32bit IEEE 754: Decimal ↗ Single Precision Floating Point Binary: -0.000 000 000 000 000 001 625 Convert the Number to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard, From a Base 10 Decimal System Number

Number -0.000 000 000 000 000 001 625(10) converted and written in 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:

|-0.000 000 000 000 000 001 625| = 0.000 000 000 000 000 001 625

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: 0.000 000 000 000 000 001 625.

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) 0.000 000 000 000 000 001 625 × 2 = 0 + 0.000 000 000 000 000 003 25;
  • 2) 0.000 000 000 000 000 003 25 × 2 = 0 + 0.000 000 000 000 000 006 5;
  • 3) 0.000 000 000 000 000 006 5 × 2 = 0 + 0.000 000 000 000 000 013;
  • 4) 0.000 000 000 000 000 013 × 2 = 0 + 0.000 000 000 000 000 026;
  • 5) 0.000 000 000 000 000 026 × 2 = 0 + 0.000 000 000 000 000 052;
  • 6) 0.000 000 000 000 000 052 × 2 = 0 + 0.000 000 000 000 000 104;
  • 7) 0.000 000 000 000 000 104 × 2 = 0 + 0.000 000 000 000 000 208;
  • 8) 0.000 000 000 000 000 208 × 2 = 0 + 0.000 000 000 000 000 416;
  • 9) 0.000 000 000 000 000 416 × 2 = 0 + 0.000 000 000 000 000 832;
  • 10) 0.000 000 000 000 000 832 × 2 = 0 + 0.000 000 000 000 001 664;
  • 11) 0.000 000 000 000 001 664 × 2 = 0 + 0.000 000 000 000 003 328;
  • 12) 0.000 000 000 000 003 328 × 2 = 0 + 0.000 000 000 000 006 656;
  • 13) 0.000 000 000 000 006 656 × 2 = 0 + 0.000 000 000 000 013 312;
  • 14) 0.000 000 000 000 013 312 × 2 = 0 + 0.000 000 000 000 026 624;
  • 15) 0.000 000 000 000 026 624 × 2 = 0 + 0.000 000 000 000 053 248;
  • 16) 0.000 000 000 000 053 248 × 2 = 0 + 0.000 000 000 000 106 496;
  • 17) 0.000 000 000 000 106 496 × 2 = 0 + 0.000 000 000 000 212 992;
  • 18) 0.000 000 000 000 212 992 × 2 = 0 + 0.000 000 000 000 425 984;
  • 19) 0.000 000 000 000 425 984 × 2 = 0 + 0.000 000 000 000 851 968;
  • 20) 0.000 000 000 000 851 968 × 2 = 0 + 0.000 000 000 001 703 936;
  • 21) 0.000 000 000 001 703 936 × 2 = 0 + 0.000 000 000 003 407 872;
  • 22) 0.000 000 000 003 407 872 × 2 = 0 + 0.000 000 000 006 815 744;
  • 23) 0.000 000 000 006 815 744 × 2 = 0 + 0.000 000 000 013 631 488;
  • 24) 0.000 000 000 013 631 488 × 2 = 0 + 0.000 000 000 027 262 976;
  • 25) 0.000 000 000 027 262 976 × 2 = 0 + 0.000 000 000 054 525 952;
  • 26) 0.000 000 000 054 525 952 × 2 = 0 + 0.000 000 000 109 051 904;
  • 27) 0.000 000 000 109 051 904 × 2 = 0 + 0.000 000 000 218 103 808;
  • 28) 0.000 000 000 218 103 808 × 2 = 0 + 0.000 000 000 436 207 616;
  • 29) 0.000 000 000 436 207 616 × 2 = 0 + 0.000 000 000 872 415 232;
  • 30) 0.000 000 000 872 415 232 × 2 = 0 + 0.000 000 001 744 830 464;
  • 31) 0.000 000 001 744 830 464 × 2 = 0 + 0.000 000 003 489 660 928;
  • 32) 0.000 000 003 489 660 928 × 2 = 0 + 0.000 000 006 979 321 856;
  • 33) 0.000 000 006 979 321 856 × 2 = 0 + 0.000 000 013 958 643 712;
  • 34) 0.000 000 013 958 643 712 × 2 = 0 + 0.000 000 027 917 287 424;
  • 35) 0.000 000 027 917 287 424 × 2 = 0 + 0.000 000 055 834 574 848;
  • 36) 0.000 000 055 834 574 848 × 2 = 0 + 0.000 000 111 669 149 696;
  • 37) 0.000 000 111 669 149 696 × 2 = 0 + 0.000 000 223 338 299 392;
  • 38) 0.000 000 223 338 299 392 × 2 = 0 + 0.000 000 446 676 598 784;
