<span>1. The definition of an additive inverse of a number is precisely that which, when added to the number, will give a sum of zero. </span>
<span>The real problem, in certain fields, is usually to show that for all numbers in that field, there exists an additive inverse. </span>
<span>Therefore, if you tell me that you have a number, and its additive inverse, and you plan to add them together, then I can tell you in advance that the sum MUST be zero. </span>
<span>2. In your question, you use the word "difference", which does not work (unless the number is zero - 0 is an integer AND a rational number, and its additive inverse is -0 which is the same as 0 - the difference would be 0 - -0 = 0). </span>
<span>For example, given the number 3, and its additive inverse -3, if you add them, you get zero: </span> <span>3 + (-3) = 0 </span>
<span>However, their "difference" will be 6 (or -6, depending which way you do the difference): </span>
<span>(because -3 is a number in the integers, then it has an additive inverse, also in the integers, of +3). </span>
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<span>A rational number is simply a number that can be expressed as the "ratio" of two integers. For example, the number 4/7 is the ratio of "four to seven". </span>
<span>It can be written as an endless decimal expansion </span> <span>0.571428571428571428....(forever), but that does not change its nature, because it CAN be written as a ratio, it is "rational". </span>
<span>Integers are rational numbers as well (because you can always write 3/1, the ratio of 3 to 1, to express the integer we call "3") </span>
<span>The additive inverse of a rational number, written as a ratio, is found by simply flipping the sign of the numerator (top) </span>
<span>The additive inverse of 4/7 is -4/7 </span>
<span>and if you ADD those two numbers together, you get zero (as per the definition of "additive inverse") </span>
<span>(4/7) + (-4/7) = 0/7 = 0 </span>
<span>If you need to "prove" it, you begin by the existence of additive inverses in the integers. </span> <span>ALL integers each have an additive inverse. </span> <span>For example, the additive inverse of 4 is -4 </span>
<span>Next, show that this (in the integers) can be applied to the rationals in this manner: </span>
<span>(4/7) + (-4/7) = ? </span> <span>common denominator, therefore you can factor out the denominator: </span>
<span>(4 + -4)/7 = ? </span> <span>Inside the bracket is the sum of an integer with its additive inverse, therefore the sum is zero </span> <span>(0)/7 = 0/7 = 0 </span>
<span>Since this is true for ALL integers, then it must also be true for ALL rational numbers.</span>
You would take random numbers preferable smaller ones such as -1,0,1,2 You would put those in the place of x Then you would square the value of x. After that you subtract 4. Whatever you come out with is your y value. Then you put the x value on the x axis and from that point count up or down to the number you need on the y axis. Together that would be your point
Our current list has 11!/2!11!/2! arrangements which we must divide into equivalence classes just as before, only this time the classes contain arrangements where only the two As are arranged, following this logic requires us to divide by arrangement of the 2 As giving (11!/2!)/2!=11!/(2!2)(11!/2!)/2!=11!/(2!2).
Repeating the process one last time for equivalence classes for arrangements of only T's leads us to divide the list once again by 2
