Do irrational numbers only have Non-terminating and non-repeating decimal.?
Yes, yes, and yes. Any radical that does not simplify to a whole number is always irrational.
Irrational numbers have decimals that are non terminating with no repeating block of digits?
TRUE. Examples are pi, e, sqrt(2), sqrt(7), etc
How do we know if non terminating non repeating number is a rational or irrational?
If you start converting a rational number r/s with 0 < r < s, then we divide 10×r by s and we get (10×r)=(q_1)s +(r_1), with 0 r/s = (q_1)/10 + (r_1)/10.s, and so (r/s) = 0.(q_1) + (r_1)/10s, We next divide 10(r_1) by s, where 10(r_1)=(q_2)s + (r_2), with 0 (r/s) = 0.(q_1)(q_2) + (r_2)/(10^2)s. In this way, either we get some remainder (r_j) as 0, in which case the decimal representation terminates or else we go on getting an ongoing sequence of quotients and remainders. But these remainders being necessarily < s, can not be all different, and in at most s+1 steps, some remainder is equal to an earlier remainder and as the divider is s each time, the whole pattern of remainders repeat and so on. Thus we are forced to get either a terminating or a recurring decimal, with its period of recurrence at most 's'.Therefore a non-terminating and non-repeating decimal always represents an irrational number.
True or false, a rational number either terminates or repeats?
Yes.. A *terminating* decimal like 1.32 = 132/100. ANY terminating decimal is a rational number. (So an irrational number certainly can't have a terminating decimal representation For most rational numbers, the decimal representation can't terminate. The decimal representation of a non-terminating rational number eventually repeats -- you get to a block of digits which repeats over & over. Conversely, such a number *must* be rational. For example, 0.123 670 670 670 670 .......... is a rational number. We can subtract the rational number 0.123 = 123/1000, & look at 0.000 670 670 670 ... For convenience, multiply this by 10^6: 670.670 670 670 670 .... = 670 + 670/10^3 + 670/10^6 + 670/10^9 + .... + 670/10^(3n) + .... = 670 ( 1 + 1/10^3 + (1/10^3)^2 + (1/10^3)^3 + .... + (1/10%3)^n + ....) This is a geometric series: 1 + r + r^2 + r^3 + .... + r^n + .. = 1/(1 -- r) provided |r| < 1. In this case, r = 1/1000; sum of series is 1/(1 - 1/1000) = 1000/999, a rational number. Then multiply by 670 to get 670,000/999. Now divide by 10^6 to get the rational number 670/(999,000) . [this business of multiplying & dividing by 10^6 was just for convenience.] 1.234567891011121314151617........... is definitely an irrational number because there is no repeating block. Actually it is a *transcendental* number: it is not the root of any polynomial with integer coefficients, such as x^6 - 13 x^4 + 2x^3 + x + 97 . ...........
Irrational numbers have decimals that either terminate or are nonterminating with a repeating block of digits?
It's false. Replace "Irrational" with "Rational" and you'd get the true statement "Rational numbers have decimals that either terminate or are nonterminating with a repeating block of digits" Irrational numbers will be nonterminating, and they won't have a repeating block of digits.
Are all irrational numbers non-terminating in their decimal expansion?
Yes, they must be. If you had a decimal representation which had a finite number of digits followed by zeroes, then that would be a rational number - you could convert that number into the form p/q where p and q are integers - just multiply by a power of ten big enough to shift all the decimal digits to the left of the decimal point.Here’s an example r = 1.399393939393399888888393then we can write this as r = 1399393939393399888888393/1000000000000000000000000so r is a rational number.If terminating decimal numbers are rational numbers, then irrational numbers must be non-terminating.