  • 39) 0.000 000 446 676 598 784 × 2 = 0 + 0.000 000 893 353 197 568;
  • 40) 0.000 000 893 353 197 568 × 2 = 0 + 0.000 001 786 706 395 136;
  • 41) 0.000 001 786 706 395 136 × 2 = 0 + 0.000 003 573 412 790 272;
  • 42) 0.000 003 573 412 790 272 × 2 = 0 + 0.000 007 146 825 580 544;
  • 43) 0.000 007 146 825 580 544 × 2 = 0 + 0.000 014 293 651 161 088;
  • 44) 0.000 014 293 651 161 088 × 2 = 0 + 0.000 028 587 302 322 176;
  • 45) 0.000 028 587 302 322 176 × 2 = 0 + 0.000 057 174 604 644 352;
  • 46) 0.000 057 174 604 644 352 × 2 = 0 + 0.000 114 349 209 288 704;
  • 47) 0.000 114 349 209 288 704 × 2 = 0 + 0.000 228 698 418 577 408;
  • 48) 0.000 228 698 418 577 408 × 2 = 0 + 0.000 457 396 837 154 816;
  • 49) 0.000 457 396 837 154 816 × 2 = 0 + 0.000 914 793 674 309 632;
  • 50) 0.000 914 793 674 309 632 × 2 = 0 + 0.001 829 587 348 619 264;
  • 51) 0.001 829 587 348 619 264 × 2 = 0 + 0.003 659 174 697 238 528;
  • 52) 0.003 659 174 697 238 528 × 2 = 0 + 0.007 318 349 394 477 056;
  • 53) 0.007 318 349 394 477 056 × 2 = 0 + 0.014 636 698 788 954 112;
  • 54) 0.014 636 698 788 954 112 × 2 = 0 + 0.029 273 397 577 908 224;
  • 55) 0.029 273 397 577 908 224 × 2 = 0 + 0.058 546 795 155 816 448;
  • 56) 0.058 546 795 155 816 448 × 2 = 0 + 0.117 093 590 311 632 896;
  • 57) 0.117 093 590 311 632 896 × 2 = 0 + 0.234 187 180 623 265 792;
  • 58) 0.234 187 180 623 265 792 × 2 = 0 + 0.468 374 361 246 531 584;
  • 59) 0.468 374 361 246 531 584 × 2 = 0 + 0.936 748 722 493 063 168;
  • 60) 0.936 748 722 493 063 168 × 2 = 1 + 0.873 497 444 986 126 336;
  • 61) 0.873 497 444 986 126 336 × 2 = 1 + 0.746 994 889 972 252 672;
  • 62) 0.746 994 889 972 252 672 × 2 = 1 + 0.493 989 779 944 505 344;
  • 63) 0.493 989 779 944 505 344 × 2 = 0 + 0.987 979 559 889 010 688;
  • 64) 0.987 979 559 889 010 688 × 2 = 1 + 0.975 959 119 778 021 376;
  • 65) 0.975 959 119 778 021 376 × 2 = 1 + 0.951 918 239 556 042 752;
  • 66) 0.951 918 239 556 042 752 × 2 = 1 + 0.903 836 479 112 085 504;
  • 67) 0.903 836 479 112 085 504 × 2 = 1 + 0.807 672 958 224 171 008;
  • 68) 0.807 672 958 224 171 008 × 2 = 1 + 0.615 345 916 448 342 016;
  • 69) 0.615 345 916 448 342 016 × 2 = 1 + 0.230 691 832 896 684 032;
  • 70) 0.230 691 832 896 684 032 × 2 = 0 + 0.461 383 665 793 368 064;
  • 71) 0.461 383 665 793 368 064 × 2 = 0 + 0.922 767 331 586 736 128;
  • 72) 0.922 767 331 586 736 128 × 2 = 1 + 0.845 534 663 173 472 256;
  • 73) 0.845 534 663 173 472 256 × 2 = 1 + 0.691 069 326 346 944 512;
  • 74) 0.691 069 326 346 944 512 × 2 = 1 + 0.382 138 652 693 889 024;
  • 75) 0.382 138 652 693 889 024 × 2 = 0 + 0.764 277 305 387 778 048;
  • 76) 0.764 277 305 387 778 048 × 2 = 1 + 0.528 554 610 775 556 096;
  • 77) 0.528 554 610 775 556 096 × 2 = 1 + 0.057 109 221 551 112 192;
  • 78) 0.057 109 221 551 112 192 × 2 = 0 + 0.114 218 443 102 224 384;
  • 79) 0.114 218 443 102 224 384 × 2 = 0 + 0.228 436 886 204 448 768;
  • 80) 0.228 436 886 204 448 768 × 2 = 0 + 0.456 873 772 408 897 536;
  • 81) 0.456 873 772 408 897 536 × 2 = 0 + 0.913 747 544 817 795 072;
  • 82) 0.913 747 544 817 795 072 × 2 = 1 + 0.827 495 089 635 590 144;
  • 83) 0.827 495 089 635 590 144 × 2 = 1 + 0.654 990 179 271 180 288;

