-0.000 000 000 742 147 676 507 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 507(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
-0.000 000 000 742 147 676 507(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:

|-0.000 000 000 742 147 676 507| = 0.000 000 000 742 147 676 507


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: 0.000 000 000 742 147 676 507.

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) 0.000 000 000 742 147 676 507 × 2 = 0 + 0.000 000 001 484 295 353 014;
  • 2) 0.000 000 001 484 295 353 014 × 2 = 0 + 0.000 000 002 968 590 706 028;
  • 3) 0.000 000 002 968 590 706 028 × 2 = 0 + 0.000 000 005 937 181 412 056;
  • 4) 0.000 000 005 937 181 412 056 × 2 = 0 + 0.000 000 011 874 362 824 112;
  • 5) 0.000 000 011 874 362 824 112 × 2 = 0 + 0.000 000 023 748 725 648 224;
  • 6) 0.000 000 023 748 725 648 224 × 2 = 0 + 0.000 000 047 497 451 296 448;
  • 7) 0.000 000 047 497 451 296 448 × 2 = 0 + 0.000 000 094 994 902 592 896;
  • 8) 0.000 000 094 994 902 592 896 × 2 = 0 + 0.000 000 189 989 805 185 792;
  • 9) 0.000 000 189 989 805 185 792 × 2 = 0 + 0.000 000 379 979 610 371 584;
  • 10) 0.000 000 379 979 610 371 584 × 2 = 0 + 0.000 000 759 959 220 743 168;
  • 11) 0.000 000 759 959 220 743 168 × 2 = 0 + 0.000 001 519 918 441 486 336;
  • 12) 0.000 001 519 918 441 486 336 × 2 = 0 + 0.000 003 039 836 882 972 672;
  • 13) 0.000 003 039 836 882 972 672 × 2 = 0 + 0.000 006 079 673 765 945 344;
  • 14) 0.000 006 079 673 765 945 344 × 2 = 0 + 0.000 012 159 347 531 890 688;
  • 15) 0.000 012 159 347 531 890 688 × 2 = 0 + 0.000 024 318 695 063 781 376;
  • 16) 0.000 024 318 695 063 781 376 × 2 = 0 + 0.000 048 637 390 127 562 752;
  • 17) 0.000 048 637 390 127 562 752 × 2 = 0 + 0.000 097 274 780 255 125 504;
  • 18) 0.000 097 274 780 255 125 504 × 2 = 0 + 0.000 194 549 560 510 251 008;
  • 19) 0.000 194 549 560 510 251 008 × 2 = 0 + 0.000 389 099 121 020 502 016;
  • 20) 0.000 389 099 121 020 502 016 × 2 = 0 + 0.000 778 198 242 041 004 032;
  • 21) 0.000 778 198 242 041 004 032 × 2 = 0 + 0.001 556 396 484 082 008 064;
  • 22) 0.001 556 396 484 082 008 064 × 2 = 0 + 0.003 112 792 968 164 016 128;
  • 23) 0.003 112 792 968 164 016 128 × 2 = 0 + 0.006 225 585 936 328 032 256;
  • 24) 0.006 225 585 936 328 032 256 × 2 = 0 + 0.012 451 171 872 656 064 512;
  • 25) 0.012 451 171 872 656 064 512 × 2 = 0 + 0.024 902 343 745 312 129 024;
  • 26) 0.024 902 343 745 312 129 024 × 2 = 0 + 0.049 804 687 490 624 258 048;
  • 27) 0.049 804 687 490 624 258 048 × 2 = 0 + 0.099 609 374 981 248 516 096;
  • 28) 0.099 609 374 981 248 516 096 × 2 = 0 + 0.199 218 749 962 497 032 192;
  • 29) 0.199 218 749 962 497 032 192 × 2 = 0 + 0.398 437 499 924 994 064 384;
  • 30) 0.398 437 499 924 994 064 384 × 2 = 0 + 0.796 874 999 849 988 128 768;
  • 31) 0.796 874 999 849 988 128 768 × 2 = 1 + 0.593 749 999 699 976 257 536;
  • 32) 0.593 749 999 699 976 257 536 × 2 = 1 + 0.187 499 999 399 952 515 072;
  • 33) 0.187 499 999 399 952 515 072 × 2 = 0 + 0.374 999 998 799 905 030 144;
