irrational. While most of the sqrts give irrational numbers as answers, perfect squares will always give you an integer result. Note: 0 is not a positive integer. 0 is neither positive nor negative.

Well, first let's define a "real number." A real number is any number that's not imaginary. It can be rational or irrational. The square root of 7 is 2.64575131... and it goes on forever. This means it cannot be written as a simple fraction. We can give an estimate of the square root of 7 in a fraction form, but this is not the exact result. So, the square root of 7 is a irrational number.Since a real number can be irrational or rational, yes, the square root of 7 is a real number.

The square root of 153 is an irrational number, so you can't give an exact answer in decimal form, but approximately it would be 12.36931688.

An irrational number cannot be written as a fraction or to an exact decimal such as the symbol for pi or the square root of two. A rational number can be written in the form of a fraction or a decimal to an exact value.

just give me an answer! K? Thanks!

No. The number pi is irrational, and if you multiply an irrational number by a non-zero rational number (in this case, -2), you will get another irrational number.As a general guideline, most calculations that involve irrational numbers will again give you an irrational number.

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An example is the square root of a number. Ex: square root of 2. This is 1 example, not the main one. Any cube root or square root which doesn't give a perfect number is an irrational number. Ex; square root and cube root of 5, since their answer will be 2.24 and 1.70 which are not perfect numbers like square roots of 25 and 64 or cube roots of 27 and 216.

1) Adding an irrational number and a rational number will always give you an irrational number. 2) Multiplying an irrational number by a non-zero rational number will always give you an irrational number.

No. For example, -root(2) + root(2) is zero, which is rational.Note that MOST calculations involving irrational numbers give you an irrational number, but there are a few exceptions.

All integers and fractions are rational numbers whereas irrational numbers can't be expressed as fractions as for example the square root of 2 can't be expressed as a fraction because it is a non-terminating decimal number.