-0.000 000 000 742 147 676 646 834 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 834(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
-0.000 000 000 742 147 676 646 834(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:

|-0.000 000 000 742 147 676 646 834| = 0.000 000 000 742 147 676 646 834


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: 0.000 000 000 742 147 676 646 834.

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) 0.000 000 000 742 147 676 646 834 × 2 = 0 + 0.000 000 001 484 295 353 293 668;
  • 2) 0.000 000 001 484 295 353 293 668 × 2 = 0 + 0.000 000 002 968 590 706 587 336;
  • 3) 0.000 000 002 968 590 706 587 336 × 2 = 0 + 0.000 000 005 937 181 413 174 672;
  • 4) 0.000 000 005 937 181 413 174 672 × 2 = 0 + 0.000 000 011 874 362 826 349 344;
  • 5) 0.000 000 011 874 362 826 349 344 × 2 = 0 + 0.000 000 023 748 725 652 698 688;
  • 6) 0.000 000 023 748 725 652 698 688 × 2 = 0 + 0.000 000 047 497 451 305 397 376;
  • 7) 0.000 000 047 497 451 305 397 376 × 2 = 0 + 0.000 000 094 994 902 610 794 752;
  • 8) 0.000 000 094 994 902 610 794 752 × 2 = 0 + 0.000 000 189 989 805 221 589 504;
  • 9) 0.000 000 189 989 805 221 589 504 × 2 = 0 + 0.000 000 379 979 610 443 179 008;
  • 10) 0.000 000 379 979 610 443 179 008 × 2 = 0 + 0.000 000 759 959 220 886 358 016;
  • 11) 0.000 000 759 959 220 886 358 016 × 2 = 0 + 0.000 001 519 918 441 772 716 032;
  • 12) 0.000 001 519 918 441 772 716 032 × 2 = 0 + 0.000 003 039 836 883 545 432 064;
  • 13) 0.000 003 039 836 883 545 432 064 × 2 = 0 + 0.000 006 079 673 767 090 864 128;
  • 14) 0.000 006 079 673 767 090 864 128 × 2 = 0 + 0.000 012 159 347 534 181 728 256;
  • 15) 0.000 012 159 347 534 181 728 256 × 2 = 0 + 0.000 024 318 695 068 363 456 512;
  • 16) 0.000 024 318 695 068 363 456 512 × 2 = 0 + 0.000 048 637 390 136 726 913 024;
  • 17) 0.000 048 637 390 136 726 913 024 × 2 = 0 + 0.000 097 274 780 273 453 826 048;
  • 18) 0.000 097 274 780 273 453 826 048 × 2 = 0 + 0.000 194 549 560 546 907 652 096;
  • 19) 0.000 194 549 560 546 907 652 096 × 2 = 0 + 0.000 389 099 121 093 815 304 192;
  • 20) 0.000 389 099 121 093 815 304 192 × 2 = 0 + 0.000 778 198 242 187 630 608 384;
  • 21) 0.000 778 198 242 187 630 608 384 × 2 = 0 + 0.001 556 396 484 375 261 216 768;
  • 22) 0.001 556 396 484 375 261 216 768 × 2 = 0 + 0.003 112 792 968 750 522 433 536;
  • 23) 0.003 112 792 968 750 522 433 536 × 2 = 0 + 0.006 225 585 937 501 044 867 072;
  • 24) 0.006 225 585 937 501 044 867 072 × 2 = 0 + 0.012 451 171 875 002 089 734 144;
  • 25) 0.012 451 171 875 002 089 734 144 × 2 = 0 + 0.024 902 343 750 004 179 468 288;
  • 26) 0.024 902 343 750 004 179 468 288 × 2 = 0 + 0.049 804 687 500 008 358 936 576;
  • 27) 0.049 804 687 500 008 358 936 576 × 2 = 0 + 0.099 609 375 000 016 717 873 152;
  • 28) 0.099 609 375 000 016 717 873 152 × 2 = 0 + 0.199 218 750 000 033 435 746 304;
  • 29) 0.199 218 750 000 033 435 746 304 × 2 = 0 + 0.398 437 500 000 066 871 492 608;
  • 30) 0.398 437 500 000 066 871 492 608 × 2 = 0 + 0.796 875 000 000 133 742 985 216;
