-0.000 000 000 742 147 676 619 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 619 6(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
-0.000 000 000 742 147 676 619 6(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:

|-0.000 000 000 742 147 676 619 6| = 0.000 000 000 742 147 676 619 6


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: 0.000 000 000 742 147 676 619 6.

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) 0.000 000 000 742 147 676 619 6 × 2 = 0 + 0.000 000 001 484 295 353 239 2;
  • 2) 0.000 000 001 484 295 353 239 2 × 2 = 0 + 0.000 000 002 968 590 706 478 4;
  • 3) 0.000 000 002 968 590 706 478 4 × 2 = 0 + 0.000 000 005 937 181 412 956 8;
  • 4) 0.000 000 005 937 181 412 956 8 × 2 = 0 + 0.000 000 011 874 362 825 913 6;
  • 5) 0.000 000 011 874 362 825 913 6 × 2 = 0 + 0.000 000 023 748 725 651 827 2;
  • 6) 0.000 000 023 748 725 651 827 2 × 2 = 0 + 0.000 000 047 497 451 303 654 4;
  • 7) 0.000 000 047 497 451 303 654 4 × 2 = 0 + 0.000 000 094 994 902 607 308 8;
  • 8) 0.000 000 094 994 902 607 308 8 × 2 = 0 + 0.000 000 189 989 805 214 617 6;
  • 9) 0.000 000 189 989 805 214 617 6 × 2 = 0 + 0.000 000 379 979 610 429 235 2;
  • 10) 0.000 000 379 979 610 429 235 2 × 2 = 0 + 0.000 000 759 959 220 858 470 4;
  • 11) 0.000 000 759 959 220 858 470 4 × 2 = 0 + 0.000 001 519 918 441 716 940 8;
  • 12) 0.000 001 519 918 441 716 940 8 × 2 = 0 + 0.000 003 039 836 883 433 881 6;
  • 13) 0.000 003 039 836 883 433 881 6 × 2 = 0 + 0.000 006 079 673 766 867 763 2;
  • 14) 0.000 006 079 673 766 867 763 2 × 2 = 0 + 0.000 012 159 347 533 735 526 4;
  • 15) 0.000 012 159 347 533 735 526 4 × 2 = 0 + 0.000 024 318 695 067 471 052 8;
  • 16) 0.000 024 318 695 067 471 052 8 × 2 = 0 + 0.000 048 637 390 134 942 105 6;
  • 17) 0.000 048 637 390 134 942 105 6 × 2 = 0 + 0.000 097 274 780 269 884 211 2;
  • 18) 0.000 097 274 780 269 884 211 2 × 2 = 0 + 0.000 194 549 560 539 768 422 4;
  • 19) 0.000 194 549 560 539 768 422 4 × 2 = 0 + 0.000 389 099 121 079 536 844 8;
  • 20) 0.000 389 099 121 079 536 844 8 × 2 = 0 + 0.000 778 198 242 159 073 689 6;
  • 21) 0.000 778 198 242 159 073 689 6 × 2 = 0 + 0.001 556 396 484 318 147 379 2;
  • 22) 0.001 556 396 484 318 147 379 2 × 2 = 0 + 0.003 112 792 968 636 294 758 4;
  • 23) 0.003 112 792 968 636 294 758 4 × 2 = 0 + 0.006 225 585 937 272 589 516 8;
  • 24) 0.006 225 585 937 272 589 516 8 × 2 = 0 + 0.012 451 171 874 545 179 033 6;
  • 25) 0.012 451 171 874 545 179 033 6 × 2 = 0 + 0.024 902 343 749 090 358 067 2;
  • 26) 0.024 902 343 749 090 358 067 2 × 2 = 0 + 0.049 804 687 498 180 716 134 4;
  • 27) 0.049 804 687 498 180 716 134 4 × 2 = 0 + 0.099 609 374 996 361 432 268 8;
  • 28) 0.099 609 374 996 361 432 268 8 × 2 = 0 + 0.199 218 749 992 722 864 537 6;
  • 29) 0.199 218 749 992 722 864 537 6 × 2 = 0 + 0.398 437 499 985 445 729 075 2;
  • 30) 0.398 437 499 985 445 729 075 2 × 2 = 0 + 0.796 874 999 970 891 458 150 4;
  • 31) 0.796 874 999 970 891 458 150 4 × 2 = 1 + 0.593 749 999 941 782 916 300 8;
  • 32) 0.593 749 999 941 782 916 300 8 × 2 = 1 + 0.187 499 999 883 565 832 601 6;
  • 33) 0.187 499 999 883 565 832 601 6 × 2 = 0 + 0.374 999 999 767 131 665 203 2;
  • 34) 0.374 999 999 767 131 665 203 2 × 2 = 0 + 0.749 999 999 534 263 330 406 4;
  • 35) 0.749 999 999 534 263 330 406 4 × 2 = 1 + 0.499 999 999 068 526 660 812 8;
  • 36) 0.499 999 999 068 526 660 812 8 × 2 = 0 + 0.999 999 998 137 053 321 625 6;
  • 37) 0.999 999 998 137 053 321 625 6 × 2 = 1 + 0.999 999 996 274 106 643 251 2;
  • 38) 0.999 999 996 274 106 643 251 2 × 2 = 1 + 0.999 999 992 548 213 286 502 4;
  • 39) 0.999 999 992 548 213 286 502 4 × 2 = 1 + 0.999 999 985 096 426 573 004 8;
  • 40) 0.999 999 985 096 426 573 004 8 × 2 = 1 + 0.999 999 970 192 853 146 009 6;
  • 41) 0.999 999 970 192 853 146 009 6 × 2 = 1 + 0.999 999 940 385 706 292 019 2;
  • 42) 0.999 999 940 385 706 292 019 2 × 2 = 1 + 0.999 999 880 771 412 584 038 4;
  • 43) 0.999 999 880 771 412 584 038 4 × 2 = 1 + 0.999 999 761 542 825 168 076 8;
  • 44) 0.999 999 761 542 825 168 076 8 × 2 = 1 + 0.999 999 523 085 650 336 153 6;
  • 45) 0.999 999 523 085 650 336 153 6 × 2 = 1 + 0.999 999 046 171 300 672 307 2;
  • 46) 0.999 999 046 171 300 672 307 2 × 2 = 1 + 0.999 998 092 342 601 344 614 4;
  • 47) 0.999 998 092 342 601 344 614 4 × 2 = 1 + 0.999 996 184 685 202 689 228 8;
  • 48) 0.999 996 184 685 202 689 228 8 × 2 = 1 + 0.999 992 369 370 405 378 457 6;
  • 49) 0.999 992 369 370 405 378 457 6 × 2 = 1 + 0.999 984 738 740 810 756 915 2;
  • 50) 0.999 984 738 740 810 756 915 2 × 2 = 1 + 0.999 969 477 481 621 513 830 4;
  • 51) 0.999 969 477 481 621 513 830 4 × 2 = 1 + 0.999 938 954 963 243 027 660 8;
  • 52) 0.999 938 954 963 243 027 660 8 × 2 = 1 + 0.999 877 909 926 486 055 321 6;
  • 53) 0.999 877 909 926 486 055 321 6 × 2 = 1 + 0.999 755 819 852 972 110 643 2;
  • 54) 0.999 755 819 852 972 110 643 2 × 2 = 1 + 0.999 511 639 705 944 221 286 4;

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:


0.000 000 000 742 147 676 619 6(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 619 6(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:


0.000 000 000 742 147 676 619 6(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 -0.000 000 000 742 147 676 619 6 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