Q: How do you find three irrational numbers between 4 5?

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The set of irrational numbers is infinitely dense. As a result there are infinitely many irrational numbers between any two numbers. So, if any irrational number, x, laid claim to be the closest irrational number to 3, it is possible to find infinitely many irrational numbers between x and 3. Consequently, the claim cannot be valid.

72 = 49 and 82 = 64. So, the square root of any integer between these two numbers, for example, sqrt(56), is irrational.

Find the difference between the two numbers, then add an irrational number between zero and one, divided by this difference, to the lower number. Such an irrational number might be pi/10, (square root of 2) / 2, etc.

Irrational numbers are infinitely dense. Between any two numbers, there are infinitely many irrational numbers. So if it was claimed that some irrational, x, was the closest irrational to 6, it is possible to find an infinite number of irrationals between 6 and x. Each one of these infinite number of irrationals would be closer to 6 than x. So the search for the nearest irrational must fail.

There may be many easier and better ways, but here's how I would do it: -- Square the first given irrational number. -- Square the second irrational number. -- Pick a nice ugly complicated decimal between the two squares. -- Take the square root of the number you picked. It's definitely between the two given numbers, and it would be a miracle if it's not irrational.

Related questions

It is proven that between two irrational numbers there's an irrational number. There's no method, you just know you can find the number.

the numbers between 0 and 1 is 0.1,0.2,0.3,0.4,0.5,0.6,0.7,0.8,0.9,0.10.

The set of irrational numbers is infinitely dense. As a result there are infinitely many irrational numbers between any two numbers. So, if any irrational number, x, laid claim to be the closest irrational number to 3, it is possible to find infinitely many irrational numbers between x and 3. Consequently, the claim cannot be valid.

72 = 49 and 82 = 64. So, the square root of any integer between these two numbers, for example, sqrt(56), is irrational.

Find the difference between the two numbers, then add an irrational number between zero and one, divided by this difference, to the lower number. Such an irrational number might be pi/10, (square root of 2) / 2, etc.

Irrational numbers are infinitely dense. Between any two numbers, there are infinitely many irrational numbers. So if it was claimed that some irrational, x, was the closest irrational to 6, it is possible to find an infinite number of irrationals between 6 and x. Each one of these infinite number of irrationals would be closer to 6 than x. So the search for the nearest irrational must fail.

0.12=1.1201001000100001 0.13=1.12101001000100001

Irrational numbers are infinitely dense. That is to say, between any two irrational (or rational) numbers there is an infinite number of irrational numbers. So, for any irrational number close to 6 it is always possible to find another that is closer; and then another that is even closer; and then another that is even closer that that, ...

Any number that can't be expressed as a fraction is irrational

There may be many easier and better ways, but here's how I would do it: -- Square the first given irrational number. -- Square the second irrational number. -- Pick a nice ugly complicated decimal between the two squares. -- Take the square root of the number you picked. It's definitely between the two given numbers, and it would be a miracle if it's not irrational.

An irrational number is expressed as a non-repeating decimal that goes on forever. Write out the enough of the decimal expansion of each number to find the first digit where the two numbers disagree. Truncate the larger number at that digit, and the result is a rational number (terminating decimal) that is between the two.

Surds are normally irrational numbers.