0.000 000 000 000 032 499 996 756 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 000 000 032 499 996 756(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
0.000 000 000 000 032 499 996 756(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. 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;

2. 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)


3. Convert to binary (base 2) the fractional part: 0.000 000 000 000 032 499 996 756.

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) 0.000 000 000 000 032 499 996 756 × 2 = 0 + 0.000 000 000 000 064 999 993 512;
  • 2) 0.000 000 000 000 064 999 993 512 × 2 = 0 + 0.000 000 000 000 129 999 987 024;
  • 3) 0.000 000 000 000 129 999 987 024 × 2 = 0 + 0.000 000 000 000 259 999 974 048;
  • 4) 0.000 000 000 000 259 999 974 048 × 2 = 0 + 0.000 000 000 000 519 999 948 096;
  • 5) 0.000 000 000 000 519 999 948 096 × 2 = 0 + 0.000 000 000 001 039 999 896 192;
  • 6) 0.000 000 000 001 039 999 896 192 × 2 = 0 + 0.000 000 000 002 079 999 792 384;
  • 7) 0.000 000 000 002 079 999 792 384 × 2 = 0 + 0.000 000 000 004 159 999 584 768;
  • 8) 0.000 000 000 004 159 999 584 768 × 2 = 0 + 0.000 000 000 008 319 999 169 536;
  • 9) 0.000 000 000 008 319 999 169 536 × 2 = 0 + 0.000 000 000 016 639 998 339 072;
  • 10) 0.000 000 000 016 639 998 339 072 × 2 = 0 + 0.000 000 000 033 279 996 678 144;
  • 11) 0.000 000 000 033 279 996 678 144 × 2 = 0 + 0.000 000 000 066 559 993 356 288;
  • 12) 0.000 000 000 066 559 993 356 288 × 2 = 0 + 0.000 000 000 133 119 986 712 576;
  • 13) 0.000 000 000 133 119 986 712 576 × 2 = 0 + 0.000 000 000 266 239 973 425 152;
  • 14) 0.000 000 000 266 239 973 425 152 × 2 = 0 + 0.000 000 000 532 479 946 850 304;
  • 15) 0.000 000 000 532 479 946 850 304 × 2 = 0 + 0.000 000 001 064 959 893 700 608;
  • 16) 0.000 000 001 064 959 893 700 608 × 2 = 0 + 0.000 000 002 129 919 787 401 216;
  • 17) 0.000 000 002 129 919 787 401 216 × 2 = 0 + 0.000 000 004 259 839 574 802 432;
  • 18) 0.000 000 004 259 839 574 802 432 × 2 = 0 + 0.000 000 008 519 679 149 604 864;
  • 19) 0.000 000 008 519 679 149 604 864 × 2 = 0 + 0.000 000 017 039 358 299 209 728;
  • 20) 0.000 000 017 039 358 299 209 728 × 2 = 0 + 0.000 000 034 078 716 598 419 456;
  • 21) 0.000 000 034 078 716 598 419 456 × 2 = 0 + 0.000 000 068 157 433 196 838 912;
  • 22) 0.000 000 068 157 433 196 838 912 × 2 = 0 + 0.000 000 136 314 866 393 677 824;
  • 23) 0.000 000 136 314 866 393 677 824 × 2 = 0 + 0.000 000 272 629 732 787 355 648;
  • 24) 0.000 000 272 629 732 787 355 648 × 2 = 0 + 0.000 000 545 259 465 574 711 296;
  • 25) 0.000 000 545 259 465 574 711 296 × 2 = 0 + 0.000 001 090 518 931 149 422 592;
  • 26) 0.000 001 090 518 931 149 422 592 × 2 = 0 + 0.000 002 181 037 862 298 845 184;
  • 27) 0.000 002 181 037 862 298 845 184 × 2 = 0 + 0.000 004 362 075 724 597 690 368;
  • 28) 0.000 004 362 075 724 597 690 368 × 2 = 0 + 0.000 008 724 151 449 195 380 736;
  • 29) 0.000 008 724 151 449 195 380 736 × 2 = 0 + 0.000 017 448 302 898 390 761 472;
  • 30) 0.000 017 448 302 898 390 761 472 × 2 = 0 + 0.000 034 896 605 796 781 522 944;
  • 31) 0.000 034 896 605 796 781 522 944 × 2 = 0 + 0.000 069 793 211 593 563 045 888;
  • 32) 0.000 069 793 211 593 563 045 888 × 2 = 0 + 0.000 139 586 423 187 126 091 776;
  • 33) 0.000 139 586 423 187 126 091 776 × 2 = 0 + 0.000 279 172 846 374 252 183 552;
  • 34) 0.000 279 172 846 374 252 183 552 × 2 = 0 + 0.000 558 345 692 748 504 367 104;
  • 35) 0.000 558 345 692 748 504 367 104 × 2 = 0 + 0.001 116 691 385 497 008 734 208;
  • 36) 0.001 116 691 385 497 008 734 208 × 2 = 0 + 0.002 233 382 770 994 017 468 416;
  • 37) 0.002 233 382 770 994 017 468 416 × 2 = 0 + 0.004 466 765 541 988 034 936 832;
  • 38) 0.004 466 765 541 988 034 936 832 × 2 = 0 + 0.008 933 531 083 976 069 873 664;
  • 39) 0.008 933 531 083 976 069 873 664 × 2 = 0 + 0.017 867 062 167 952 139 747 328;
  • 40) 0.017 867 062 167 952 139 747 328 × 2 = 0 + 0.035 734 124 335 904 279 494 656;
  • 41) 0.035 734 124 335 904 279 494 656 × 2 = 0 + 0.071 468 248 671 808 558 989 312;
  • 42) 0.071 468 248 671 808 558 989 312 × 2 = 0 + 0.142 936 497 343 617 117 978 624;
  • 43) 0.142 936 497 343 617 117 978 624 × 2 = 0 + 0.285 872 994 687 234 235 957 248;
  • 44) 0.285 872 994 687 234 235 957 248 × 2 = 0 + 0.571 745 989 374 468 471 914 496;
  • 45) 0.571 745 989 374 468 471 914 496 × 2 = 1 + 0.143 491 978 748 936 943 828 992;
  • 46) 0.143 491 978 748 936 943 828 992 × 2 = 0 + 0.286 983 957 497 873 887 657 984;
  • 47) 0.286 983 957 497 873 887 657 984 × 2 = 0 + 0.573 967 914 995 747 775 315 968;
  • 48) 0.573 967 914 995 747 775 315 968 × 2 = 1 + 0.147 935 829 991 495 550 631 936;
  • 49) 0.147 935 829 991 495 550 631 936 × 2 = 0 + 0.295 871 659 982 991 101 263 872;
  • 50) 0.295 871 659 982 991 101 263 872 × 2 = 0 + 0.591 743 319 965 982 202 527 744;
  • 51) 0.591 743 319 965 982 202 527 744 × 2 = 1 + 0.183 486 639 931 964 405 055 488;
  • 52) 0.183 486 639 931 964 405 055 488 × 2 = 0 + 0.366 973 279 863 928 810 110 976;
  • 53) 0.366 973 279 863 928 810 110 976 × 2 = 0 + 0.733 946 559 727 857 620 221 952;
  • 54) 0.733 946 559 727 857 620 221 952 × 2 = 1 + 0.467 893 119 455 715 240 443 904;
  • 55) 0.467 893 119 455 715 240 443 904 × 2 = 0 + 0.935 786 238 911 430 480 887 808;
  • 56) 0.935 786 238 911 430 480 887 808 × 2 = 1 + 0.871 572 477 822 860 961 775 616;
  • 57) 0.871 572 477 822 860 961 775 616 × 2 = 1 + 0.743 144 955 645 721 923 551 232;
  • 58) 0.743 144 955 645 721 923 551 232 × 2 = 1 + 0.486 289 911 291 443 847 102 464;
  • 59) 0.486 289 911 291 443 847 102 464 × 2 = 0 + 0.972 579 822 582 887 694 204 928;
  • 60) 0.972 579 822 582 887 694 204 928 × 2 = 1 + 0.945 159 645 165 775 388 409 856;
  • 61) 0.945 159 645 165 775 388 409 856 × 2 = 1 + 0.890 319 290 331 550 776 819 712;
  • 62) 0.890 319 290 331 550 776 819 712 × 2 = 1 + 0.780 638 580 663 101 553 639 424;
  • 63) 0.780 638 580 663 101 553 639 424 × 2 = 1 + 0.561 277 161 326 203 107 278 848;
  • 64) 0.561 277 161 326 203 107 278 848 × 2 = 1 + 0.122 554 322 652 406 214 557 696;
  • 65) 0.122 554 322 652 406 214 557 696 × 2 = 0 + 0.245 108 645 304 812 429 115 392;
  • 66) 0.245 108 645 304 812 429 115 392 × 2 = 0 + 0.490 217 290 609 624 858 230 784;
  • 67) 0.490 217 290 609 624 858 230 784 × 2 = 0 + 0.980 434 581 219 249 716 461 568;
  • 68) 0.980 434 581 219 249 716 461 568 × 2 = 1 + 0.960 869 162 438 499 432 923 136;

