Mixing
For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational...
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For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$
How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.
proof-writing rational-numbers
asked Jan 29 at 18:57
macymacy
526
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add a comment |
$begingroup$
For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$
How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.
proof-writing rational-numbers
asked Jan 29 at 18:57
macymacy
526
$endgroup$
$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06
add a comment |
$begingroup$
For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$
How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.
proof-writing rational-numbers
asked Jan 29 at 18:57
macymacy
526
$endgroup$
For all integers $w, x, y, z$ with $wneq{y}$ and $wz-xyneq0$, prove that there exists a unique rational number $r$ such that $(wr+x)div(yr+z)=1$
How do I prove uniqueness? I know to show that there exists a number all I need to do is use an example.
proof-writing rational-numbers
proof-writing rational-numbers
asked Jan 29 at 18:57
macymacy
526
asked Jan 29 at 18:57
macymacy
526
asked Jan 29 at 18:57
macymacy
526
asked Jan 29 at 18:57
macymacy
526
asked Jan 29 at 18:57
macymacy
526
526
$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06
add a comment |
$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06
$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06
$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06
add a comment |
2 Answers
2
active
oldest
votes
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One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.
In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).
Hope I was useful.
answered Jan 29 at 19:06
maxbpmaxbp
1467
$endgroup$
add a comment |
$begingroup$
To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
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add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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active
oldest
votes
$begingroup$
One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.
In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).
Hope I was useful.
answered Jan 29 at 19:06
maxbpmaxbp
1467
$endgroup$
add a comment |
$begingroup$
One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.
In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).
Hope I was useful.
answered Jan 29 at 19:06
maxbpmaxbp
1467
$endgroup$
add a comment |
$begingroup$
One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.
In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).
Hope I was useful.
answered Jan 29 at 19:06
maxbpmaxbp
1467
$endgroup$
One way to prove uniqueness, and is a way that can be used here, is to find a closed expression for the value you are searching.
In this case by isolating r (and being careful with divisions by 0, i.e. justifying that what you are dividing is not 0) you will get a closed expression for r. Meaning that the solution is unique (it can only be that exact value).
Hope I was useful.
answered Jan 29 at 19:06
maxbpmaxbp
1467
answered Jan 29 at 19:06
maxbpmaxbp
1467
answered Jan 29 at 19:06
maxbpmaxbp
1467
answered Jan 29 at 19:06
maxbpmaxbp
1467
1467
add a comment |
add a comment |
$begingroup$
To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
$endgroup$
add a comment |
$begingroup$
To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
$endgroup$
add a comment |
$begingroup$
To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
$endgroup$
To prove uniqueness, assume $$(wr + x)/(yr+z)=(ws+x)/(ys+z).$$ Cross-multiplying, $$(wr+x)(ys+z)=(ws+x)(yr+z),$$ so $$wrys+wrz+xys+xz=wsyr+wsz+xyr+xz,$$ so $$wrz+xys=wsz+xyr,$$ so $$wrz-wsz=xyr-xys,$$ i.e., $$(wz-xy)r=(wz-xy)s.$$ Under the assumption $wz-xy ne 0$, this means $r=s$.
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
answered Jan 29 at 19:13
J. W. TannerJ. W. Tanner
4,2361320
4,2361320
add a comment |
add a comment |
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$begingroup$
Assume $(wr + x)/(yr+z)=(ws+x)/(ys+z)$ and then show $r=s$ under the assumptions.
$endgroup$
– J. W. Tanner
Jan 29 at 19:06