-0.000 000 047 684 674 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 047 684 674(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
-0.000 000 047 684 674(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. Start with the positive version of the number:

|-0.000 000 047 684 674| = 0.000 000 047 684 674


2. 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;

3. 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)


4. Convert to binary (base 2) the fractional part: 0.000 000 047 684 674.

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) 0.000 000 047 684 674 × 2 = 0 + 0.000 000 095 369 348;
  • 2) 0.000 000 095 369 348 × 2 = 0 + 0.000 000 190 738 696;
  • 3) 0.000 000 190 738 696 × 2 = 0 + 0.000 000 381 477 392;
  • 4) 0.000 000 381 477 392 × 2 = 0 + 0.000 000 762 954 784;
  • 5) 0.000 000 762 954 784 × 2 = 0 + 0.000 001 525 909 568;
  • 6) 0.000 001 525 909 568 × 2 = 0 + 0.000 003 051 819 136;
  • 7) 0.000 003 051 819 136 × 2 = 0 + 0.000 006 103 638 272;
  • 8) 0.000 006 103 638 272 × 2 = 0 + 0.000 012 207 276 544;
  • 9) 0.000 012 207 276 544 × 2 = 0 + 0.000 024 414 553 088;
  • 10) 0.000 024 414 553 088 × 2 = 0 + 0.000 048 829 106 176;
  • 11) 0.000 048 829 106 176 × 2 = 0 + 0.000 097 658 212 352;
  • 12) 0.000 097 658 212 352 × 2 = 0 + 0.000 195 316 424 704;
  • 13) 0.000 195 316 424 704 × 2 = 0 + 0.000 390 632 849 408;
  • 14) 0.000 390 632 849 408 × 2 = 0 + 0.000 781 265 698 816;
  • 15) 0.000 781 265 698 816 × 2 = 0 + 0.001 562 531 397 632;
  • 16) 0.001 562 531 397 632 × 2 = 0 + 0.003 125 062 795 264;
  • 17) 0.003 125 062 795 264 × 2 = 0 + 0.006 250 125 590 528;
  • 18) 0.006 250 125 590 528 × 2 = 0 + 0.012 500 251 181 056;
  • 19) 0.012 500 251 181 056 × 2 = 0 + 0.025 000 502 362 112;
  • 20) 0.025 000 502 362 112 × 2 = 0 + 0.050 001 004 724 224;
  • 21) 0.050 001 004 724 224 × 2 = 0 + 0.100 002 009 448 448;
  • 22) 0.100 002 009 448 448 × 2 = 0 + 0.200 004 018 896 896;
  • 23) 0.200 004 018 896 896 × 2 = 0 + 0.400 008 037 793 792;
  • 24) 0.400 008 037 793 792 × 2 = 0 + 0.800 016 075 587 584;
  • 25) 0.800 016 075 587 584 × 2 = 1 + 0.600 032 151 175 168;
  • 26) 0.600 032 151 175 168 × 2 = 1 + 0.200 064 302 350 336;
  • 27) 0.200 064 302 350 336 × 2 = 0 + 0.400 128 604 700 672;
  • 28) 0.400 128 604 700 672 × 2 = 0 + 0.800 257 209 401 344;
  • 29) 0.800 257 209 401 344 × 2 = 1 + 0.600 514 418 802 688;
  • 30) 0.600 514 418 802 688 × 2 = 1 + 0.201 028 837 605 376;
  • 31) 0.201 028 837 605 376 × 2 = 0 + 0.402 057 675 210 752;
  • 32) 0.402 057 675 210 752 × 2 = 0 + 0.804 115 350 421 504;
  • 33) 0.804 115 350 421 504 × 2 = 1 + 0.608 230 700 843 008;
  • 34) 0.608 230 700 843 008 × 2 = 1 + 0.216 461 401 686 016;
  • 35) 0.216 461 401 686 016 × 2 = 0 + 0.432 922 803 372 032;
  • 36) 0.432 922 803 372 032 × 2 = 0 + 0.865 845 606 744 064;
  • 37) 0.865 845 606 744 064 × 2 = 1 + 0.731 691 213 488 128;
  • 38) 0.731 691 213 488 128 × 2 = 1 + 0.463 382 426 976 256;
  • 39) 0.463 382 426 976 256 × 2 = 0 + 0.926 764 853 952 512;
  • 40) 0.926 764 853 952 512 × 2 = 1 + 0.853 529 707 905 024;
  • 41) 0.853 529 707 905 024 × 2 = 1 + 0.707 059 415 810 048;
  • 42) 0.707 059 415 810 048 × 2 = 1 + 0.414 118 831 620 096;
  • 43) 0.414 118 831 620 096 × 2 = 0 + 0.828 237 663 240 192;
  • 44) 0.828 237 663 240 192 × 2 = 1 + 0.656 475 326 480 384;
  • 45) 0.656 475 326 480 384 × 2 = 1 + 0.312 950 652 960 768;
  • 46) 0.312 950 652 960 768 × 2 = 0 + 0.625 901 305 921 536;
  • 47) 0.625 901 305 921 536 × 2 = 1 + 0.251 802 611 843 072;
  • 48) 0.251 802 611 843 072 × 2 = 0 + 0.503 605 223 686 144;

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).


5. 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:


0.000 000 047 684 674(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1101 1010(2)

6. Positive number before normalization:

0.000 000 047 684 674(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1101 1010(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 25 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 047 684 674(10) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1101 1010(2) =

0.0000 0000 0000 0000 0000 0000 1100 1100 1100 1101 1101 1010(2) × 20 =

1.1001 1001 1001 1011 1011 010(2) × 2-25


8. 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 1 (a negative number)

Exponent (unadjusted): -25

Mantissa (not normalized):
1.1001 1001 1001 1011 1011 010


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-25 + 2(8-1) - 1 =

(-25 + 127)(10) =

102(10)


10. 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;
  • 102 ÷ 2 = 51 + 0;
  • 51 ÷ 2 = 25 + 1;
  • 25 ÷ 2 = 12 + 1;
  • 12 ÷ 2 = 6 + 0;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

102(10) =

0110 0110(2)


12. 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. 100 1100 1100 1101 1101 1010 =

100 1100 1100 1101 1101 1010


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

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

Exponent (8 bits) =
0110 0110

Mantissa (23 bits) =
100 1100 1100 1101 1101 1010


Decimal number -0.000 000 047 684 674 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0110 - 100 1100 1100 1101 1101 1010

Share

» Convert decimal -0.000 000 047 684 61 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 09, 2026 [September]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: September - by our visitors

» Improve your social skills: Why Do Colleagues From Other Teams Have Better Results? Gemini Notebook YouTube Video of 11.31 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