# How to convert decimal with scientefic notation, to binary?

I have number $$-1e35$$, and I'm supposed to convert it to binary. The answer is : $$-1.10101001010110100101101...\text{e}–117.$$

I can't figure out how to get this! and how we can calculate numbers partially as you see here. Originally it is a programming practice, but what I care about is the mathematical part.

I tried to solve it like this: $$\log 10^{35}=\log 2^x.$$ The result is: $$116.666$$.

My question is not exactly how to convert. I know how to convert smaller numbers, but I am confused with scientific notation, and how to represent part of the number.

• Why is the binary exponent negative? Commented Feb 3, 2016 at 18:36

When you represent a number in decimal scientific notation, the base of the exponent is $$10$$, so $$1E35=1\cdot 10^{35}$$. In binary, the base is $$2$$, so you are trying to solve $$1\cdot 10^{35}=m2^e$$ where $$1 \lt m \lt 10_2$$ is the mantissa and $$e$$ is an integer exponent. To find $$e$$ we can take logs: $$1 \cdot 10^{35} = m2^e\\ 35 \log_2(10)=e+ \log_2 (m)\\e=\lfloor35 \log_2(10)\rfloor=116_{10}$$ Now to get the mantissa you can just subtract off powers of $$2$$ $$10^{35}-2^{116}=16923250263442757943512058732478464\\ \lfloor \log_2(16923250263442757943512058732478464)\rfloor=113$$ so we start with $$1.001_2$$ because the exponent dropped by $$3$$ $$16923250263442757943512058732478464-2^{113}=6538656546373102686451066074038272\\ \lfloor \log_2(6538656546373102686451066074038272)\rfloor=112$$ so our mantissa becomes $$1.0011_2$$ and so on until you get tired, your word fills up, or you get to the end. The final answer is $$1.001101000010011000010111001011000111010011011000001000101011100001111000111111101_2E116$$
• Yes, binary means base 2. I took the floor to get $116$. We have $2^{116} \lt 10^{35} \lt 2^{117}$ and want $1 \lt m \lt 2$. Commented Feb 3, 2016 at 20:12
• @RossMillikan I obtain $e = 35 \log_2 10 - \log_2 m$, then it is not clear why $e=\lfloor35 \log_2(10)\rfloor=116_{10}$, please, can you explain better? Thanks! Commented May 25, 2020 at 17:06
• @JB-Franco: Take the base $10$ log of the first centered equation, which gives $35=e\log_{10}2+\log_{10}m$ Now use $\log_{10}2=\frac 1{\log_210}$ to move that term over and take the floor to get rid of $\log_{10}m$, which is between $0$ and $1$ Commented May 25, 2020 at 17:12