  • 34) 0.374 999 998 799 905 030 144 × 2 = 0 + 0.749 999 997 599 810 060 288;
  • 35) 0.749 999 997 599 810 060 288 × 2 = 1 + 0.499 999 995 199 620 120 576;
  • 36) 0.499 999 995 199 620 120 576 × 2 = 0 + 0.999 999 990 399 240 241 152;
  • 37) 0.999 999 990 399 240 241 152 × 2 = 1 + 0.999 999 980 798 480 482 304;
  • 38) 0.999 999 980 798 480 482 304 × 2 = 1 + 0.999 999 961 596 960 964 608;
  • 39) 0.999 999 961 596 960 964 608 × 2 = 1 + 0.999 999 923 193 921 929 216;
  • 40) 0.999 999 923 193 921 929 216 × 2 = 1 + 0.999 999 846 387 843 858 432;
  • 41) 0.999 999 846 387 843 858 432 × 2 = 1 + 0.999 999 692 775 687 716 864;
  • 42) 0.999 999 692 775 687 716 864 × 2 = 1 + 0.999 999 385 551 375 433 728;
  • 43) 0.999 999 385 551 375 433 728 × 2 = 1 + 0.999 998 771 102 750 867 456;
  • 44) 0.999 998 771 102 750 867 456 × 2 = 1 + 0.999 997 542 205 501 734 912;
  • 45) 0.999 997 542 205 501 734 912 × 2 = 1 + 0.999 995 084 411 003 469 824;
  • 46) 0.999 995 084 411 003 469 824 × 2 = 1 + 0.999 990 168 822 006 939 648;
  • 47) 0.999 990 168 822 006 939 648 × 2 = 1 + 0.999 980 337 644 013 879 296;
  • 48) 0.999 980 337 644 013 879 296 × 2 = 1 + 0.999 960 675 288 027 758 592;
  • 49) 0.999 960 675 288 027 758 592 × 2 = 1 + 0.999 921 350 576 055 517 184;
  • 50) 0.999 921 350 576 055 517 184 × 2 = 1 + 0.999 842 701 152 111 034 368;
  • 51) 0.999 842 701 152 111 034 368 × 2 = 1 + 0.999 685 402 304 222 068 736;
  • 52) 0.999 685 402 304 222 068 736 × 2 = 1 + 0.999 370 804 608 444 137 472;
  • 53) 0.999 370 804 608 444 137 472 × 2 = 1 + 0.998 741 609 216 888 274 944;
  • 54) 0.998 741 609 216 888 274 944 × 2 = 1 + 0.997 483 218 433 776 549 888;

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:


0.000 000 000 742 147 676 507(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 507(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2)

7. Normalize the binary representation of the number.

Shift the decimal mark 31 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 000 742 147 676 507(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0010 1111 1111 1111 1111 11(2) × 20 =

1.1001 0111 1111 1111 1111 111(2) × 2-31


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): -31

Mantissa (not normalized):
1.1001 0111 1111 1111 1111 111


9. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

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

-31 + 2(8-1) - 1 =

(-31 + 127)(10) =

96(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;
  • 96 ÷ 2 = 48 + 0;
  • 48 ÷ 2 = 24 + 0;
  • 24 ÷ 2 = 12 + 0;
  • 12 ÷ 2 = 6 + 0;
  • 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) =

96(10) =

0110 0000(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. 100 1011 1111 1111 1111 1111 =

100 1011 1111 1111 1111 1111


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 0000

Mantissa (23 bits) =
100 1011 1111 1111 1111 1111


Decimal number -0.000 000 000 742 147 676 507 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

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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