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).


4. 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:


0.000 000 000 000 032 499 996 756(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0010 0101 1101 1111 0001(2)

5. Positive number before normalization:

0.000 000 000 000 032 499 996 756(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0010 0101 1101 1111 0001(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 45 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 000 032 499 996 756(10) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0010 0101 1101 1111 0001(2) =

0.0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1001 0010 0101 1101 1111 0001(2) × 20 =

1.0010 0100 1011 1011 1110 001(2) × 2-45


7. 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 0 (a positive number)

Exponent (unadjusted): -45

Mantissa (not normalized):
1.0010 0100 1011 1011 1110 001


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-45 + 2(8-1) - 1 =

(-45 + 127)(10) =

82(10)


9. 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;
  • 82 ÷ 2 = 41 + 0;
  • 41 ÷ 2 = 20 + 1;
  • 20 ÷ 2 = 10 + 0;
  • 10 ÷ 2 = 5 + 0;
  • 5 ÷ 2 = 2 + 1;
  • 2 ÷ 2 = 1 + 0;
  • 1 ÷ 2 = 0 + 1;

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

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


Exponent (adjusted) =

82(10) =

0101 0010(2)


11. 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. 001 0010 0101 1101 1111 0001 =

001 0010 0101 1101 1111 0001


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

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0101 0010

Mantissa (23 bits) =
001 0010 0101 1101 1111 0001


Decimal number 0.000 000 000 000 032 499 996 756 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0101 0010 - 001 0010 0101 1101 1111 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