-0.000 000 000 000 000 000 000 044 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 000 000 000 000 044(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
-0.000 000 000 000 000 000 000 044(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:

|-0.000 000 000 000 000 000 000 044| = 0.000 000 000 000 000 000 000 044


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: 0.000 000 000 000 000 000 000 044.

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) 0.000 000 000 000 000 000 000 044 × 2 = 0 + 0.000 000 000 000 000 000 000 088;
  • 2) 0.000 000 000 000 000 000 000 088 × 2 = 0 + 0.000 000 000 000 000 000 000 176;
  • 3) 0.000 000 000 000 000 000 000 176 × 2 = 0 + 0.000 000 000 000 000 000 000 352;
  • 4) 0.000 000 000 000 000 000 000 352 × 2 = 0 + 0.000 000 000 000 000 000 000 704;
  • 5) 0.000 000 000 000 000 000 000 704 × 2 = 0 + 0.000 000 000 000 000 000 001 408;
  • 6) 0.000 000 000 000 000 000 001 408 × 2 = 0 + 0.000 000 000 000 000 000 002 816;
  • 7) 0.000 000 000 000 000 000 002 816 × 2 = 0 + 0.000 000 000 000 000 000 005 632;
  • 8) 0.000 000 000 000 000 000 005 632 × 2 = 0 + 0.000 000 000 000 000 000 011 264;
  • 9) 0.000 000 000 000 000 000 011 264 × 2 = 0 + 0.000 000 000 000 000 000 022 528;
  • 10) 0.000 000 000 000 000 000 022 528 × 2 = 0 + 0.000 000 000 000 000 000 045 056;
  • 11) 0.000 000 000 000 000 000 045 056 × 2 = 0 + 0.000 000 000 000 000 000 090 112;
  • 12) 0.000 000 000 000 000 000 090 112 × 2 = 0 + 0.000 000 000 000 000 000 180 224;
  • 13) 0.000 000 000 000 000 000 180 224 × 2 = 0 + 0.000 000 000 000 000 000 360 448;
  • 14) 0.000 000 000 000 000 000 360 448 × 2 = 0 + 0.000 000 000 000 000 000 720 896;
  • 15) 0.000 000 000 000 000 000 720 896 × 2 = 0 + 0.000 000 000 000 000 001 441 792;
  • 16) 0.000 000 000 000 000 001 441 792 × 2 = 0 + 0.000 000 000 000 000 002 883 584;
  • 17) 0.000 000 000 000 000 002 883 584 × 2 = 0 + 0.000 000 000 000 000 005 767 168;
  • 18) 0.000 000 000 000 000 005 767 168 × 2 = 0 + 0.000 000 000 000 000 011 534 336;
  • 19) 0.000 000 000 000 000 011 534 336 × 2 = 0 + 0.000 000 000 000 000 023 068 672;
  • 20) 0.000 000 000 000 000 023 068 672 × 2 = 0 + 0.000 000 000 000 000 046 137 344;
  • 21) 0.000 000 000 000 000 046 137 344 × 2 = 0 + 0.000 000 000 000 000 092 274 688;
  • 22) 0.000 000 000 000 000 092 274 688 × 2 = 0 + 0.000 000 000 000 000 184 549 376;
  • 23) 0.000 000 000 000 000 184 549 376 × 2 = 0 + 0.000 000 000 000 000 369 098 752;
  • 24) 0.000 000 000 000 000 369 098 752 × 2 = 0 + 0.000 000 000 000 000 738 197 504;
  • 25) 0.000 000 000 000 000 738 197 504 × 2 = 0 + 0.000 000 000 000 001 476 395 008;
  • 26) 0.000 000 000 000 001 476 395 008 × 2 = 0 + 0.000 000 000 000 002 952 790 016;
  • 27) 0.000 000 000 000 002 952 790 016 × 2 = 0 + 0.000 000 000 000 005 905 580 032;
  • 28) 0.000 000 000 000 005 905 580 032 × 2 = 0 + 0.000 000 000 000 011 811 160 064;
  • 29) 0.000 000 000 000 011 811 160 064 × 2 = 0 + 0.000 000 000 000 023 622 320 128;
  • 30) 0.000 000 000 000 023 622 320 128 × 2 = 0 + 0.000 000 000 000 047 244 640 256;
  • 31) 0.000 000 000 000 047 244 640 256 × 2 = 0 + 0.000 000 000 000 094 489 280 512;
  • 32) 0.000 000 000 000 094 489 280 512 × 2 = 0 + 0.000 000 000 000 188 978 561 024;
  • 33) 0.000 000 000 000 188 978 561 024 × 2 = 0 + 0.000 000 000 000 377 957 122 048;
  • 34) 0.000 000 000 000 377 957 122 048 × 2 = 0 + 0.000 000 000 000 755 914 244 096;
  • 35) 0.000 000 000 000 755 914 244 096 × 2 = 0 + 0.000 000 000 001 511 828 488 192;
  • 36) 0.000 000 000 001 511 828 488 192 × 2 = 0 + 0.000 000 000 003 023 656 976 384;
  • 37) 0.000 000 000 003 023 656 976 384 × 2 = 0 + 0.000 000 000 006 047 313 952 768;
  • 38) 0.000 000 000 006 047 313 952 768 × 2 = 0 + 0.000 000 000 012 094 627 905 536;
  • 39) 0.000 000 000 012 094 627 905 536 × 2 = 0 + 0.000 000 000 024 189 255 811 072;
  • 40) 0.000 000 000 024 189 255 811 072 × 2 = 0 + 0.000 000 000 048 378 511 622 144;
  • 41) 0.000 000 000 048 378 511 622 144 × 2 = 0 + 0.000 000 000 096 757 023 244 288;
  • 42) 0.000 000 000 096 757 023 244 288 × 2 = 0 + 0.000 000 000 193 514 046 488 576;
  • 43) 0.000 000 000 193 514 046 488 576 × 2 = 0 + 0.000 000 000 387 028 092 977 152;
  • 44) 0.000 000 000 387 028 092 977 152 × 2 = 0 + 0.000 000 000 774 056 185 954 304;
  • 45) 0.000 000 000 774 056 185 954 304 × 2 = 0 + 0.000 000 001 548 112 371 908 608;
  • 46) 0.000 000 001 548 112 371 908 608 × 2 = 0 + 0.000 000 003 096 224 743 817 216;
