-0.000 000 000 742 147 676 627 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 627(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
-0.000 000 000 742 147 676 627(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:

|-0.000 000 000 742 147 676 627| = 0.000 000 000 742 147 676 627


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: 0.000 000 000 742 147 676 627.

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) 0.000 000 000 742 147 676 627 × 2 = 0 + 0.000 000 001 484 295 353 254;
  • 2) 0.000 000 001 484 295 353 254 × 2 = 0 + 0.000 000 002 968 590 706 508;
  • 3) 0.000 000 002 968 590 706 508 × 2 = 0 + 0.000 000 005 937 181 413 016;
  • 4) 0.000 000 005 937 181 413 016 × 2 = 0 + 0.000 000 011 874 362 826 032;
  • 5) 0.000 000 011 874 362 826 032 × 2 = 0 + 0.000 000 023 748 725 652 064;
  • 6) 0.000 000 023 748 725 652 064 × 2 = 0 + 0.000 000 047 497 451 304 128;
  • 7) 0.000 000 047 497 451 304 128 × 2 = 0 + 0.000 000 094 994 902 608 256;
  • 8) 0.000 000 094 994 902 608 256 × 2 = 0 + 0.000 000 189 989 805 216 512;
  • 9) 0.000 000 189 989 805 216 512 × 2 = 0 + 0.000 000 379 979 610 433 024;
  • 10) 0.000 000 379 979 610 433 024 × 2 = 0 + 0.000 000 759 959 220 866 048;
  • 11) 0.000 000 759 959 220 866 048 × 2 = 0 + 0.000 001 519 918 441 732 096;
  • 12) 0.000 001 519 918 441 732 096 × 2 = 0 + 0.000 003 039 836 883 464 192;
  • 13) 0.000 003 039 836 883 464 192 × 2 = 0 + 0.000 006 079 673 766 928 384;
  • 14) 0.000 006 079 673 766 928 384 × 2 = 0 + 0.000 012 159 347 533 856 768;
  • 15) 0.000 012 159 347 533 856 768 × 2 = 0 + 0.000 024 318 695 067 713 536;
  • 16) 0.000 024 318 695 067 713 536 × 2 = 0 + 0.000 048 637 390 135 427 072;
  • 17) 0.000 048 637 390 135 427 072 × 2 = 0 + 0.000 097 274 780 270 854 144;
  • 18) 0.000 097 274 780 270 854 144 × 2 = 0 + 0.000 194 549 560 541 708 288;
  • 19) 0.000 194 549 560 541 708 288 × 2 = 0 + 0.000 389 099 121 083 416 576;
  • 20) 0.000 389 099 121 083 416 576 × 2 = 0 + 0.000 778 198 242 166 833 152;
  • 21) 0.000 778 198 242 166 833 152 × 2 = 0 + 0.001 556 396 484 333 666 304;
  • 22) 0.001 556 396 484 333 666 304 × 2 = 0 + 0.003 112 792 968 667 332 608;
  • 23) 0.003 112 792 968 667 332 608 × 2 = 0 + 0.006 225 585 937 334 665 216;
  • 24) 0.006 225 585 937 334 665 216 × 2 = 0 + 0.012 451 171 874 669 330 432;
  • 25) 0.012 451 171 874 669 330 432 × 2 = 0 + 0.024 902 343 749 338 660 864;
  • 26) 0.024 902 343 749 338 660 864 × 2 = 0 + 0.049 804 687 498 677 321 728;
  • 27) 0.049 804 687 498 677 321 728 × 2 = 0 + 0.099 609 374 997 354 643 456;
  • 28) 0.099 609 374 997 354 643 456 × 2 = 0 + 0.199 218 749 994 709 286 912;
  • 29) 0.199 218 749 994 709 286 912 × 2 = 0 + 0.398 437 499 989 418 573 824;
  • 30) 0.398 437 499 989 418 573 824 × 2 = 0 + 0.796 874 999 978 837 147 648;
  • 31) 0.796 874 999 978 837 147 648 × 2 = 1 + 0.593 749 999 957 674 295 296;
  • 32) 0.593 749 999 957 674 295 296 × 2 = 1 + 0.187 499 999 915 348 590 592;
  • 33) 0.187 499 999 915 348 590 592 × 2 = 0 + 0.374 999 999 830 697 181 184;
  • 34) 0.374 999 999 830 697 181 184 × 2 = 0 + 0.749 999 999 661 394 362 368;
  • 35) 0.749 999 999 661 394 362 368 × 2 = 1 + 0.499 999 999 322 788 724 736;
  • 36) 0.499 999 999 322 788 724 736 × 2 = 0 + 0.999 999 998 645 577 449 472;
  • 37) 0.999 999 998 645 577 449 472 × 2 = 1 + 0.999 999 997 291 154 898 944;
  • 38) 0.999 999 997 291 154 898 944 × 2 = 1 + 0.999 999 994 582 309 797 888;
  • 39) 0.999 999 994 582 309 797 888 × 2 = 1 + 0.999 999 989 164 619 595 776;
  • 40) 0.999 999 989 164 619 595 776 × 2 = 1 + 0.999 999 978 329 239 191 552;
  • 41) 0.999 999 978 329 239 191 552 × 2 = 1 + 0.999 999 956 658 478 383 104;
  • 42) 0.999 999 956 658 478 383 104 × 2 = 1 + 0.999 999 913 316 956 766 208;
  • 43) 0.999 999 913 316 956 766 208 × 2 = 1 + 0.999 999 826 633 913 532 416;
  • 44) 0.999 999 826 633 913 532 416 × 2 = 1 + 0.999 999 653 267 827 064 832;
  • 45) 0.999 999 653 267 827 064 832 × 2 = 1 + 0.999 999 306 535 654 129 664;
  • 46) 0.999 999 306 535 654 129 664 × 2 = 1 + 0.999 998 613 071 308 259 328;
  • 47) 0.999 998 613 071 308 259 328 × 2 = 1 + 0.999 997 226 142 616 518 656;
  • 48) 0.999 997 226 142 616 518 656 × 2 = 1 + 0.999 994 452 285 233 037 312;
  • 49) 0.999 994 452 285 233 037 312 × 2 = 1 + 0.999 988 904 570 466 074 624;
  • 50) 0.999 988 904 570 466 074 624 × 2 = 1 + 0.999 977 809 140 932 149 248;
  • 51) 0.999 977 809 140 932 149 248 × 2 = 1 + 0.999 955 618 281 864 298 496;
  • 52) 0.999 955 618 281 864 298 496 × 2 = 1 + 0.999 911 236 563 728 596 992;
  • 53) 0.999 911 236 563 728 596 992 × 2 = 1 + 0.999 822 473 127 457 193 984;
  • 54) 0.999 822 473 127 457 193 984 × 2 = 1 + 0.999 644 946 254 914 387 968;

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:


0.000 000 000 742 147 676 627(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 627(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:


0.000 000 000 742 147 676 627(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 -0.000 000 000 742 147 676 627 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