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Dealing with adding, subtracting, multiplying, dividing, and taking powers of negative numbers.

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Complete the following without the use of a calculator:

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$\\var{small} - \\var{big}$ = [[0]]

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When you are doing a subtraction that will result in a negative (that is, you are 'taking away' more than you have), many people reverse the order of the subtraction to make it easier and then put a negative in front of the answer.

For example, to determine $\\var{small}-\\var{big}$. We can see the answer will be a negative number (since $\\var{big}$ is bigger than $\\var{small}$). Let's do the related subtraction $\\var{big}-\\var{small}$ first. Here are two common methods:

Subtract $\\var{small}$ from $\\var{big}$ which we automatically know is $\\var{-ans}$ since $\\var{big}=\\var{-ans}+\\var{small}$. which is $\\var{-ans}$ since $\\var{big-10}=\\var{small}+\\var{big-10-small}$. by breaking $\\var{small}$ up into $\\var{big-10}$ and $\\var{small-big+10}$ and doing $\\var{big}-\\var{big-10}-\\var{small-big+10}=10-\\var{small-big+10}=\\var{-ans}$.
You could determine how many you need to add to $\\var{small}$ to get $\\var{big}$, that is determine $\\var{small}+\\var{-ans}=\\var{big}$.

Regardless, we find that $\\var{big}-\\var{small}=\\var{-ans}$ and so our answer to the original question is $\\var{small}-\\var{big}=\\var{ans}$.

Why does this work? If you have seen brackets before the following explains it

\\[\\var{small}-\\var{big}=-(\\var{big}-\\var{small})\\]

If this doesn't help, recall a subtraction tells you the difference between two numbers, it is the distance between two numbers on the number line, if you swap the numbers around in the subtraction you will cover the same distance on the number line just heading in the opposition direction. So if you know the answer will be negative, you can swap the numbers around, do the easier subtraction and then put a negative in front of your answer.

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Complete the following without the use of a calculator:

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$\\var{n[0]}\\var{n[1]}$ = [[0]]

Visualising the number line is quite helpful for these questions.

We can think of being at $\\var{n[0]}$ on the number line and then moving another $\\var{-n[1]}$ to the left. Now we are at $\\var{ans1}$.

Alternatively, imagine being in an elevator $\\var{-n[0]}$ floors below ground and then going another $\\var{-n[1]}$ floors further down. You would be $\\var{-ans1}$ floors below ground.

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$\\var{n[2]}+(\\var{n[3]})$ = [[0]]

This is actually the same process as the last question. We can use the number line/elevator or we can think of a negative number as a debt.

We can think of $\\var{n[2]}$ as being in debt by \\${-n[2]}. Suppose we then add another debt of \\${-n[3]}. Now we are in debt by \\${-ans2}. That is, $\\var{n[2]}+(\\var{n[3]})=\\var{ans2}$.

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$\\var{n[4]}+\\var{n[5]}+(\\var{n[6]})$ = [[0]]

This is actually the same process as the last question. We can use the number line/elevator or we can think of a negative number as a debt.

We can think of $\\var{n[4]}$ as being in debt by \\${-n[4]}. Suppose we then add another debt of \\${-n[5]}. Now we are in debt by \\${-n[4]-n[5]}. Then we add another debt of \\${-n[6]} and we are in debt by \\${-ans3}. That is, $\\var{n[4]}+\\var{n[5]}+(\\var{n[6]})=\\var{ans3}$

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Complete the following without the use of a calculator:

$\\var{n[0]}-(\\var{n[1]})$ = [[0]]

Thinking about negative numbers as debt can help you understand these questions.

We can think of $\\var{n[0]}$ as being in debt by \\${-n[0]}. Suppose we then 'take away' another debt of \\${-n[1]}, that is, we pay back \\${-n[1]}. Now we are only in debt actually in front by \\${abs(ans1)}. That is, $\\var{n[0]}-(\\var{n[1]})=\\var{ans1}$.

Another way is to think of the minus sign as going in the reverse direction.

Suppose on the number line we are at $\\var{n[0]}$, and now instead of going another $\\var{-n[1]}$ to the left (which we would do if we were doing $\\var{n[0]}+(\\var{n[1]})$) we have to go the opposite way, we have to move $\\var{-n[1]}$ to the right (which is the same as $\\var{n[0]}+\\var{-n[1]}$, and we end up at $\\var{ans1}$.

In general two negative symbols next to each other are the same as a positive symbol.

\\[++=+\\]

\\[+-=-\\]

\\[-+=-\\]

\\[--=+\\]

Any even number of negative symbols will be the same as a positive symbol. For example,

\\[-----=+-=-\\]

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$\\var{n[2]}-\\var{n[3]}$ = [[0]]

This is actually the same process as the last question. The lack of brackets doesn't affect this question.