  • 47) 0.000 000 003 096 224 743 817 216 × 2 = 0 + 0.000 000 006 192 449 487 634 432;
  • 48) 0.000 000 006 192 449 487 634 432 × 2 = 0 + 0.000 000 012 384 898 975 268 864;
  • 49) 0.000 000 012 384 898 975 268 864 × 2 = 0 + 0.000 000 024 769 797 950 537 728;
  • 50) 0.000 000 024 769 797 950 537 728 × 2 = 0 + 0.000 000 049 539 595 901 075 456;
  • 51) 0.000 000 049 539 595 901 075 456 × 2 = 0 + 0.000 000 099 079 191 802 150 912;
  • 52) 0.000 000 099 079 191 802 150 912 × 2 = 0 + 0.000 000 198 158 383 604 301 824;
  • 53) 0.000 000 198 158 383 604 301 824 × 2 = 0 + 0.000 000 396 316 767 208 603 648;
  • 54) 0.000 000 396 316 767 208 603 648 × 2 = 0 + 0.000 000 792 633 534 417 207 296;
  • 55) 0.000 000 792 633 534 417 207 296 × 2 = 0 + 0.000 001 585 267 068 834 414 592;
  • 56) 0.000 001 585 267 068 834 414 592 × 2 = 0 + 0.000 003 170 534 137 668 829 184;
  • 57) 0.000 003 170 534 137 668 829 184 × 2 = 0 + 0.000 006 341 068 275 337 658 368;
  • 58) 0.000 006 341 068 275 337 658 368 × 2 = 0 + 0.000 012 682 136 550 675 316 736;
  • 59) 0.000 012 682 136 550 675 316 736 × 2 = 0 + 0.000 025 364 273 101 350 633 472;
  • 60) 0.000 025 364 273 101 350 633 472 × 2 = 0 + 0.000 050 728 546 202 701 266 944;
  • 61) 0.000 050 728 546 202 701 266 944 × 2 = 0 + 0.000 101 457 092 405 402 533 888;
  • 62) 0.000 101 457 092 405 402 533 888 × 2 = 0 + 0.000 202 914 184 810 805 067 776;
  • 63) 0.000 202 914 184 810 805 067 776 × 2 = 0 + 0.000 405 828 369 621 610 135 552;
  • 64) 0.000 405 828 369 621 610 135 552 × 2 = 0 + 0.000 811 656 739 243 220 271 104;
  • 65) 0.000 811 656 739 243 220 271 104 × 2 = 0 + 0.001 623 313 478 486 440 542 208;
  • 66) 0.001 623 313 478 486 440 542 208 × 2 = 0 + 0.003 246 626 956 972 881 084 416;
  • 67) 0.003 246 626 956 972 881 084 416 × 2 = 0 + 0.006 493 253 913 945 762 168 832;
  • 68) 0.006 493 253 913 945 762 168 832 × 2 = 0 + 0.012 986 507 827 891 524 337 664;
  • 69) 0.012 986 507 827 891 524 337 664 × 2 = 0 + 0.025 973 015 655 783 048 675 328;
  • 70) 0.025 973 015 655 783 048 675 328 × 2 = 0 + 0.051 946 031 311 566 097 350 656;
  • 71) 0.051 946 031 311 566 097 350 656 × 2 = 0 + 0.103 892 062 623 132 194 701 312;
  • 72) 0.103 892 062 623 132 194 701 312 × 2 = 0 + 0.207 784 125 246 264 389 402 624;
  • 73) 0.207 784 125 246 264 389 402 624 × 2 = 0 + 0.415 568 250 492 528 778 805 248;
  • 74) 0.415 568 250 492 528 778 805 248 × 2 = 0 + 0.831 136 500 985 057 557 610 496;
  • 75) 0.831 136 500 985 057 557 610 496 × 2 = 1 + 0.662 273 001 970 115 115 220 992;
  • 76) 0.662 273 001 970 115 115 220 992 × 2 = 1 + 0.324 546 003 940 230 230 441 984;
  • 77) 0.324 546 003 940 230 230 441 984 × 2 = 0 + 0.649 092 007 880 460 460 883 968;
  • 78) 0.649 092 007 880 460 460 883 968 × 2 = 1 + 0.298 184 015 760 920 921 767 936;
  • 79) 0.298 184 015 760 920 921 767 936 × 2 = 0 + 0.596 368 031 521 841 843 535 872;
  • 80) 0.596 368 031 521 841 843 535 872 × 2 = 1 + 0.192 736 063 043 683 687 071 744;
  • 81) 0.192 736 063 043 683 687 071 744 × 2 = 0 + 0.385 472 126 087 367 374 143 488;
  • 82) 0.385 472 126 087 367 374 143 488 × 2 = 0 + 0.770 944 252 174 734 748 286 976;
  • 83) 0.770 944 252 174 734 748 286 976 × 2 = 1 + 0.541 888 504 349 469 496 573 952;
  • 84) 0.541 888 504 349 469 496 573 952 × 2 = 1 + 0.083 777 008 698 938 993 147 904;
  • 85) 0.083 777 008 698 938 993 147 904 × 2 = 0 + 0.167 554 017 397 877 986 295 808;
  • 86) 0.167 554 017 397 877 986 295 808 × 2 = 0 + 0.335 108 034 795 755 972 591 616;
  • 87) 0.335 108 034 795 755 972 591 616 × 2 = 0 + 0.670 216 069 591 511 945 183 232;
  • 88) 0.670 216 069 591 511 945 183 232 × 2 = 1 + 0.340 432 139 183 023 890 366 464;
  • 89) 0.340 432 139 183 023 890 366 464 × 2 = 0 + 0.680 864 278 366 047 780 732 928;
  • 90) 0.680 864 278 366 047 780 732 928 × 2 = 1 + 0.361 728 556 732 095 561 465 856;
  • 91) 0.361 728 556 732 095 561 465 856 × 2 = 0 + 0.723 457 113 464 191 122 931 712;
  • 92) 0.723 457 113 464 191 122 931 712 × 2 = 1 + 0.446 914 226 928 382 245 863 424;
  • 93) 0.446 914 226 928 382 245 863 424 × 2 = 0 + 0.893 828 453 856 764 491 726 848;
  • 94) 0.893 828 453 856 764 491 726 848 × 2 = 1 + 0.787 656 907 713 528 983 453 696;
  • 95) 0.787 656 907 713 528 983 453 696 × 2 = 1 + 0.575 313 815 427 057 966 907 392;
  • 96) 0.575 313 815 427 057 966 907 392 × 2 = 1 + 0.150 627 630 854 115 933 814 784;
  • 97) 0.150 627 630 854 115 933 814 784 × 2 = 0 + 0.301 255 261 708 231 867 629 568;
  • 98) 0.301 255 261 708 231 867 629 568 × 2 = 0 + 0.602 510 523 416 463 735 259 136;

