is 7.4 irrational?
7 answers
No, 7.4 is a rational number. It can be expressed as a fraction 74/10, which is a ratio of two integers.
Find an irrational number between 7 and 8. Explain your answer.
One irrational number between 7 and 8 is √2 + 7. To understand why it's irrational, we first need to establish that √2 is irrational.
Assume √2 is rational, meaning it can be expressed as a fraction p/q, where p and q are integers with no common factors other than 1 and q is not equal to 0. By squaring both sides of the equation √2 = p/q, we get 2 = (p^2)/(q^2), which can further be simplified to p^2 = 2*(q^2).
From this equation, we deduce that p^2 is even since it is equal to 2 times a whole number (q^2). Therefore, p must also be even because the square of an odd number is odd. Consequently, we can write p as 2k, where k is an integer.
Substituting 2k for p in the previous equation, we have (2k)^2 = 2*(q^2), which simplifies to 4k^2 = 2*(q^2) and further reduces to 2k^2 = q^2. This implies that q^2 is even since it is equal to 2 times a whole number (k^2). By extension, q must also be even.
However, we initially stated that p and q have no common factors other than 1, which is a contradiction since we deduced that p and q are both divisible by 2. Therefore, our assumption that √2 is rational is false, and √2 is, in fact, irrational.
Now, to find an irrational number between 7 and 8, we add √2 (an irrational number) to 7. This gives us √2 + 7, which is an irrational number between 7 and 8.
Assume √2 is rational, meaning it can be expressed as a fraction p/q, where p and q are integers with no common factors other than 1 and q is not equal to 0. By squaring both sides of the equation √2 = p/q, we get 2 = (p^2)/(q^2), which can further be simplified to p^2 = 2*(q^2).
From this equation, we deduce that p^2 is even since it is equal to 2 times a whole number (q^2). Therefore, p must also be even because the square of an odd number is odd. Consequently, we can write p as 2k, where k is an integer.
Substituting 2k for p in the previous equation, we have (2k)^2 = 2*(q^2), which simplifies to 4k^2 = 2*(q^2) and further reduces to 2k^2 = q^2. This implies that q^2 is even since it is equal to 2 times a whole number (k^2). By extension, q must also be even.
However, we initially stated that p and q have no common factors other than 1, which is a contradiction since we deduced that p and q are both divisible by 2. Therefore, our assumption that √2 is rational is false, and √2 is, in fact, irrational.
Now, to find an irrational number between 7 and 8, we add √2 (an irrational number) to 7. This gives us √2 + 7, which is an irrational number between 7 and 8.
can you give me a more simple answer?
Certainly! One simple example of an irrational number between 7 and 8 is 7.5. This number cannot be expressed as a fraction and has an infinite non-repeating decimal expansion. Therefore, it is irrational.
is 7.5 irrational?
Apologies for the confusion in my previous response. No, 7.5 is not an irrational number. It can be expressed as a fraction 15/2, making it a rational number. Irreducibility refers to numbers that cannot be expressed as a fraction, while rational numbers can be expressed as a ratio of two integers.
7 answers