When one or both of the integers is/are zero.a*b=0 if a=0, b=0, or both a and b are equal to 0. In other words, if one or both integers are zero.
Because a is rational, there exist integers m and n such that a=m/n. Because b is rational, there exist integers p and q such that b=p/q. Consider a+b. a+b=(m/n)+(p/q)=(mq/nq)+(pn/mq)=(mq+pn)/(nq). (mq+pn) is an integer because the product of two integers is an integer, and the sum of two integers is an integer. nq is an integer since the product of two integers is an integer. Because a+b equals the quotient of two integers, a+b is rational.
I assume you meant "sum". The answer is 60. 60=4 (integers) * 15 (average value of the integers). average of 4 integers = 15 --> (a + b + c + d)/4 = 15 --> (a + b + c + d) = 15 * 4 --> (a + b + c + d) = 60
Yes it is : a + b = b + a for all integers a and b. In fact , if an operation is called addition you can bet that it is commutative. It would be perverse to call an non-commutative operation addition.
NoA rational number is a one that can be written as a fraction i.e a/b. where a and be are integers (whole numbers)Considera/b and c/d. Where a b c and d are integers and as such rational numbersa/b + c/d = (ad + bd)/cdad, bd and cd will all be integers and as such a/b + c/d will always be rational
No, they are not because fractions can be negative also. fractions aren't integers
Rational fractions of the form a/b where both a and b are integers, b > 0 and, in its simplified form, the denominator is not 1.
The two integers are A and A+40 or, equivalently, B and B-40.
Subtracting an integer is the same as adding the additive inverse. In symbols: a - b = a + (-b), where "-b" is the additive inverse (the opposite) of b.
These are rational number that are not integers. In their simplest form, they are of the form a/b where a and b are integers and b is not 0 nor 1.
When one or both of the integers is/are zero.a*b=0 if a=0, b=0, or both a and b are equal to 0. In other words, if one or both integers are zero.
No. All integers are rational numbers with no fractional part-that is, they can be written as A/B such that B goes into A evenly.
The associative property states that, for the sum of three or more integers the order in which the summation in carried out does not make a difference to the answer. Thus, for any three integers, A, B and C: (A + B) + C = A + (B + C) and so, without ambiguity, we can write either as A + B + C. Note that A + B need not be the same as B + A. The order of the integers DOES matter. It is the order of the summing that does not.
Because a is rational, there exist integers m and n such that a=m/n. Because b is rational, there exist integers p and q such that b=p/q. Consider a+b. a+b=(m/n)+(p/q)=(mq/nq)+(pn/mq)=(mq+pn)/(nq). (mq+pn) is an integer because the product of two integers is an integer, and the sum of two integers is an integer. nq is an integer since the product of two integers is an integer. Because a+b equals the quotient of two integers, a+b is rational.
That a = ±b
A number that can be written in the form a/b whre a and b are integers is called a rational number.
Two integers A and B are graphed on a number line. If A is less than B is A always less than B?