Thinking about negative numbers as debt can help you understand these questions.

We can think of $\\var{n[2]}$ as being in debt by \\${-n[2]}. Suppose we then 'take away' another debt of \\${-n[3]}, that is, we pay back \\${-n[3]}. Now we are only in debt actually in front by \\${abs(ans2)}. That is, $\\var{n[2]}-(\\var{n[3]})=\\var{ans2}$.

Another way is to think of the minus sign as going in the reverse direction.

Suppose on the number line we are at $\\var{n[2]}$, and now instead of going another $\\var{-n[3]}$ to the left (which we would do if we were doing $\\var{n[2]}+(\\var{n[3]})$) we have to go the opposite way, we have to move $\\var{-n[3]}$ to the right (which is the same as $\\var{n[2]}+\\var{-n[3]}$, and we end up at $\\var{ans2}$.

In general two negative symbols next to each other is the same as a positive symbol.

\\[++=+\\]

\\[+-=-\\]

\\[-+=-\\]

\\[--=+\\]

Any even number of negative symbols will be the same as a positive symbol. For example,

\\[-----=+-=-\\]

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Complete the following without the use of a calculator:

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$(\\var{n[0]})\\times(\\var{n[1]})$ = [[0]]

Consider $(\\var{n[0]})\\times (\\var{n[1]})$. Let's move the negatives out the front to get $--\\var{-n[0]}\\times \\var{-n[1]}$, we can just do the multiplication and get $--\\var{ans1}$, but this is the same as $\\var{ans1}$.

In essence, we work out the numbers and the signs separately.

In general two negatives multiplied results in a positive.

\\[+\\times+=+\\]

\\[+\\times-=-\\]

\\[-\\times+=-\\]

\\[-\\times-=+\\]

Any even number of negative symbols will be the same as a positive symbol and any odd number of negative symbols will be the same as a negative symbol. For example,

\\[-\\times-\\times-\\times-\\times-=+\\times-=-\\]

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$\\var{n[2]}\\times (\\var{n[3]})$ = [[0]]

This is actually the same process as the last question. The lack of brackets doesn't affect this question.

Consider $(\\var{n[2]})\\times (\\var{n[3]})$. Let's move the negatives out the front to get $--\\var{-n[2]}\\times \\var{-n[3]}$, we can just do the multiplication and get $--\\var{ans2}$, but this is the same as $\\var{ans2}$.

In essence, we work out the numbers and the signs separately.

In general two negatives multiplied results in a positive.

\\[+\\times+=+\\]

\\[+\\times-=-\\]

\\[-\\times+=-\\]

\\[-\\times-=+\\]

Any even number of negative symbols will be the same as a positive symbol and any odd number of negative symbols will be the same as a negative symbol. For example,

\\[-\\times-\\times-\\times-\\times-=+\\times-=-\\]

"}], "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{ans2}", "maxValue": "{ans2}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": false, "customName": "", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "

$(\\var{m[0]})(\\var{m[1]})(\\var{m[2]})$ = [[0]]

Recall that $(\\var{m[0]})(\\var{m[1]})(\\var{m[2]})$ is a way of writing $(\\var{m[0]})\\times(\\var{m[1]})\\times(\\var{m[2]})$.

Let's move the negatives out the front to get $---\\var{-m[0]}\\times \\var{-m[1]}\\times \\var{-m[2]}$, we can just do the multiplication and get $---\\var{-ans3}$, but this is the same as $\\var{ans3}$.

In essence, we work out the numbers and the signs separately.

In general two negatives multiplied results in a positive.

\\[+\\times+=+\\]

\\[+\\times-=-\\]

\\[-\\times+=-\\]

\\[-\\times-=+\\]

Any even number of negative symbols will be the same as a positive symbol and any odd number of negative symbols will be the same as a negative symbol. For example,

\\[-\\times-\\times-\\times-\\times-=+\\times-=-\\]

"}], "gaps": [{"type": "numberentry", "useCustomName": false, "customName": "", "marks": 1, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "{ans3}", "maxValue": "{ans3}", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "displayAnswer": "", "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"], "correctAnswerStyle": "plain"}], "sortAnswers": false}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Negatives: dividing negative numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "tags": ["Dividing", "dividing", "division", "negative numbers", "Negative Numbers", "negatives"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Complete the following without the use of a calculator:

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Which of the following are equal to $(\\var{n[0]})\\div(\\var{d[0]})$?