  • 31) 0.796 875 000 000 133 742 985 216 × 2 = 1 + 0.593 750 000 000 267 485 970 432;
  • 32) 0.593 750 000 000 267 485 970 432 × 2 = 1 + 0.187 500 000 000 534 971 940 864;
  • 33) 0.187 500 000 000 534 971 940 864 × 2 = 0 + 0.375 000 000 001 069 943 881 728;
  • 34) 0.375 000 000 001 069 943 881 728 × 2 = 0 + 0.750 000 000 002 139 887 763 456;
  • 35) 0.750 000 000 002 139 887 763 456 × 2 = 1 + 0.500 000 000 004 279 775 526 912;
  • 36) 0.500 000 000 004 279 775 526 912 × 2 = 1 + 0.000 000 000 008 559 551 053 824;
  • 37) 0.000 000 000 008 559 551 053 824 × 2 = 0 + 0.000 000 000 017 119 102 107 648;
  • 38) 0.000 000 000 017 119 102 107 648 × 2 = 0 + 0.000 000 000 034 238 204 215 296;
  • 39) 0.000 000 000 034 238 204 215 296 × 2 = 0 + 0.000 000 000 068 476 408 430 592;
  • 40) 0.000 000 000 068 476 408 430 592 × 2 = 0 + 0.000 000 000 136 952 816 861 184;
  • 41) 0.000 000 000 136 952 816 861 184 × 2 = 0 + 0.000 000 000 273 905 633 722 368;
  • 42) 0.000 000 000 273 905 633 722 368 × 2 = 0 + 0.000 000 000 547 811 267 444 736;
  • 43) 0.000 000 000 547 811 267 444 736 × 2 = 0 + 0.000 000 001 095 622 534 889 472;
  • 44) 0.000 000 001 095 622 534 889 472 × 2 = 0 + 0.000 000 002 191 245 069 778 944;
  • 45) 0.000 000 002 191 245 069 778 944 × 2 = 0 + 0.000 000 004 382 490 139 557 888;
  • 46) 0.000 000 004 382 490 139 557 888 × 2 = 0 + 0.000 000 008 764 980 279 115 776;
  • 47) 0.000 000 008 764 980 279 115 776 × 2 = 0 + 0.000 000 017 529 960 558 231 552;
  • 48) 0.000 000 017 529 960 558 231 552 × 2 = 0 + 0.000 000 035 059 921 116 463 104;
  • 49) 0.000 000 035 059 921 116 463 104 × 2 = 0 + 0.000 000 070 119 842 232 926 208;
  • 50) 0.000 000 070 119 842 232 926 208 × 2 = 0 + 0.000 000 140 239 684 465 852 416;
  • 51) 0.000 000 140 239 684 465 852 416 × 2 = 0 + 0.000 000 280 479 368 931 704 832;
  • 52) 0.000 000 280 479 368 931 704 832 × 2 = 0 + 0.000 000 560 958 737 863 409 664;
  • 53) 0.000 000 560 958 737 863 409 664 × 2 = 0 + 0.000 001 121 917 475 726 819 328;
  • 54) 0.000 001 121 917 475 726 819 328 × 2 = 0 + 0.000 002 243 834 951 453 638 656;

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:


0.000 000 000 742 147 676 646 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2)

6. Positive number before normalization:

0.000 000 000 742 147 676 646 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(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:


0.000 000 000 742 147 676 646 834(10) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) =

0.0000 0000 0000 0000 0000 0000 0000 0011 0011 0000 0000 0000 0000 00(2) × 20 =

1.1001 1000 0000 0000 0000 000(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 1000 0000 0000 0000 000


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 1100 0000 0000 0000 0000 =

100 1100 0000 0000 0000 0000


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 1100 0000 0000 0000 0000


Decimal number -0.000 000 000 742 147 676 646 834 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1100 0000 0000 0000 0000

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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