-0.000 000 000 742 147 676 146 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 146(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
-0.000 000 000 742 147 676 146(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:

|-0.000 000 000 742 147 676 146| = 0.000 000 000 742 147 676 146


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: 0.000 000 000 742 147 676 146.

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) 0.000 000 000 742 147 676 146 × 2 = 0 + 0.000 000 001 484 295 352 292;
  • 2) 0.000 000 001 484 295 352 292 × 2 = 0 + 0.000 000 002 968 590 704 584;
  • 3) 0.000 000 002 968 590 704 584 × 2 = 0 + 0.000 000 005 937 181 409 168;
  • 4) 0.000 000 005 937 181 409 168 × 2 = 0 + 0.000 000 011 874 362 818 336;
  • 5) 0.000 000 011 874 362 818 336 × 2 = 0 + 0.000 000 023 748 725 636 672;
  • 6) 0.000 000 023 748 725 636 672 × 2 = 0 + 0.000 000 047 497 451 273 344;
  • 7) 0.000 000 047 497 451 273 344 × 2 = 0 + 0.000 000 094 994 902 546 688;
  • 8) 0.000 000 094 994 902 546 688 × 2 = 0 + 0.000 000 189 989 805 093 376;
  • 9) 0.000 000 189 989 805 093 376 × 2 = 0 + 0.000 000 379 979 610 186 752;
  • 10) 0.000 000 379 979 610 186 752 × 2 = 0 + 0.000 000 759 959 220 373 504;
  • 11) 0.000 000 759 959 220 373 504 × 2 = 0 + 0.000 001 519 918 440 747 008;
  • 12) 0.000 001 519 918 440 747 008 × 2 = 0 + 0.000 003 039 836 881 494 016;
  • 13) 0.000 003 039 836 881 494 016 × 2 = 0 + 0.000 006 079 673 762 988 032;
  • 14) 0.000 006 079 673 762 988 032 × 2 = 0 + 0.000 012 159 347 525 976 064;
  • 15) 0.000 012 159 347 525 976 064 × 2 = 0 + 0.000 024 318 695 051 952 128;
  • 16) 0.000 024 318 695 051 952 128 × 2 = 0 + 0.000 048 637 390 103 904 256;
  • 17) 0.000 048 637 390 103 904 256 × 2 = 0 + 0.000 097 274 780 207 808 512;
  • 18) 0.000 097 274 780 207 808 512 × 2 = 0 + 0.000 194 549 560 415 617 024;
  • 19) 0.000 194 549 560 415 617 024 × 2 = 0 + 0.000 389 099 120 831 234 048;
  • 20) 0.000 389 099 120 831 234 048 × 2 = 0 + 0.000 778 198 241 662 468 096;
  • 21) 0.000 778 198 241 662 468 096 × 2 = 0 + 0.001 556 396 483 324 936 192;
  • 22) 0.001 556 396 483 324 936 192 × 2 = 0 + 0.003 112 792 966 649 872 384;
  • 23) 0.003 112 792 966 649 872 384 × 2 = 0 + 0.006 225 585 933 299 744 768;
  • 24) 0.006 225 585 933 299 744 768 × 2 = 0 + 0.012 451 171 866 599 489 536;
  • 25) 0.012 451 171 866 599 489 536 × 2 = 0 + 0.024 902 343 733 198 979 072;
  • 26) 0.024 902 343 733 198 979 072 × 2 = 0 + 0.049 804 687 466 397 958 144;
  • 27) 0.049 804 687 466 397 958 144 × 2 = 0 + 0.099 609 374 932 795 916 288;
  • 28) 0.099 609 374 932 795 916 288 × 2 = 0 + 0.199 218 749 865 591 832 576;
  • 29) 0.199 218 749 865 591 832 576 × 2 = 0 + 0.398 437 499 731 183 665 152;
  • 30) 0.398 437 499 731 183 665 152 × 2 = 0 + 0.796 874 999 462 367 330 304;
  • 31) 0.796 874 999 462 367 330 304 × 2 = 1 + 0.593 749 998 924 734 660 608;
  • 32) 0.593 749 998 924 734 660 608 × 2 = 1 + 0.187 499 997 849 469 321 216;
  • 33) 0.187 499 997 849 469 321 216 × 2 = 0 + 0.374 999 995 698 938 642 432;
  • 34) 0.374 999 995 698 938 642 432 × 2 = 0 + 0.749 999 991 397 877 284 864;
  • 35) 0.749 999 991 397 877 284 864 × 2 = 1 + 0.499 999 982 795 754 569 728;
  • 36) 0.499 999 982 795 754 569 728 × 2 = 0 + 0.999 999 965 591 509 139 456;
  • 37) 0.999 999 965 591 509 139 456 × 2 = 1 + 0.999 999 931 183 018 278 912;
  • 38) 0.999 999 931 183 018 278 912 × 2 = 1 + 0.999 999 862 366 036 557 824;
  • 39) 0.999 999 862 366 036 557 824 × 2 = 1 + 0.999 999 724 732 073 115 648;
  • 40) 0.999 999 724 732 073 115 648 × 2 = 1 + 0.999 999 449 464 146 231 296;
  • 41) 0.999 999 449 464 146 231 296 × 2 = 1 + 0.999 998 898 928 292 462 592;
  • 42) 0.999 998 898 928 292 462 592 × 2 = 1 + 0.999 997 797 856 584 925 184;
  • 43) 0.999 997 797 856 584 925 184 × 2 = 1 + 0.999 995 595 713 169 850 368;
  • 44) 0.999 995 595 713 169 850 368 × 2 = 1 + 0.999 991 191 426 339 700 736;
  • 45) 0.999 991 191 426 339 700 736 × 2 = 1 + 0.999 982 382 852 679 401 472;
  • 46) 0.999 982 382 852 679 401 472 × 2 = 1 + 0.999 964 765 705 358 802 944;
  • 47) 0.999 964 765 705 358 802 944 × 2 = 1 + 0.999 929 531 410 717 605 888;
  • 48) 0.999 929 531 410 717 605 888 × 2 = 1 + 0.999 859 062 821 435 211 776;
  • 49) 0.999 859 062 821 435 211 776 × 2 = 1 + 0.999 718 125 642 870 423 552;
  • 50) 0.999 718 125 642 870 423 552 × 2 = 1 + 0.999 436 251 285 740 847 104;
  • 51) 0.999 436 251 285 740 847 104 × 2 = 1 + 0.998 872 502 571 481 694 208;
  • 52) 0.998 872 502 571 481 694 208 × 2 = 1 + 0.997 745 005 142 963 388 416;
  • 53) 0.997 745 005 142 963 388 416 × 2 = 1 + 0.995 490 010 285 926 776 832;
  • 54) 0.995 490 010 285 926 776 832 × 2 = 1 + 0.990 980 020 571 853 553 664;

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:


0.000 000 000 742 147 676 146(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 146(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:


0.000 000 000 742 147 676 146(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 -0.000 000 000 742 147 676 146 converted to 32 bit single precision IEEE 754 binary floating point representation:

1 - 0110 0000 - 100 1011 1111 1111 1111 1111

Share

» Convert decimal -0.000 000 000 742 147 676 181 to 32 bit single precision IEEE 754 binary floating point representation standard

» Calculations Performed by Our Visitors: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Data organized on a Monthly Basis

» Month 05, 2026 [May]: Decimal Numbers Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard. Calculations performed during the month of: May - by our visitors

» Improve your social skills: Are you using communication by design, or by default? NotebookLM YouTube Video of 6.33 minutes

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