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:


0.000 000 000 000 000 000 000 044(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0001 0101 0111 00(2)

6. Positive number before normalization:

0.000 000 000 000 000 000 000 044(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0001 0101 0111 00(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 75 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 000 000 000 044(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0001 0101 0111 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0011 0101 0011 0001 0101 0111 00(2) × 20 =

1.1010 1001 1000 1010 1011 100(2) × 2-75


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): -75

Mantissa (not normalized):
1.1010 1001 1000 1010 1011 100


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-75 + 2(8-1) - 1 =

(-75 + 127)(10) =

52(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;
  • 52 ÷ 2 = 26 + 0;
  • 26 ÷ 2 = 13 + 0;
  • 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) =

52(10) =

0011 0100(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. 101 0100 1100 0101 0101 1100 =

101 0100 1100 0101 0101 1100


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) =
0011 0100

Mantissa (23 bits) =
101 0100 1100 0101 0101 1100


Decimal number -0.000 000 000 000 000 000 000 044 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0011 0100 - 101 0100 1100 0101 0101 1100

Share

» Convert decimal -0.000 000 000 000 000 000 000 083 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 08, 2026 [August]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: August - by our visitors

» Improve your social skills: Are You Using Communication by Design, or by Default? Gemini Notebook YouTube Video of 6.33 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