-0.000 000 349 613 904 800 000 238 907 548 936 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 349 613 904 800 000 238 907 548 936(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
-0.000 000 349 613 904 800 000 238 907 548 936(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:

|-0.000 000 349 613 904 800 000 238 907 548 936| = 0.000 000 349 613 904 800 000 238 907 548 936


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: 0.000 000 349 613 904 800 000 238 907 548 936.

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) 0.000 000 349 613 904 800 000 238 907 548 936 × 2 = 0 + 0.000 000 699 227 809 600 000 477 815 097 872;
  • 2) 0.000 000 699 227 809 600 000 477 815 097 872 × 2 = 0 + 0.000 001 398 455 619 200 000 955 630 195 744;
  • 3) 0.000 001 398 455 619 200 000 955 630 195 744 × 2 = 0 + 0.000 002 796 911 238 400 001 911 260 391 488;
  • 4) 0.000 002 796 911 238 400 001 911 260 391 488 × 2 = 0 + 0.000 005 593 822 476 800 003 822 520 782 976;
  • 5) 0.000 005 593 822 476 800 003 822 520 782 976 × 2 = 0 + 0.000 011 187 644 953 600 007 645 041 565 952;
  • 6) 0.000 011 187 644 953 600 007 645 041 565 952 × 2 = 0 + 0.000 022 375 289 907 200 015 290 083 131 904;
  • 7) 0.000 022 375 289 907 200 015 290 083 131 904 × 2 = 0 + 0.000 044 750 579 814 400 030 580 166 263 808;
  • 8) 0.000 044 750 579 814 400 030 580 166 263 808 × 2 = 0 + 0.000 089 501 159 628 800 061 160 332 527 616;
  • 9) 0.000 089 501 159 628 800 061 160 332 527 616 × 2 = 0 + 0.000 179 002 319 257 600 122 320 665 055 232;
  • 10) 0.000 179 002 319 257 600 122 320 665 055 232 × 2 = 0 + 0.000 358 004 638 515 200 244 641 330 110 464;
  • 11) 0.000 358 004 638 515 200 244 641 330 110 464 × 2 = 0 + 0.000 716 009 277 030 400 489 282 660 220 928;
  • 12) 0.000 716 009 277 030 400 489 282 660 220 928 × 2 = 0 + 0.001 432 018 554 060 800 978 565 320 441 856;
  • 13) 0.001 432 018 554 060 800 978 565 320 441 856 × 2 = 0 + 0.002 864 037 108 121 601 957 130 640 883 712;
  • 14) 0.002 864 037 108 121 601 957 130 640 883 712 × 2 = 0 + 0.005 728 074 216 243 203 914 261 281 767 424;
  • 15) 0.005 728 074 216 243 203 914 261 281 767 424 × 2 = 0 + 0.011 456 148 432 486 407 828 522 563 534 848;
  • 16) 0.011 456 148 432 486 407 828 522 563 534 848 × 2 = 0 + 0.022 912 296 864 972 815 657 045 127 069 696;
  • 17) 0.022 912 296 864 972 815 657 045 127 069 696 × 2 = 0 + 0.045 824 593 729 945 631 314 090 254 139 392;
  • 18) 0.045 824 593 729 945 631 314 090 254 139 392 × 2 = 0 + 0.091 649 187 459 891 262 628 180 508 278 784;
  • 19) 0.091 649 187 459 891 262 628 180 508 278 784 × 2 = 0 + 0.183 298 374 919 782 525 256 361 016 557 568;
  • 20) 0.183 298 374 919 782 525 256 361 016 557 568 × 2 = 0 + 0.366 596 749 839 565 050 512 722 033 115 136;
  • 21) 0.366 596 749 839 565 050 512 722 033 115 136 × 2 = 0 + 0.733 193 499 679 130 101 025 444 066 230 272;
  • 22) 0.733 193 499 679 130 101 025 444 066 230 272 × 2 = 1 + 0.466 386 999 358 260 202 050 888 132 460 544;
  • 23) 0.466 386 999 358 260 202 050 888 132 460 544 × 2 = 0 + 0.932 773 998 716 520 404 101 776 264 921 088;
  • 24) 0.932 773 998 716 520 404 101 776 264 921 088 × 2 = 1 + 0.865 547 997 433 040 808 203 552 529 842 176;
  • 25) 0.865 547 997 433 040 808 203 552 529 842 176 × 2 = 1 + 0.731 095 994 866 081 616 407 105 059 684 352;
  • 26) 0.731 095 994 866 081 616 407 105 059 684 352 × 2 = 1 + 0.462 191 989 732 163 232 814 210 119 368 704;
