-0.000 000 000 742 147 676 624 5 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 624 5(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
-0.000 000 000 742 147 676 624 5(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:

|-0.000 000 000 742 147 676 624 5| = 0.000 000 000 742 147 676 624 5


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: 0.000 000 000 742 147 676 624 5.

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) 0.000 000 000 742 147 676 624 5 × 2 = 0 + 0.000 000 001 484 295 353 249;
  • 2) 0.000 000 001 484 295 353 249 × 2 = 0 + 0.000 000 002 968 590 706 498;
  • 3) 0.000 000 002 968 590 706 498 × 2 = 0 + 0.000 000 005 937 181 412 996;
  • 4) 0.000 000 005 937 181 412 996 × 2 = 0 + 0.000 000 011 874 362 825 992;
  • 5) 0.000 000 011 874 362 825 992 × 2 = 0 + 0.000 000 023 748 725 651 984;
  • 6) 0.000 000 023 748 725 651 984 × 2 = 0 + 0.000 000 047 497 451 303 968;
  • 7) 0.000 000 047 497 451 303 968 × 2 = 0 + 0.000 000 094 994 902 607 936;
  • 8) 0.000 000 094 994 902 607 936 × 2 = 0 + 0.000 000 189 989 805 215 872;
  • 9) 0.000 000 189 989 805 215 872 × 2 = 0 + 0.000 000 379 979 610 431 744;
  • 10) 0.000 000 379 979 610 431 744 × 2 = 0 + 0.000 000 759 959 220 863 488;
  • 11) 0.000 000 759 959 220 863 488 × 2 = 0 + 0.000 001 519 918 441 726 976;
  • 12) 0.000 001 519 918 441 726 976 × 2 = 0 + 0.000 003 039 836 883 453 952;
  • 13) 0.000 003 039 836 883 453 952 × 2 = 0 + 0.000 006 079 673 766 907 904;
  • 14) 0.000 006 079 673 766 907 904 × 2 = 0 + 0.000 012 159 347 533 815 808;
  • 15) 0.000 012 159 347 533 815 808 × 2 = 0 + 0.000 024 318 695 067 631 616;
  • 16) 0.000 024 318 695 067 631 616 × 2 = 0 + 0.000 048 637 390 135 263 232;
  • 17) 0.000 048 637 390 135 263 232 × 2 = 0 + 0.000 097 274 780 270 526 464;
  • 18) 0.000 097 274 780 270 526 464 × 2 = 0 + 0.000 194 549 560 541 052 928;
  • 19) 0.000 194 549 560 541 052 928 × 2 = 0 + 0.000 389 099 121 082 105 856;
  • 20) 0.000 389 099 121 082 105 856 × 2 = 0 + 0.000 778 198 242 164 211 712;
  • 21) 0.000 778 198 242 164 211 712 × 2 = 0 + 0.001 556 396 484 328 423 424;
  • 22) 0.001 556 396 484 328 423 424 × 2 = 0 + 0.003 112 792 968 656 846 848;
  • 23) 0.003 112 792 968 656 846 848 × 2 = 0 + 0.006 225 585 937 313 693 696;
  • 24) 0.006 225 585 937 313 693 696 × 2 = 0 + 0.012 451 171 874 627 387 392;
  • 25) 0.012 451 171 874 627 387 392 × 2 = 0 + 0.024 902 343 749 254 774 784;
  • 26) 0.024 902 343 749 254 774 784 × 2 = 0 + 0.049 804 687 498 509 549 568;
  • 27) 0.049 804 687 498 509 549 568 × 2 = 0 + 0.099 609 374 997 019 099 136;
  • 28) 0.099 609 374 997 019 099 136 × 2 = 0 + 0.199 218 749 994 038 198 272;
  • 29) 0.199 218 749 994 038 198 272 × 2 = 0 + 0.398 437 499 988 076 396 544;
  • 30) 0.398 437 499 988 076 396 544 × 2 = 0 + 0.796 874 999 976 152 793 088;
  • 31) 0.796 874 999 976 152 793 088 × 2 = 1 + 0.593 749 999 952 305 586 176;
  • 32) 0.593 749 999 952 305 586 176 × 2 = 1 + 0.187 499 999 904 611 172 352;
  • 33) 0.187 499 999 904 611 172 352 × 2 = 0 + 0.374 999 999 809 222 344 704;
  • 34) 0.374 999 999 809 222 344 704 × 2 = 0 + 0.749 999 999 618 444 689 408;
  • 35) 0.749 999 999 618 444 689 408 × 2 = 1 + 0.499 999 999 236 889 378 816;
  • 36) 0.499 999 999 236 889 378 816 × 2 = 0 + 0.999 999 998 473 778 757 632;
  • 37) 0.999 999 998 473 778 757 632 × 2 = 1 + 0.999 999 996 947 557 515 264;
  • 38) 0.999 999 996 947 557 515 264 × 2 = 1 + 0.999 999 993 895 115 030 528;
  • 39) 0.999 999 993 895 115 030 528 × 2 = 1 + 0.999 999 987 790 230 061 056;
  • 40) 0.999 999 987 790 230 061 056 × 2 = 1 + 0.999 999 975 580 460 122 112;
  • 41) 0.999 999 975 580 460 122 112 × 2 = 1 + 0.999 999 951 160 920 244 224;
  • 42) 0.999 999 951 160 920 244 224 × 2 = 1 + 0.999 999 902 321 840 488 448;
  • 43) 0.999 999 902 321 840 488 448 × 2 = 1 + 0.999 999 804 643 680 976 896;
  • 44) 0.999 999 804 643 680 976 896 × 2 = 1 + 0.999 999 609 287 361 953 792;
  • 45) 0.999 999 609 287 361 953 792 × 2 = 1 + 0.999 999 218 574 723 907 584;
  • 46) 0.999 999 218 574 723 907 584 × 2 = 1 + 0.999 998 437 149 447 815 168;
  • 47) 0.999 998 437 149 447 815 168 × 2 = 1 + 0.999 996 874 298 895 630 336;
  • 48) 0.999 996 874 298 895 630 336 × 2 = 1 + 0.999 993 748 597 791 260 672;
  • 49) 0.999 993 748 597 791 260 672 × 2 = 1 + 0.999 987 497 195 582 521 344;
  • 50) 0.999 987 497 195 582 521 344 × 2 = 1 + 0.999 974 994 391 165 042 688;
  • 51) 0.999 974 994 391 165 042 688 × 2 = 1 + 0.999 949 988 782 330 085 376;
  • 52) 0.999 949 988 782 330 085 376 × 2 = 1 + 0.999 899 977 564 660 170 752;
  • 53) 0.999 899 977 564 660 170 752 × 2 = 1 + 0.999 799 955 129 320 341 504;
  • 54) 0.999 799 955 129 320 341 504 × 2 = 1 + 0.999 599 910 258 640 683 008;

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:


0.000 000 000 742 147 676 624 5(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 624 5(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:


0.000 000 000 742 147 676 624 5(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 -0.000 000 000 742 147 676 624 5 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