Measurement
- Thread starter skinski43
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skinski43 said:I really need help with this proof. The statement I need to prove is prove that if a:b=a':b' then (a+a')b+b')=a:b. I can get to the (a+a')
Thanks
Please show us excatly what do you get - and how.
Hello, skinski43!
\(\displaystyle \text{We have: }\;\frac{a}{b} \:=\:\frac{a'}{b'} \quad\Rightarrow\quad ab' \:=\:a'b \quad\Rightarrow\quad \frac{a'}{a} \:=\:\frac{b'}{b}\)
\(\displaystyle \text{Add 1 to both sides: }\:\frac{a'}{a} + 1 \:=\:\frac{b'}{b} + 1\)
\(\displaystyle \text{Therefore: }\:\frac{a'+a}{a} \:=\:\frac{b'+b}{b} \quad\Rightarrow\quad (a+a') : a \:=\b+b') : b\)
\(\displaystyle \text{if }a:b=a':b'\text{, then: }(a+a'): (b+b')\:=\:a:b\)
I can get to the (a+a')
but I am having trouble getting a:b.[/size]
\(\displaystyle \text{We have: }\;\frac{a}{b} \:=\:\frac{a'}{b'} \quad\Rightarrow\quad ab' \:=\:a'b \quad\Rightarrow\quad \frac{a'}{a} \:=\:\frac{b'}{b}\)
\(\displaystyle \text{Add 1 to both sides: }\:\frac{a'}{a} + 1 \:=\:\frac{b'}{b} + 1\)
\(\displaystyle \text{Therefore: }\:\frac{a'+a}{a} \:=\:\frac{b'+b}{b} \quad\Rightarrow\quad (a+a') : a \:=\
b+b') : b\)