  • 27) 0.462 191 989 732 163 232 814 210 119 368 704 × 2 = 0 + 0.924 383 979 464 326 465 628 420 238 737 408;
  • 28) 0.924 383 979 464 326 465 628 420 238 737 408 × 2 = 1 + 0.848 767 958 928 652 931 256 840 477 474 816;
  • 29) 0.848 767 958 928 652 931 256 840 477 474 816 × 2 = 1 + 0.697 535 917 857 305 862 513 680 954 949 632;
  • 30) 0.697 535 917 857 305 862 513 680 954 949 632 × 2 = 1 + 0.395 071 835 714 611 725 027 361 909 899 264;
  • 31) 0.395 071 835 714 611 725 027 361 909 899 264 × 2 = 0 + 0.790 143 671 429 223 450 054 723 819 798 528;
  • 32) 0.790 143 671 429 223 450 054 723 819 798 528 × 2 = 1 + 0.580 287 342 858 446 900 109 447 639 597 056;
  • 33) 0.580 287 342 858 446 900 109 447 639 597 056 × 2 = 1 + 0.160 574 685 716 893 800 218 895 279 194 112;
  • 34) 0.160 574 685 716 893 800 218 895 279 194 112 × 2 = 0 + 0.321 149 371 433 787 600 437 790 558 388 224;
  • 35) 0.321 149 371 433 787 600 437 790 558 388 224 × 2 = 0 + 0.642 298 742 867 575 200 875 581 116 776 448;
  • 36) 0.642 298 742 867 575 200 875 581 116 776 448 × 2 = 1 + 0.284 597 485 735 150 401 751 162 233 552 896;
  • 37) 0.284 597 485 735 150 401 751 162 233 552 896 × 2 = 0 + 0.569 194 971 470 300 803 502 324 467 105 792;
  • 38) 0.569 194 971 470 300 803 502 324 467 105 792 × 2 = 1 + 0.138 389 942 940 601 607 004 648 934 211 584;
  • 39) 0.138 389 942 940 601 607 004 648 934 211 584 × 2 = 0 + 0.276 779 885 881 203 214 009 297 868 423 168;
  • 40) 0.276 779 885 881 203 214 009 297 868 423 168 × 2 = 0 + 0.553 559 771 762 406 428 018 595 736 846 336;
  • 41) 0.553 559 771 762 406 428 018 595 736 846 336 × 2 = 1 + 0.107 119 543 524 812 856 037 191 473 692 672;
  • 42) 0.107 119 543 524 812 856 037 191 473 692 672 × 2 = 0 + 0.214 239 087 049 625 712 074 382 947 385 344;
  • 43) 0.214 239 087 049 625 712 074 382 947 385 344 × 2 = 0 + 0.428 478 174 099 251 424 148 765 894 770 688;
  • 44) 0.428 478 174 099 251 424 148 765 894 770 688 × 2 = 0 + 0.856 956 348 198 502 848 297 531 789 541 376;
  • 45) 0.856 956 348 198 502 848 297 531 789 541 376 × 2 = 1 + 0.713 912 696 397 005 696 595 063 579 082 752;

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:


0.000 000 349 613 904 800 000 238 907 548 936(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2)

6. Positive number before normalization:

0.000 000 349 613 904 800 000 238 907 548 936(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 22 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 349 613 904 800 000 238 907 548 936(10) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2) =

0.0000 0000 0000 0000 0000 0101 1101 1101 1001 0100 1000 1(2) × 20 =

1.0111 0111 0110 0101 0010 001(2) × 2-22


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): -22

Mantissa (not normalized):
1.0111 0111 0110 0101 0010 001


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-22 + 2(8-1) - 1 =

(-22 + 127)(10) =

105(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;
  • 105 ÷ 2 = 52 + 1;
  • 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) =

105(10) =

0110 1001(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. 011 1011 1011 0010 1001 0001 =

011 1011 1011 0010 1001 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 1001

Mantissa (23 bits) =
011 1011 1011 0010 1001 0001


Decimal number -0.000 000 349 613 904 800 000 238 907 548 936 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 1001 - 011 1011 1011 0010 1001 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