-0.000 000 000 742 147 676 632 8 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 632 8(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
-0.000 000 000 742 147 676 632 8(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:

|-0.000 000 000 742 147 676 632 8| = 0.000 000 000 742 147 676 632 8


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: 0.000 000 000 742 147 676 632 8.

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) 0.000 000 000 742 147 676 632 8 × 2 = 0 + 0.000 000 001 484 295 353 265 6;
  • 2) 0.000 000 001 484 295 353 265 6 × 2 = 0 + 0.000 000 002 968 590 706 531 2;
  • 3) 0.000 000 002 968 590 706 531 2 × 2 = 0 + 0.000 000 005 937 181 413 062 4;
  • 4) 0.000 000 005 937 181 413 062 4 × 2 = 0 + 0.000 000 011 874 362 826 124 8;
  • 5) 0.000 000 011 874 362 826 124 8 × 2 = 0 + 0.000 000 023 748 725 652 249 6;
  • 6) 0.000 000 023 748 725 652 249 6 × 2 = 0 + 0.000 000 047 497 451 304 499 2;
  • 7) 0.000 000 047 497 451 304 499 2 × 2 = 0 + 0.000 000 094 994 902 608 998 4;
  • 8) 0.000 000 094 994 902 608 998 4 × 2 = 0 + 0.000 000 189 989 805 217 996 8;
  • 9) 0.000 000 189 989 805 217 996 8 × 2 = 0 + 0.000 000 379 979 610 435 993 6;
  • 10) 0.000 000 379 979 610 435 993 6 × 2 = 0 + 0.000 000 759 959 220 871 987 2;
  • 11) 0.000 000 759 959 220 871 987 2 × 2 = 0 + 0.000 001 519 918 441 743 974 4;
  • 12) 0.000 001 519 918 441 743 974 4 × 2 = 0 + 0.000 003 039 836 883 487 948 8;
  • 13) 0.000 003 039 836 883 487 948 8 × 2 = 0 + 0.000 006 079 673 766 975 897 6;
  • 14) 0.000 006 079 673 766 975 897 6 × 2 = 0 + 0.000 012 159 347 533 951 795 2;
  • 15) 0.000 012 159 347 533 951 795 2 × 2 = 0 + 0.000 024 318 695 067 903 590 4;
  • 16) 0.000 024 318 695 067 903 590 4 × 2 = 0 + 0.000 048 637 390 135 807 180 8;
  • 17) 0.000 048 637 390 135 807 180 8 × 2 = 0 + 0.000 097 274 780 271 614 361 6;
  • 18) 0.000 097 274 780 271 614 361 6 × 2 = 0 + 0.000 194 549 560 543 228 723 2;
  • 19) 0.000 194 549 560 543 228 723 2 × 2 = 0 + 0.000 389 099 121 086 457 446 4;
  • 20) 0.000 389 099 121 086 457 446 4 × 2 = 0 + 0.000 778 198 242 172 914 892 8;
  • 21) 0.000 778 198 242 172 914 892 8 × 2 = 0 + 0.001 556 396 484 345 829 785 6;
  • 22) 0.001 556 396 484 345 829 785 6 × 2 = 0 + 0.003 112 792 968 691 659 571 2;
  • 23) 0.003 112 792 968 691 659 571 2 × 2 = 0 + 0.006 225 585 937 383 319 142 4;
  • 24) 0.006 225 585 937 383 319 142 4 × 2 = 0 + 0.012 451 171 874 766 638 284 8;
  • 25) 0.012 451 171 874 766 638 284 8 × 2 = 0 + 0.024 902 343 749 533 276 569 6;
  • 26) 0.024 902 343 749 533 276 569 6 × 2 = 0 + 0.049 804 687 499 066 553 139 2;
  • 27) 0.049 804 687 499 066 553 139 2 × 2 = 0 + 0.099 609 374 998 133 106 278 4;
  • 28) 0.099 609 374 998 133 106 278 4 × 2 = 0 + 0.199 218 749 996 266 212 556 8;
  • 29) 0.199 218 749 996 266 212 556 8 × 2 = 0 + 0.398 437 499 992 532 425 113 6;
  • 30) 0.398 437 499 992 532 425 113 6 × 2 = 0 + 0.796 874 999 985 064 850 227 2;
  • 31) 0.796 874 999 985 064 850 227 2 × 2 = 1 + 0.593 749 999 970 129 700 454 4;
  • 32) 0.593 749 999 970 129 700 454 4 × 2 = 1 + 0.187 499 999 940 259 400 908 8;
  • 33) 0.187 499 999 940 259 400 908 8 × 2 = 0 + 0.374 999 999 880 518 801 817 6;
  • 34) 0.374 999 999 880 518 801 817 6 × 2 = 0 + 0.749 999 999 761 037 603 635 2;
  • 35) 0.749 999 999 761 037 603 635 2 × 2 = 1 + 0.499 999 999 522 075 207 270 4;
  • 36) 0.499 999 999 522 075 207 270 4 × 2 = 0 + 0.999 999 999 044 150 414 540 8;
  • 37) 0.999 999 999 044 150 414 540 8 × 2 = 1 + 0.999 999 998 088 300 829 081 6;
  • 38) 0.999 999 998 088 300 829 081 6 × 2 = 1 + 0.999 999 996 176 601 658 163 2;
  • 39) 0.999 999 996 176 601 658 163 2 × 2 = 1 + 0.999 999 992 353 203 316 326 4;
  • 40) 0.999 999 992 353 203 316 326 4 × 2 = 1 + 0.999 999 984 706 406 632 652 8;
  • 41) 0.999 999 984 706 406 632 652 8 × 2 = 1 + 0.999 999 969 412 813 265 305 6;
  • 42) 0.999 999 969 412 813 265 305 6 × 2 = 1 + 0.999 999 938 825 626 530 611 2;
  • 43) 0.999 999 938 825 626 530 611 2 × 2 = 1 + 0.999 999 877 651 253 061 222 4;
  • 44) 0.999 999 877 651 253 061 222 4 × 2 = 1 + 0.999 999 755 302 506 122 444 8;
  • 45) 0.999 999 755 302 506 122 444 8 × 2 = 1 + 0.999 999 510 605 012 244 889 6;
  • 46) 0.999 999 510 605 012 244 889 6 × 2 = 1 + 0.999 999 021 210 024 489 779 2;
  • 47) 0.999 999 021 210 024 489 779 2 × 2 = 1 + 0.999 998 042 420 048 979 558 4;
  • 48) 0.999 998 042 420 048 979 558 4 × 2 = 1 + 0.999 996 084 840 097 959 116 8;
  • 49) 0.999 996 084 840 097 959 116 8 × 2 = 1 + 0.999 992 169 680 195 918 233 6;
  • 50) 0.999 992 169 680 195 918 233 6 × 2 = 1 + 0.999 984 339 360 391 836 467 2;
  • 51) 0.999 984 339 360 391 836 467 2 × 2 = 1 + 0.999 968 678 720 783 672 934 4;
  • 52) 0.999 968 678 720 783 672 934 4 × 2 = 1 + 0.999 937 357 441 567 345 868 8;
  • 53) 0.999 937 357 441 567 345 868 8 × 2 = 1 + 0.999 874 714 883 134 691 737 6;
  • 54) 0.999 874 714 883 134 691 737 6 × 2 = 1 + 0.999 749 429 766 269 383 475 2;

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:


0.000 000 000 742 147 676 632 8(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 632 8(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:


0.000 000 000 742 147 676 632 8(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 -0.000 000 000 742 147 676 632 8 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