Multiplication and division by addition and subtraction
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I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?
You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.
functions
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up vote
1
down vote
favorite
I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?
You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.
functions
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?
You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.
functions
I know that this sounds like the stupidest question possible, but I had to ask, is it possible to express any multiplication and division solely by addition and subtraction? Such as... 7*0.3 or 7/0.3?
You can express sine and cosine by a Taylor series so I wondered this purely out of curiosity. My searching didn't seem to come up with an answer anywhere though hence I'm asking.
functions
functions
asked Aug 30 at 4:26
John Ernest
83
83
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2 Answers
2
active
oldest
votes
up vote
0
down vote
accepted
Well,first let's look at the rules. For multiplication,
if, $x,yin mathbbQ$
$$x.y=underbrace(x+x+x+x+...+x)_texty times$$
For division, if
$$ x/y=n$$
$$Then,x=y.n$$
$$means,x-underbrace(y+y+y+...+y)_textn times=0$$
Now,lets work on your given numbers.
Here, $$7*0.3=7*dfrac310$$
That means we have to multiply 7, 3 times.than yiels $implies$
$$7*3=7+7+7=21$$
Now, we have $$dfrac2110$$
Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
$$21-(10+10)=1$$
1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$
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up vote
0
down vote
Yes it is possible
$$xcdot y=
underbracex+x+cdots+x _text$y$ times$$
if $$fracxy=n$$
$$0=x-underbrace(y+y+cdots+y)_text $n$ times$$
$fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0
But here is one limitation $x,y$ belongs to Rational Numbers
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Well,first let's look at the rules. For multiplication,
if, $x,yin mathbbQ$
$$x.y=underbrace(x+x+x+x+...+x)_texty times$$
For division, if
$$ x/y=n$$
$$Then,x=y.n$$
$$means,x-underbrace(y+y+y+...+y)_textn times=0$$
Now,lets work on your given numbers.
Here, $$7*0.3=7*dfrac310$$
That means we have to multiply 7, 3 times.than yiels $implies$
$$7*3=7+7+7=21$$
Now, we have $$dfrac2110$$
Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
$$21-(10+10)=1$$
1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$
add a comment |Â
up vote
0
down vote
accepted
Well,first let's look at the rules. For multiplication,
if, $x,yin mathbbQ$
$$x.y=underbrace(x+x+x+x+...+x)_texty times$$
For division, if
$$ x/y=n$$
$$Then,x=y.n$$
$$means,x-underbrace(y+y+y+...+y)_textn times=0$$
Now,lets work on your given numbers.
Here, $$7*0.3=7*dfrac310$$
That means we have to multiply 7, 3 times.than yiels $implies$
$$7*3=7+7+7=21$$
Now, we have $$dfrac2110$$
Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
$$21-(10+10)=1$$
1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Well,first let's look at the rules. For multiplication,
if, $x,yin mathbbQ$
$$x.y=underbrace(x+x+x+x+...+x)_texty times$$
For division, if
$$ x/y=n$$
$$Then,x=y.n$$
$$means,x-underbrace(y+y+y+...+y)_textn times=0$$
Now,lets work on your given numbers.
Here, $$7*0.3=7*dfrac310$$
That means we have to multiply 7, 3 times.than yiels $implies$
$$7*3=7+7+7=21$$
Now, we have $$dfrac2110$$
Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
$$21-(10+10)=1$$
1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$
Well,first let's look at the rules. For multiplication,
if, $x,yin mathbbQ$
$$x.y=underbrace(x+x+x+x+...+x)_texty times$$
For division, if
$$ x/y=n$$
$$Then,x=y.n$$
$$means,x-underbrace(y+y+y+...+y)_textn times=0$$
Now,lets work on your given numbers.
Here, $$7*0.3=7*dfrac310$$
That means we have to multiply 7, 3 times.than yiels $implies$
$$7*3=7+7+7=21$$
Now, we have $$dfrac2110$$
Now,how many 10 do have a sum of 21,clearly it is more than 2 but not perfect 2.let's add 10, 2 times first.so we get,
$$21-(10+10)=1$$
1 is our remainder.now,let's think this 1 as 10.now we have to add another 10 to get this and now the remainder is 0. But as this was my imagination to think it as 10,i have to put the number behind a "decimal(point)" sign.so the answer is $2.1$
edited Aug 30 at 5:11
answered Aug 30 at 5:05
Rakibul Islam Prince
3398
3398
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up vote
0
down vote
Yes it is possible
$$xcdot y=
underbracex+x+cdots+x _text$y$ times$$
if $$fracxy=n$$
$$0=x-underbrace(y+y+cdots+y)_text $n$ times$$
$fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0
But here is one limitation $x,y$ belongs to Rational Numbers
add a comment |Â
up vote
0
down vote
Yes it is possible
$$xcdot y=
underbracex+x+cdots+x _text$y$ times$$
if $$fracxy=n$$
$$0=x-underbrace(y+y+cdots+y)_text $n$ times$$
$fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0
But here is one limitation $x,y$ belongs to Rational Numbers
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes it is possible
$$xcdot y=
underbracex+x+cdots+x _text$y$ times$$
if $$fracxy=n$$
$$0=x-underbrace(y+y+cdots+y)_text $n$ times$$
$fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0
But here is one limitation $x,y$ belongs to Rational Numbers
Yes it is possible
$$xcdot y=
underbracex+x+cdots+x _text$y$ times$$
if $$fracxy=n$$
$$0=x-underbrace(y+y+cdots+y)_text $n$ times$$
$fracxy$ means how many time we have to subtract $y$ from $x$ so that $x$ becomes 0
But here is one limitation $x,y$ belongs to Rational Numbers
answered Aug 30 at 4:42
Deepesh Meena
3,2492824
3,2492824
add a comment |Â
add a comment |Â
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