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...)


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:


0.000 000 000 000 000 001 625(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1101 1111 1001 1101 1000 011(2)


6. Positive number before normalization:

0.000 000 000 000 000 001 625(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1101 1111 1001 1101 1000 011(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 60 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 001 625(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1101 1111 1001 1101 1000 011(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0001 1101 1111 1001 1101 1000 011(2) × 20 =

1.1101 1111 1001 1101 1000 011(2) × 2-60


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): -60

Mantissa (not normalized):
1.1101 1111 1001 1101 1000 011


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-60 + 2(8-1) - 1 =

(-60 + 127)(10) =

67(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;
  • 67 ÷ 2 = 33 + 1;
  • 33 ÷ 2 = 16 + 1;
  • 16 ÷ 2 = 8 + 0;
  • 8 ÷ 2 = 4 + 0;
  • 4 ÷ 2 = 2 + 0;
  • 2 ÷ 2 = 1 + 0;
  • 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) =

67(10) =

0100 0011(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 1111 1100 1110 1100 0011 =

110 1111 1100 1110 1100 0011


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) =
0100 0011

Mantissa (23 bits) =
110 1111 1100 1110 1100 0011


The base ten decimal number -0.000 000 000 000 000 001 625 converted and written in 32 bit single precision IEEE 754 binary floating point representation:
1 - 0100 0011 - 110 1111 1100 1110 1100 0011

More operations with decimal numbers converted to 32 bit single precision IEEE 754 binary floating point representation:

» Number -0.000 000 000 000 000 001 626 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

» Number -0.000 000 000 000 000 001 624 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point representation = ?

The latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

Number 110 990 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number 10 100 001 111 010 111 051 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number -15 184.687 79 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number 0.066 406 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number 0.015 872 68 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number 375 274 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number 43 949 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number -6.14 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number -5 555.64 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
Number 24.752 39 converted from decimal system (written in base ten) to 32 bit single precision IEEE 754 binary floating point representation standard May 12 01:12 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

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