// Numbas version: exam_results_page_options {"name": "Negative Numbers", "feedback": {"intro": "", "showtotalmark": true, "advicethreshold": 0, "allowrevealanswer": true, "feedbackmessages": [], "showanswerstate": true, "showactualmark": true}, "timing": {"allowPause": true, "timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "allQuestions": true, "shuffleQuestions": true, "percentPass": "40", "duration": 0, "pickQuestions": "12", "navigation": {"onleave": {"action": "warnifunattempted", "message": "

You haven't finished answering this! Are you sure you want to leave?

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This is to test your knowledge of adding, subtracting, multiplying and dividing negative numbers.

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Complete the following without the use of a calculator:

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

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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:

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

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

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

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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-=-\\]

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

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

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