-0.000 000 000 742 147 676 646 698 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 646 698(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
-0.000 000 000 742 147 676 646 698(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:

|-0.000 000 000 742 147 676 646 698| = 0.000 000 000 742 147 676 646 698


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: 0.000 000 000 742 147 676 646 698.

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) 0.000 000 000 742 147 676 646 698 × 2 = 0 + 0.000 000 001 484 295 353 293 396;
  • 2) 0.000 000 001 484 295 353 293 396 × 2 = 0 + 0.000 000 002 968 590 706 586 792;
  • 3) 0.000 000 002 968 590 706 586 792 × 2 = 0 + 0.000 000 005 937 181 413 173 584;
  • 4) 0.000 000 005 937 181 413 173 584 × 2 = 0 + 0.000 000 011 874 362 826 347 168;
  • 5) 0.000 000 011 874 362 826 347 168 × 2 = 0 + 0.000 000 023 748 725 652 694 336;
  • 6) 0.000 000 023 748 725 652 694 336 × 2 = 0 + 0.000 000 047 497 451 305 388 672;
  • 7) 0.000 000 047 497 451 305 388 672 × 2 = 0 + 0.000 000 094 994 902 610 777 344;
  • 8) 0.000 000 094 994 902 610 777 344 × 2 = 0 + 0.000 000 189 989 805 221 554 688;
  • 9) 0.000 000 189 989 805 221 554 688 × 2 = 0 + 0.000 000 379 979 610 443 109 376;
  • 10) 0.000 000 379 979 610 443 109 376 × 2 = 0 + 0.000 000 759 959 220 886 218 752;
  • 11) 0.000 000 759 959 220 886 218 752 × 2 = 0 + 0.000 001 519 918 441 772 437 504;
  • 12) 0.000 001 519 918 441 772 437 504 × 2 = 0 + 0.000 003 039 836 883 544 875 008;
  • 13) 0.000 003 039 836 883 544 875 008 × 2 = 0 + 0.000 006 079 673 767 089 750 016;
  • 14) 0.000 006 079 673 767 089 750 016 × 2 = 0 + 0.000 012 159 347 534 179 500 032;
  • 15) 0.000 012 159 347 534 179 500 032 × 2 = 0 + 0.000 024 318 695 068 359 000 064;
  • 16) 0.000 024 318 695 068 359 000 064 × 2 = 0 + 0.000 048 637 390 136 718 000 128;
  • 17) 0.000 048 637 390 136 718 000 128 × 2 = 0 + 0.000 097 274 780 273 436 000 256;
  • 18) 0.000 097 274 780 273 436 000 256 × 2 = 0 + 0.000 194 549 560 546 872 000 512;
  • 19) 0.000 194 549 560 546 872 000 512 × 2 = 0 + 0.000 389 099 121 093 744 001 024;
  • 20) 0.000 389 099 121 093 744 001 024 × 2 = 0 + 0.000 778 198 242 187 488 002 048;
  • 21) 0.000 778 198 242 187 488 002 048 × 2 = 0 + 0.001 556 396 484 374 976 004 096;
  • 22) 0.001 556 396 484 374 976 004 096 × 2 = 0 + 0.003 112 792 968 749 952 008 192;
  • 23) 0.003 112 792 968 749 952 008 192 × 2 = 0 + 0.006 225 585 937 499 904 016 384;
  • 24) 0.006 225 585 937 499 904 016 384 × 2 = 0 + 0.012 451 171 874 999 808 032 768;
  • 25) 0.012 451 171 874 999 808 032 768 × 2 = 0 + 0.024 902 343 749 999 616 065 536;
  • 26) 0.024 902 343 749 999 616 065 536 × 2 = 0 + 0.049 804 687 499 999 232 131 072;
  • 27) 0.049 804 687 499 999 232 131 072 × 2 = 0 + 0.099 609 374 999 998 464 262 144;
  • 28) 0.099 609 374 999 998 464 262 144 × 2 = 0 + 0.199 218 749 999 996 928 524 288;
  • 29) 0.199 218 749 999 996 928 524 288 × 2 = 0 + 0.398 437 499 999 993 857 048 576;
  • 30) 0.398 437 499 999 993 857 048 576 × 2 = 0 + 0.796 874 999 999 987 714 097 152;
  • 31) 0.796 874 999 999 987 714 097 152 × 2 = 1 + 0.593 749 999 999 975 428 194 304;
  • 32) 0.593 749 999 999 975 428 194 304 × 2 = 1 + 0.187 499 999 999 950 856 388 608;
  • 33) 0.187 499 999 999 950 856 388 608 × 2 = 0 + 0.374 999 999 999 901 712 777 216;
  • 34) 0.374 999 999 999 901 712 777 216 × 2 = 0 + 0.749 999 999 999 803 425 554 432;
  • 35) 0.749 999 999 999 803 425 554 432 × 2 = 1 + 0.499 999 999 999 606 851 108 864;
  • 36) 0.499 999 999 999 606 851 108 864 × 2 = 0 + 0.999 999 999 999 213 702 217 728;
  • 37) 0.999 999 999 999 213 702 217 728 × 2 = 1 + 0.999 999 999 998 427 404 435 456;
  • 38) 0.999 999 999 998 427 404 435 456 × 2 = 1 + 0.999 999 999 996 854 808 870 912;
  • 39) 0.999 999 999 996 854 808 870 912 × 2 = 1 + 0.999 999 999 993 709 617 741 824;
  • 40) 0.999 999 999 993 709 617 741 824 × 2 = 1 + 0.999 999 999 987 419 235 483 648;
  • 41) 0.999 999 999 987 419 235 483 648 × 2 = 1 + 0.999 999 999 974 838 470 967 296;
  • 42) 0.999 999 999 974 838 470 967 296 × 2 = 1 + 0.999 999 999 949 676 941 934 592;
  • 43) 0.999 999 999 949 676 941 934 592 × 2 = 1 + 0.999 999 999 899 353 883 869 184;
  • 44) 0.999 999 999 899 353 883 869 184 × 2 = 1 + 0.999 999 999 798 707 767 738 368;
  • 45) 0.999 999 999 798 707 767 738 368 × 2 = 1 + 0.999 999 999 597 415 535 476 736;
  • 46) 0.999 999 999 597 415 535 476 736 × 2 = 1 + 0.999 999 999 194 831 070 953 472;
  • 47) 0.999 999 999 194 831 070 953 472 × 2 = 1 + 0.999 999 998 389 662 141 906 944;
  • 48) 0.999 999 998 389 662 141 906 944 × 2 = 1 + 0.999 999 996 779 324 283 813 888;
  • 49) 0.999 999 996 779 324 283 813 888 × 2 = 1 + 0.999 999 993 558 648 567 627 776;
  • 50) 0.999 999 993 558 648 567 627 776 × 2 = 1 + 0.999 999 987 117 297 135 255 552;
  • 51) 0.999 999 987 117 297 135 255 552 × 2 = 1 + 0.999 999 974 234 594 270 511 104;
  • 52) 0.999 999 974 234 594 270 511 104 × 2 = 1 + 0.999 999 948 469 188 541 022 208;
  • 53) 0.999 999 948 469 188 541 022 208 × 2 = 1 + 0.999 999 896 938 377 082 044 416;
  • 54) 0.999 999 896 938 377 082 044 416 × 2 = 1 + 0.999 999 793 876 754 164 088 832;

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:


0.000 000 000 742 147 676 646 698(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 646 698(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:


0.000 000 000 742 147 676 646 698(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 -0.000 000 000 742 147 676 646 698 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