0.000 000 007 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 007(10) to 32 bit single precision IEEE 754 binary floating point representation standard (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

What are the steps to convert decimal number
0.000 000 007(10) to 32 bit single precision IEEE 754 binary floating point representation (1 bit for sign, 8 bits for exponent, 23 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 007.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 007 × 2 = 0 + 0.000 000 014;
  • 2) 0.000 000 014 × 2 = 0 + 0.000 000 028;
  • 3) 0.000 000 028 × 2 = 0 + 0.000 000 056;
  • 4) 0.000 000 056 × 2 = 0 + 0.000 000 112;
  • 5) 0.000 000 112 × 2 = 0 + 0.000 000 224;
  • 6) 0.000 000 224 × 2 = 0 + 0.000 000 448;
  • 7) 0.000 000 448 × 2 = 0 + 0.000 000 896;
  • 8) 0.000 000 896 × 2 = 0 + 0.000 001 792;
  • 9) 0.000 001 792 × 2 = 0 + 0.000 003 584;
  • 10) 0.000 003 584 × 2 = 0 + 0.000 007 168;
  • 11) 0.000 007 168 × 2 = 0 + 0.000 014 336;
  • 12) 0.000 014 336 × 2 = 0 + 0.000 028 672;
  • 13) 0.000 028 672 × 2 = 0 + 0.000 057 344;
  • 14) 0.000 057 344 × 2 = 0 + 0.000 114 688;
  • 15) 0.000 114 688 × 2 = 0 + 0.000 229 376;
  • 16) 0.000 229 376 × 2 = 0 + 0.000 458 752;
  • 17) 0.000 458 752 × 2 = 0 + 0.000 917 504;
  • 18) 0.000 917 504 × 2 = 0 + 0.001 835 008;
  • 19) 0.001 835 008 × 2 = 0 + 0.003 670 016;
  • 20) 0.003 670 016 × 2 = 0 + 0.007 340 032;
  • 21) 0.007 340 032 × 2 = 0 + 0.014 680 064;
  • 22) 0.014 680 064 × 2 = 0 + 0.029 360 128;
  • 23) 0.029 360 128 × 2 = 0 + 0.058 720 256;
  • 24) 0.058 720 256 × 2 = 0 + 0.117 440 512;
  • 25) 0.117 440 512 × 2 = 0 + 0.234 881 024;
  • 26) 0.234 881 024 × 2 = 0 + 0.469 762 048;
  • 27) 0.469 762 048 × 2 = 0 + 0.939 524 096;
  • 28) 0.939 524 096 × 2 = 1 + 0.879 048 192;
  • 29) 0.879 048 192 × 2 = 1 + 0.758 096 384;
  • 30) 0.758 096 384 × 2 = 1 + 0.516 192 768;
  • 31) 0.516 192 768 × 2 = 1 + 0.032 385 536;
  • 32) 0.032 385 536 × 2 = 0 + 0.064 771 072;
  • 33) 0.064 771 072 × 2 = 0 + 0.129 542 144;
  • 34) 0.129 542 144 × 2 = 0 + 0.259 084 288;
  • 35) 0.259 084 288 × 2 = 0 + 0.518 168 576;
  • 36) 0.518 168 576 × 2 = 1 + 0.036 337 152;
  • 37) 0.036 337 152 × 2 = 0 + 0.072 674 304;
  • 38) 0.072 674 304 × 2 = 0 + 0.145 348 608;
  • 39) 0.145 348 608 × 2 = 0 + 0.290 697 216;
  • 40) 0.290 697 216 × 2 = 0 + 0.581 394 432;
  • 41) 0.581 394 432 × 2 = 1 + 0.162 788 864;
  • 42) 0.162 788 864 × 2 = 0 + 0.325 577 728;
  • 43) 0.325 577 728 × 2 = 0 + 0.651 155 456;
  • 44) 0.651 155 456 × 2 = 1 + 0.302 310 912;
  • 45) 0.302 310 912 × 2 = 0 + 0.604 621 824;
  • 46) 0.604 621 824 × 2 = 1 + 0.209 243 648;
  • 47) 0.209 243 648 × 2 = 0 + 0.418 487 296;
  • 48) 0.418 487 296 × 2 = 0 + 0.836 974 592;
  • 49) 0.836 974 592 × 2 = 1 + 0.673 949 184;
  • 50) 0.673 949 184 × 2 = 1 + 0.347 898 368;
  • 51) 0.347 898 368 × 2 = 0 + 0.695 796 736;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 007(10) =

0.0000 0000 0000 0000 0000 0000 0001 1110 0001 0000 1001 0100 110(2)

5. Positive number before normalization:

0.000 000 007(10) =

0.0000 0000 0000 0000 0000 0000 0001 1110 0001 0000 1001 0100 110(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 28 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 007(10) =

0.0000 0000 0000 0000 0000 0000 0001 1110 0001 0000 1001 0100 110(2) =

0.0000 0000 0000 0000 0000 0000 0001 1110 0001 0000 1001 0100 110(2) × 20 =

1.1110 0001 0000 1001 0100 110(2) × 2-28


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -28

Mantissa (not normalized):
1.1110 0001 0000 1001 0100 110


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-28 + 2(8-1) - 1 =

(-28 + 127)(10) =

99(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 99 ÷ 2 = 49 + 1;
  • 49 ÷ 2 = 24 + 1;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

99(10) =

0110 0011(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 111 0000 1000 0100 1010 0110 =

111 0000 1000 0100 1010 0110


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 0011

Mantissa (23 bits) =
111 0000 1000 0100 1010 0110


Decimal number 0.000 000 007 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0110 0011 - 111 0000 1000 0100 1010 0110

Share

» Convert decimal -0.000 000 004 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: The Art of Strategic Ambiguity and Concealed Intentions. NotebookLM YouTube Video of 6.59 minutes

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =
    1 - 1000 0011 - 100 1010 1100 0110 1010 0111