-0.000 000 000 742 147 676 327 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 327(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
-0.000 000 000 742 147 676 327(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:

|-0.000 000 000 742 147 676 327| = 0.000 000 000 742 147 676 327


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: 0.000 000 000 742 147 676 327.

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) 0.000 000 000 742 147 676 327 × 2 = 0 + 0.000 000 001 484 295 352 654;
  • 2) 0.000 000 001 484 295 352 654 × 2 = 0 + 0.000 000 002 968 590 705 308;
  • 3) 0.000 000 002 968 590 705 308 × 2 = 0 + 0.000 000 005 937 181 410 616;
  • 4) 0.000 000 005 937 181 410 616 × 2 = 0 + 0.000 000 011 874 362 821 232;
  • 5) 0.000 000 011 874 362 821 232 × 2 = 0 + 0.000 000 023 748 725 642 464;
  • 6) 0.000 000 023 748 725 642 464 × 2 = 0 + 0.000 000 047 497 451 284 928;
  • 7) 0.000 000 047 497 451 284 928 × 2 = 0 + 0.000 000 094 994 902 569 856;
  • 8) 0.000 000 094 994 902 569 856 × 2 = 0 + 0.000 000 189 989 805 139 712;
  • 9) 0.000 000 189 989 805 139 712 × 2 = 0 + 0.000 000 379 979 610 279 424;
  • 10) 0.000 000 379 979 610 279 424 × 2 = 0 + 0.000 000 759 959 220 558 848;
  • 11) 0.000 000 759 959 220 558 848 × 2 = 0 + 0.000 001 519 918 441 117 696;
  • 12) 0.000 001 519 918 441 117 696 × 2 = 0 + 0.000 003 039 836 882 235 392;
  • 13) 0.000 003 039 836 882 235 392 × 2 = 0 + 0.000 006 079 673 764 470 784;
  • 14) 0.000 006 079 673 764 470 784 × 2 = 0 + 0.000 012 159 347 528 941 568;
  • 15) 0.000 012 159 347 528 941 568 × 2 = 0 + 0.000 024 318 695 057 883 136;
  • 16) 0.000 024 318 695 057 883 136 × 2 = 0 + 0.000 048 637 390 115 766 272;
  • 17) 0.000 048 637 390 115 766 272 × 2 = 0 + 0.000 097 274 780 231 532 544;
  • 18) 0.000 097 274 780 231 532 544 × 2 = 0 + 0.000 194 549 560 463 065 088;
  • 19) 0.000 194 549 560 463 065 088 × 2 = 0 + 0.000 389 099 120 926 130 176;
  • 20) 0.000 389 099 120 926 130 176 × 2 = 0 + 0.000 778 198 241 852 260 352;
  • 21) 0.000 778 198 241 852 260 352 × 2 = 0 + 0.001 556 396 483 704 520 704;
  • 22) 0.001 556 396 483 704 520 704 × 2 = 0 + 0.003 112 792 967 409 041 408;
  • 23) 0.003 112 792 967 409 041 408 × 2 = 0 + 0.006 225 585 934 818 082 816;
  • 24) 0.006 225 585 934 818 082 816 × 2 = 0 + 0.012 451 171 869 636 165 632;
  • 25) 0.012 451 171 869 636 165 632 × 2 = 0 + 0.024 902 343 739 272 331 264;
  • 26) 0.024 902 343 739 272 331 264 × 2 = 0 + 0.049 804 687 478 544 662 528;
  • 27) 0.049 804 687 478 544 662 528 × 2 = 0 + 0.099 609 374 957 089 325 056;
  • 28) 0.099 609 374 957 089 325 056 × 2 = 0 + 0.199 218 749 914 178 650 112;
  • 29) 0.199 218 749 914 178 650 112 × 2 = 0 + 0.398 437 499 828 357 300 224;
  • 30) 0.398 437 499 828 357 300 224 × 2 = 0 + 0.796 874 999 656 714 600 448;
  • 31) 0.796 874 999 656 714 600 448 × 2 = 1 + 0.593 749 999 313 429 200 896;
  • 32) 0.593 749 999 313 429 200 896 × 2 = 1 + 0.187 499 998 626 858 401 792;
  • 33) 0.187 499 998 626 858 401 792 × 2 = 0 + 0.374 999 997 253 716 803 584;
  • 34) 0.374 999 997 253 716 803 584 × 2 = 0 + 0.749 999 994 507 433 607 168;
  • 35) 0.749 999 994 507 433 607 168 × 2 = 1 + 0.499 999 989 014 867 214 336;
  • 36) 0.499 999 989 014 867 214 336 × 2 = 0 + 0.999 999 978 029 734 428 672;
  • 37) 0.999 999 978 029 734 428 672 × 2 = 1 + 0.999 999 956 059 468 857 344;
  • 38) 0.999 999 956 059 468 857 344 × 2 = 1 + 0.999 999 912 118 937 714 688;
  • 39) 0.999 999 912 118 937 714 688 × 2 = 1 + 0.999 999 824 237 875 429 376;
  • 40) 0.999 999 824 237 875 429 376 × 2 = 1 + 0.999 999 648 475 750 858 752;
  • 41) 0.999 999 648 475 750 858 752 × 2 = 1 + 0.999 999 296 951 501 717 504;
  • 42) 0.999 999 296 951 501 717 504 × 2 = 1 + 0.999 998 593 903 003 435 008;
  • 43) 0.999 998 593 903 003 435 008 × 2 = 1 + 0.999 997 187 806 006 870 016;
  • 44) 0.999 997 187 806 006 870 016 × 2 = 1 + 0.999 994 375 612 013 740 032;
  • 45) 0.999 994 375 612 013 740 032 × 2 = 1 + 0.999 988 751 224 027 480 064;
  • 46) 0.999 988 751 224 027 480 064 × 2 = 1 + 0.999 977 502 448 054 960 128;
  • 47) 0.999 977 502 448 054 960 128 × 2 = 1 + 0.999 955 004 896 109 920 256;
  • 48) 0.999 955 004 896 109 920 256 × 2 = 1 + 0.999 910 009 792 219 840 512;
  • 49) 0.999 910 009 792 219 840 512 × 2 = 1 + 0.999 820 019 584 439 681 024;
  • 50) 0.999 820 019 584 439 681 024 × 2 = 1 + 0.999 640 039 168 879 362 048;
  • 51) 0.999 640 039 168 879 362 048 × 2 = 1 + 0.999 280 078 337 758 724 096;
  • 52) 0.999 280 078 337 758 724 096 × 2 = 1 + 0.998 560 156 675 517 448 192;
  • 53) 0.998 560 156 675 517 448 192 × 2 = 1 + 0.997 120 313 351 034 896 384;
  • 54) 0.997 120 313 351 034 896 384 × 2 = 1 + 0.994 240 626 702 069 792 768;

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:


0.000 000 000 742 147 676 327(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 327(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:


0.000 000 000 742 147 676 327(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 -0.000 000 000 742 147 676 327 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