-0.000 000 000 742 147 676 646 578 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 578(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
-0.000 000 000 742 147 676 646 578(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:

|-0.000 000 000 742 147 676 646 578| = 0.000 000 000 742 147 676 646 578


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: 0.000 000 000 742 147 676 646 578.

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) 0.000 000 000 742 147 676 646 578 × 2 = 0 + 0.000 000 001 484 295 353 293 156;
  • 2) 0.000 000 001 484 295 353 293 156 × 2 = 0 + 0.000 000 002 968 590 706 586 312;
  • 3) 0.000 000 002 968 590 706 586 312 × 2 = 0 + 0.000 000 005 937 181 413 172 624;
  • 4) 0.000 000 005 937 181 413 172 624 × 2 = 0 + 0.000 000 011 874 362 826 345 248;
  • 5) 0.000 000 011 874 362 826 345 248 × 2 = 0 + 0.000 000 023 748 725 652 690 496;
  • 6) 0.000 000 023 748 725 652 690 496 × 2 = 0 + 0.000 000 047 497 451 305 380 992;
  • 7) 0.000 000 047 497 451 305 380 992 × 2 = 0 + 0.000 000 094 994 902 610 761 984;
  • 8) 0.000 000 094 994 902 610 761 984 × 2 = 0 + 0.000 000 189 989 805 221 523 968;
  • 9) 0.000 000 189 989 805 221 523 968 × 2 = 0 + 0.000 000 379 979 610 443 047 936;
  • 10) 0.000 000 379 979 610 443 047 936 × 2 = 0 + 0.000 000 759 959 220 886 095 872;
  • 11) 0.000 000 759 959 220 886 095 872 × 2 = 0 + 0.000 001 519 918 441 772 191 744;
  • 12) 0.000 001 519 918 441 772 191 744 × 2 = 0 + 0.000 003 039 836 883 544 383 488;
  • 13) 0.000 003 039 836 883 544 383 488 × 2 = 0 + 0.000 006 079 673 767 088 766 976;
  • 14) 0.000 006 079 673 767 088 766 976 × 2 = 0 + 0.000 012 159 347 534 177 533 952;
  • 15) 0.000 012 159 347 534 177 533 952 × 2 = 0 + 0.000 024 318 695 068 355 067 904;
  • 16) 0.000 024 318 695 068 355 067 904 × 2 = 0 + 0.000 048 637 390 136 710 135 808;
  • 17) 0.000 048 637 390 136 710 135 808 × 2 = 0 + 0.000 097 274 780 273 420 271 616;
  • 18) 0.000 097 274 780 273 420 271 616 × 2 = 0 + 0.000 194 549 560 546 840 543 232;
  • 19) 0.000 194 549 560 546 840 543 232 × 2 = 0 + 0.000 389 099 121 093 681 086 464;
  • 20) 0.000 389 099 121 093 681 086 464 × 2 = 0 + 0.000 778 198 242 187 362 172 928;
  • 21) 0.000 778 198 242 187 362 172 928 × 2 = 0 + 0.001 556 396 484 374 724 345 856;
  • 22) 0.001 556 396 484 374 724 345 856 × 2 = 0 + 0.003 112 792 968 749 448 691 712;
  • 23) 0.003 112 792 968 749 448 691 712 × 2 = 0 + 0.006 225 585 937 498 897 383 424;
  • 24) 0.006 225 585 937 498 897 383 424 × 2 = 0 + 0.012 451 171 874 997 794 766 848;
  • 25) 0.012 451 171 874 997 794 766 848 × 2 = 0 + 0.024 902 343 749 995 589 533 696;
  • 26) 0.024 902 343 749 995 589 533 696 × 2 = 0 + 0.049 804 687 499 991 179 067 392;
  • 27) 0.049 804 687 499 991 179 067 392 × 2 = 0 + 0.099 609 374 999 982 358 134 784;
  • 28) 0.099 609 374 999 982 358 134 784 × 2 = 0 + 0.199 218 749 999 964 716 269 568;
  • 29) 0.199 218 749 999 964 716 269 568 × 2 = 0 + 0.398 437 499 999 929 432 539 136;
  • 30) 0.398 437 499 999 929 432 539 136 × 2 = 0 + 0.796 874 999 999 858 865 078 272;
  • 31) 0.796 874 999 999 858 865 078 272 × 2 = 1 + 0.593 749 999 999 717 730 156 544;
  • 32) 0.593 749 999 999 717 730 156 544 × 2 = 1 + 0.187 499 999 999 435 460 313 088;
  • 33) 0.187 499 999 999 435 460 313 088 × 2 = 0 + 0.374 999 999 998 870 920 626 176;
  • 34) 0.374 999 999 998 870 920 626 176 × 2 = 0 + 0.749 999 999 997 741 841 252 352;
  • 35) 0.749 999 999 997 741 841 252 352 × 2 = 1 + 0.499 999 999 995 483 682 504 704;
  • 36) 0.499 999 999 995 483 682 504 704 × 2 = 0 + 0.999 999 999 990 967 365 009 408;
  • 37) 0.999 999 999 990 967 365 009 408 × 2 = 1 + 0.999 999 999 981 934 730 018 816;
  • 38) 0.999 999 999 981 934 730 018 816 × 2 = 1 + 0.999 999 999 963 869 460 037 632;
  • 39) 0.999 999 999 963 869 460 037 632 × 2 = 1 + 0.999 999 999 927 738 920 075 264;
  • 40) 0.999 999 999 927 738 920 075 264 × 2 = 1 + 0.999 999 999 855 477 840 150 528;
  • 41) 0.999 999 999 855 477 840 150 528 × 2 = 1 + 0.999 999 999 710 955 680 301 056;
  • 42) 0.999 999 999 710 955 680 301 056 × 2 = 1 + 0.999 999 999 421 911 360 602 112;
  • 43) 0.999 999 999 421 911 360 602 112 × 2 = 1 + 0.999 999 998 843 822 721 204 224;
  • 44) 0.999 999 998 843 822 721 204 224 × 2 = 1 + 0.999 999 997 687 645 442 408 448;
  • 45) 0.999 999 997 687 645 442 408 448 × 2 = 1 + 0.999 999 995 375 290 884 816 896;
  • 46) 0.999 999 995 375 290 884 816 896 × 2 = 1 + 0.999 999 990 750 581 769 633 792;
  • 47) 0.999 999 990 750 581 769 633 792 × 2 = 1 + 0.999 999 981 501 163 539 267 584;
  • 48) 0.999 999 981 501 163 539 267 584 × 2 = 1 + 0.999 999 963 002 327 078 535 168;
  • 49) 0.999 999 963 002 327 078 535 168 × 2 = 1 + 0.999 999 926 004 654 157 070 336;
  • 50) 0.999 999 926 004 654 157 070 336 × 2 = 1 + 0.999 999 852 009 308 314 140 672;
  • 51) 0.999 999 852 009 308 314 140 672 × 2 = 1 + 0.999 999 704 018 616 628 281 344;
  • 52) 0.999 999 704 018 616 628 281 344 × 2 = 1 + 0.999 999 408 037 233 256 562 688;
  • 53) 0.999 999 408 037 233 256 562 688 × 2 = 1 + 0.999 998 816 074 466 513 125 376;
  • 54) 0.999 998 816 074 466 513 125 376 × 2 = 1 + 0.999 997 632 148 933 026 250 752;

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:


0.000 000 000 742 147 676 646 578(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 578(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:


0.000 000 000 742 147 676 646 578(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 -0.000 000 000 742 147 676 646 578 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