-0.000 013 504 178 241 419 553 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 013 504 178 241 419 553(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
-0.000 013 504 178 241 419 553(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:

|-0.000 013 504 178 241 419 553| = 0.000 013 504 178 241 419 553


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: 0.000 013 504 178 241 419 553.

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) 0.000 013 504 178 241 419 553 × 2 = 0 + 0.000 027 008 356 482 839 106;
  • 2) 0.000 027 008 356 482 839 106 × 2 = 0 + 0.000 054 016 712 965 678 212;
  • 3) 0.000 054 016 712 965 678 212 × 2 = 0 + 0.000 108 033 425 931 356 424;
  • 4) 0.000 108 033 425 931 356 424 × 2 = 0 + 0.000 216 066 851 862 712 848;
  • 5) 0.000 216 066 851 862 712 848 × 2 = 0 + 0.000 432 133 703 725 425 696;
  • 6) 0.000 432 133 703 725 425 696 × 2 = 0 + 0.000 864 267 407 450 851 392;
  • 7) 0.000 864 267 407 450 851 392 × 2 = 0 + 0.001 728 534 814 901 702 784;
  • 8) 0.001 728 534 814 901 702 784 × 2 = 0 + 0.003 457 069 629 803 405 568;
  • 9) 0.003 457 069 629 803 405 568 × 2 = 0 + 0.006 914 139 259 606 811 136;
  • 10) 0.006 914 139 259 606 811 136 × 2 = 0 + 0.013 828 278 519 213 622 272;
  • 11) 0.013 828 278 519 213 622 272 × 2 = 0 + 0.027 656 557 038 427 244 544;
  • 12) 0.027 656 557 038 427 244 544 × 2 = 0 + 0.055 313 114 076 854 489 088;
  • 13) 0.055 313 114 076 854 489 088 × 2 = 0 + 0.110 626 228 153 708 978 176;
  • 14) 0.110 626 228 153 708 978 176 × 2 = 0 + 0.221 252 456 307 417 956 352;
  • 15) 0.221 252 456 307 417 956 352 × 2 = 0 + 0.442 504 912 614 835 912 704;
  • 16) 0.442 504 912 614 835 912 704 × 2 = 0 + 0.885 009 825 229 671 825 408;
  • 17) 0.885 009 825 229 671 825 408 × 2 = 1 + 0.770 019 650 459 343 650 816;
  • 18) 0.770 019 650 459 343 650 816 × 2 = 1 + 0.540 039 300 918 687 301 632;
  • 19) 0.540 039 300 918 687 301 632 × 2 = 1 + 0.080 078 601 837 374 603 264;
  • 20) 0.080 078 601 837 374 603 264 × 2 = 0 + 0.160 157 203 674 749 206 528;
  • 21) 0.160 157 203 674 749 206 528 × 2 = 0 + 0.320 314 407 349 498 413 056;
  • 22) 0.320 314 407 349 498 413 056 × 2 = 0 + 0.640 628 814 698 996 826 112;
  • 23) 0.640 628 814 698 996 826 112 × 2 = 1 + 0.281 257 629 397 993 652 224;
  • 24) 0.281 257 629 397 993 652 224 × 2 = 0 + 0.562 515 258 795 987 304 448;
  • 25) 0.562 515 258 795 987 304 448 × 2 = 1 + 0.125 030 517 591 974 608 896;
  • 26) 0.125 030 517 591 974 608 896 × 2 = 0 + 0.250 061 035 183 949 217 792;
  • 27) 0.250 061 035 183 949 217 792 × 2 = 0 + 0.500 122 070 367 898 435 584;
  • 28) 0.500 122 070 367 898 435 584 × 2 = 1 + 0.000 244 140 735 796 871 168;
  • 29) 0.000 244 140 735 796 871 168 × 2 = 0 + 0.000 488 281 471 593 742 336;
  • 30) 0.000 488 281 471 593 742 336 × 2 = 0 + 0.000 976 562 943 187 484 672;
  • 31) 0.000 976 562 943 187 484 672 × 2 = 0 + 0.001 953 125 886 374 969 344;
  • 32) 0.001 953 125 886 374 969 344 × 2 = 0 + 0.003 906 251 772 749 938 688;
  • 33) 0.003 906 251 772 749 938 688 × 2 = 0 + 0.007 812 503 545 499 877 376;
  • 34) 0.007 812 503 545 499 877 376 × 2 = 0 + 0.015 625 007 090 999 754 752;
  • 35) 0.015 625 007 090 999 754 752 × 2 = 0 + 0.031 250 014 181 999 509 504;
  • 36) 0.031 250 014 181 999 509 504 × 2 = 0 + 0.062 500 028 363 999 019 008;
  • 37) 0.062 500 028 363 999 019 008 × 2 = 0 + 0.125 000 056 727 998 038 016;
  • 38) 0.125 000 056 727 998 038 016 × 2 = 0 + 0.250 000 113 455 996 076 032;
  • 39) 0.250 000 113 455 996 076 032 × 2 = 0 + 0.500 000 226 911 992 152 064;
  • 40) 0.500 000 226 911 992 152 064 × 2 = 1 + 0.000 000 453 823 984 304 128;

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:


0.000 013 504 178 241 419 553(10) =

0.0000 0000 0000 0000 1110 0010 1001 0000 0000 0001(2)

6. Positive number before normalization:

0.000 013 504 178 241 419 553(10) =

0.0000 0000 0000 0000 1110 0010 1001 0000 0000 0001(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 17 positions to the right, so that only one non zero digit remains to the left of it:


0.000 013 504 178 241 419 553(10) =

0.0000 0000 0000 0000 1110 0010 1001 0000 0000 0001(2) =

0.0000 0000 0000 0000 1110 0010 1001 0000 0000 0001(2) × 20 =

1.1100 0101 0010 0000 0000 001(2) × 2-17


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): -17

Mantissa (not normalized):
1.1100 0101 0010 0000 0000 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-17 + 2(8-1) - 1 =

(-17 + 127)(10) =

110(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;
  • 110 ÷ 2 = 55 + 0;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 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) =

110(10) =

0110 1110(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. 110 0010 1001 0000 0000 0001 =

110 0010 1001 0000 0000 0001


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 1110

Mantissa (23 bits) =
110 0010 1001 0000 0000 0001


Decimal number -0.000 013 504 178 241 419 553 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1110 - 110 0010 1001 0000 0000 0001

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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