-0.000 000 000 742 147 676 624 9 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 624 9(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 9(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 9| = 0.000 000 000 742 147 676 624 9


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 9.

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 9 × 2 = 0 + 0.000 000 001 484 295 353 249 8;
  • 2) 0.000 000 001 484 295 353 249 8 × 2 = 0 + 0.000 000 002 968 590 706 499 6;
  • 3) 0.000 000 002 968 590 706 499 6 × 2 = 0 + 0.000 000 005 937 181 412 999 2;
  • 4) 0.000 000 005 937 181 412 999 2 × 2 = 0 + 0.000 000 011 874 362 825 998 4;
  • 5) 0.000 000 011 874 362 825 998 4 × 2 = 0 + 0.000 000 023 748 725 651 996 8;
  • 6) 0.000 000 023 748 725 651 996 8 × 2 = 0 + 0.000 000 047 497 451 303 993 6;
  • 7) 0.000 000 047 497 451 303 993 6 × 2 = 0 + 0.000 000 094 994 902 607 987 2;
  • 8) 0.000 000 094 994 902 607 987 2 × 2 = 0 + 0.000 000 189 989 805 215 974 4;
  • 9) 0.000 000 189 989 805 215 974 4 × 2 = 0 + 0.000 000 379 979 610 431 948 8;
  • 10) 0.000 000 379 979 610 431 948 8 × 2 = 0 + 0.000 000 759 959 220 863 897 6;
  • 11) 0.000 000 759 959 220 863 897 6 × 2 = 0 + 0.000 001 519 918 441 727 795 2;
  • 12) 0.000 001 519 918 441 727 795 2 × 2 = 0 + 0.000 003 039 836 883 455 590 4;
  • 13) 0.000 003 039 836 883 455 590 4 × 2 = 0 + 0.000 006 079 673 766 911 180 8;
  • 14) 0.000 006 079 673 766 911 180 8 × 2 = 0 + 0.000 012 159 347 533 822 361 6;
  • 15) 0.000 012 159 347 533 822 361 6 × 2 = 0 + 0.000 024 318 695 067 644 723 2;
  • 16) 0.000 024 318 695 067 644 723 2 × 2 = 0 + 0.000 048 637 390 135 289 446 4;
  • 17) 0.000 048 637 390 135 289 446 4 × 2 = 0 + 0.000 097 274 780 270 578 892 8;
  • 18) 0.000 097 274 780 270 578 892 8 × 2 = 0 + 0.000 194 549 560 541 157 785 6;
  • 19) 0.000 194 549 560 541 157 785 6 × 2 = 0 + 0.000 389 099 121 082 315 571 2;
  • 20) 0.000 389 099 121 082 315 571 2 × 2 = 0 + 0.000 778 198 242 164 631 142 4;
  • 21) 0.000 778 198 242 164 631 142 4 × 2 = 0 + 0.001 556 396 484 329 262 284 8;
  • 22) 0.001 556 396 484 329 262 284 8 × 2 = 0 + 0.003 112 792 968 658 524 569 6;
  • 23) 0.003 112 792 968 658 524 569 6 × 2 = 0 + 0.006 225 585 937 317 049 139 2;
  • 24) 0.006 225 585 937 317 049 139 2 × 2 = 0 + 0.012 451 171 874 634 098 278 4;
  • 25) 0.012 451 171 874 634 098 278 4 × 2 = 0 + 0.024 902 343 749 268 196 556 8;
  • 26) 0.024 902 343 749 268 196 556 8 × 2 = 0 + 0.049 804 687 498 536 393 113 6;
  • 27) 0.049 804 687 498 536 393 113 6 × 2 = 0 + 0.099 609 374 997 072 786 227 2;
  • 28) 0.099 609 374 997 072 786 227 2 × 2 = 0 + 0.199 218 749 994 145 572 454 4;
  • 29) 0.199 218 749 994 145 572 454 4 × 2 = 0 + 0.398 437 499 988 291 144 908 8;
  • 30) 0.398 437 499 988 291 144 908 8 × 2 = 0 + 0.796 874 999 976 582 289 817 6;
  • 31) 0.796 874 999 976 582 289 817 6 × 2 = 1 + 0.593 749 999 953 164 579 635 2;
  • 32) 0.593 749 999 953 164 579 635 2 × 2 = 1 + 0.187 499 999 906 329 159 270 4;
  • 33) 0.187 499 999 906 329 159 270 4 × 2 = 0 + 0.374 999 999 812 658 318 540 8;
  • 34) 0.374 999 999 812 658 318 540 8 × 2 = 0 + 0.749 999 999 625 316 637 081 6;
  • 35) 0.749 999 999 625 316 637 081 6 × 2 = 1 + 0.499 999 999 250 633 274 163 2;
  • 36) 0.499 999 999 250 633 274 163 2 × 2 = 0 + 0.999 999 998 501 266 548 326 4;
  • 37) 0.999 999 998 501 266 548 326 4 × 2 = 1 + 0.999 999 997 002 533 096 652 8;
  • 38) 0.999 999 997 002 533 096 652 8 × 2 = 1 + 0.999 999 994 005 066 193 305 6;
  • 39) 0.999 999 994 005 066 193 305 6 × 2 = 1 + 0.999 999 988 010 132 386 611 2;
  • 40) 0.999 999 988 010 132 386 611 2 × 2 = 1 + 0.999 999 976 020 264 773 222 4;
  • 41) 0.999 999 976 020 264 773 222 4 × 2 = 1 + 0.999 999 952 040 529 546 444 8;
  • 42) 0.999 999 952 040 529 546 444 8 × 2 = 1 + 0.999 999 904 081 059 092 889 6;
  • 43) 0.999 999 904 081 059 092 889 6 × 2 = 1 + 0.999 999 808 162 118 185 779 2;
  • 44) 0.999 999 808 162 118 185 779 2 × 2 = 1 + 0.999 999 616 324 236 371 558 4;
  • 45) 0.999 999 616 324 236 371 558 4 × 2 = 1 + 0.999 999 232 648 472 743 116 8;
  • 46) 0.999 999 232 648 472 743 116 8 × 2 = 1 + 0.999 998 465 296 945 486 233 6;
  • 47) 0.999 998 465 296 945 486 233 6 × 2 = 1 + 0.999 996 930 593 890 972 467 2;
  • 48) 0.999 996 930 593 890 972 467 2 × 2 = 1 + 0.999 993 861 187 781 944 934 4;
  • 49) 0.999 993 861 187 781 944 934 4 × 2 = 1 + 0.999 987 722 375 563 889 868 8;
  • 50) 0.999 987 722 375 563 889 868 8 × 2 = 1 + 0.999 975 444 751 127 779 737 6;
  • 51) 0.999 975 444 751 127 779 737 6 × 2 = 1 + 0.999 950 889 502 255 559 475 2;
  • 52) 0.999 950 889 502 255 559 475 2 × 2 = 1 + 0.999 901 779 004 511 118 950 4;
  • 53) 0.999 901 779 004 511 118 950 4 × 2 = 1 + 0.999 803 558 009 022 237 900 8;
  • 54) 0.999 803 558 009 022 237 900 8 × 2 = 1 + 0.999 607 116 018 044 475 801 6;

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 9(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 9(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 9(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 9 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