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Dividing negative numbers can be done by pretending the numbers are positive, doing the division, and then determining whether the actual answer should be positive or negative.

It is also crucial to realise division can be written as a fraction.

Consider $(\\var{n[0]})\\div(\\var{d[0]})$. Let's move the negatives out the front to get $--\\var{-n[0]}\\div\\var{-d[0]}$, we can just do the division and get $--\\frac{\\var{-n[0]}}{\\var{-d[0]}}$, but this is the same as $\\frac{\\var{-n[0]}}{\\var{-d[0]}}$.

In general two negatives divided results in a positive.

\\[\\frac{+}{+}=+\\]

\\[\\frac{+}{-}=-\\]

\\[\\frac{-}{+}=-\\]

\\[\\frac{-}{-}=+\\]

This should not be surprising given that division is just multiplying by the reciprocal.

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Which of the following are equal to $\\var{n[1]}\\div\\var{-d[1]}$?

Dividing negative numbers can be done by pretending the numbers are positive, doing the division, and then determining whether the actual answer should be positive or negative.

It is also crucial to realise division can be written as a fraction.

Consider $\\var{n[1]}\\div\\var{-d[1]}$. This will be the same as $-\\frac{\\var{-n[1]}}{\\var{-d[1]}}$, and $\\frac{\\var{n[1]}}{\\var{-d[1]}}$, and even the strange looking $\\frac{\\var{-n[1]}}{\\var{d[1]}}$.

In general two negatives divided results in a positive.

\\[\\frac{+}{+}=+\\]

\\[\\frac{+}{-}=-\\]

\\[\\frac{-}{+}=-\\]

\\[\\frac{-}{-}=+\\]

This should not be surprising given that division is just multiplying by the reciprocal.

"}], "minMarks": 0, "maxMarks": 0, "shuffleChoices": true, "displayType": "checkbox", "displayColumns": 0, "minAnswers": 0, "maxAnswers": 0, "warningType": "none", "showCellAnswerState": true, "markingMethod": "sum ticked cells", "choices": "{choices_b}", "matrix": "[-1,-1,-1,1,1,1]"}], "partsMode": "all", "maxMarks": 0, "objectives": [], "penalties": [], "objectiveVisibility": "always", "penaltyVisibility": "always"}, {"name": "Negatives: powers of negative numbers", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Ben Brawn", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/605/"}], "tags": ["exponents", "indices", "negative numbers", "Negative Numbers", "negatives", "powers"], "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "

Complete the following without the use of a calculator:

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$\\var{n[0]}^2$ = [[0]]

Here the square is acting only on the number and not on the negative sign. So you square the number first and then take the negative. That is, $\\var{n[0]}^2=-(\\var{-n[0]})^2=-(\\var{-n[0]}\\times\\var{-n[0]})=-\\var{-ans1}$.

Note

\\[\\var{n[0]}^2\\ne (\\var{n[0]})^2\\]

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$(\\var{n[1]})^2$ = [[0]]

Here we have the square acting on a negative number. When two negatives are multiplied together the result is a positive number.

This means when a negative number is squared the result is positive.

In particular, $(\\var{n[1]})^2=(\\var{n[1]})\\times(\\var{n[1]})=\\var{ans2}$.

Note

\\[\\var{n[1]}^2\\ne (\\var{n[1]})^2\\]

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$-(\\var{n[1]})^2$ = [[0]]

Deal with the squaring of the negative number first (which results in a positive) and then deal with the negative outside the brackets.

That is, $-(\\var{n[1]})^2=-(\\var{n[1]}\\times\\var{n[1]})=-(\\var{ans2})=\\var{-ans2}$

Note

\\[-(\\var{n[1]})^2\\ne \\var{-n[1]}^2\\]

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$(\\var{m})^3$ = [[0]]

We have $(\\var{m})^3$. Recall this is the same as $(\\var{m})\\times(\\var{m})\\times(\\var{m})$. Here we have three negative symbols, two give a positive and the remaining negative makes the result negative, that is,

\\[(\\var{m})^3=\\var{ans3}\\]

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$(-1)^\\var{p}$ = [[0]]

Any even power of a negative number will have an even number of negatives multiplied together. This will then result in a positive number.

Any odd power of a negative number will have an odd number of negatives multiplied together. This will then result in a negative number.

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$\\displaystyle\\var{n[0]}-(\\var{n[1]})--\\var{n[2]}\\times(\\var{n[3]})+\\left(\\frac{\\var{n[4]*n[4]}}{\\var{n[4]}}\\right)+\\var{mult1}(-1)^\\var{power}+\\var{mult2}(-1)^\\var{power+1}$ = [[0]]

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Complete the following without the use of a calculator:

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