-0.000 000 000 742 147 676 611 4 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal -0.000 000 000 742 147 676 611 4(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
-0.000 000 000 742 147 676 611 4(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:

|-0.000 000 000 742 147 676 611 4| = 0.000 000 000 742 147 676 611 4


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: 0.000 000 000 742 147 676 611 4.

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) 0.000 000 000 742 147 676 611 4 × 2 = 0 + 0.000 000 001 484 295 353 222 8;
  • 2) 0.000 000 001 484 295 353 222 8 × 2 = 0 + 0.000 000 002 968 590 706 445 6;
  • 3) 0.000 000 002 968 590 706 445 6 × 2 = 0 + 0.000 000 005 937 181 412 891 2;
  • 4) 0.000 000 005 937 181 412 891 2 × 2 = 0 + 0.000 000 011 874 362 825 782 4;
  • 5) 0.000 000 011 874 362 825 782 4 × 2 = 0 + 0.000 000 023 748 725 651 564 8;
  • 6) 0.000 000 023 748 725 651 564 8 × 2 = 0 + 0.000 000 047 497 451 303 129 6;
  • 7) 0.000 000 047 497 451 303 129 6 × 2 = 0 + 0.000 000 094 994 902 606 259 2;
  • 8) 0.000 000 094 994 902 606 259 2 × 2 = 0 + 0.000 000 189 989 805 212 518 4;
  • 9) 0.000 000 189 989 805 212 518 4 × 2 = 0 + 0.000 000 379 979 610 425 036 8;
  • 10) 0.000 000 379 979 610 425 036 8 × 2 = 0 + 0.000 000 759 959 220 850 073 6;
  • 11) 0.000 000 759 959 220 850 073 6 × 2 = 0 + 0.000 001 519 918 441 700 147 2;
  • 12) 0.000 001 519 918 441 700 147 2 × 2 = 0 + 0.000 003 039 836 883 400 294 4;
  • 13) 0.000 003 039 836 883 400 294 4 × 2 = 0 + 0.000 006 079 673 766 800 588 8;
  • 14) 0.000 006 079 673 766 800 588 8 × 2 = 0 + 0.000 012 159 347 533 601 177 6;
  • 15) 0.000 012 159 347 533 601 177 6 × 2 = 0 + 0.000 024 318 695 067 202 355 2;
  • 16) 0.000 024 318 695 067 202 355 2 × 2 = 0 + 0.000 048 637 390 134 404 710 4;
  • 17) 0.000 048 637 390 134 404 710 4 × 2 = 0 + 0.000 097 274 780 268 809 420 8;
  • 18) 0.000 097 274 780 268 809 420 8 × 2 = 0 + 0.000 194 549 560 537 618 841 6;
  • 19) 0.000 194 549 560 537 618 841 6 × 2 = 0 + 0.000 389 099 121 075 237 683 2;
  • 20) 0.000 389 099 121 075 237 683 2 × 2 = 0 + 0.000 778 198 242 150 475 366 4;
  • 21) 0.000 778 198 242 150 475 366 4 × 2 = 0 + 0.001 556 396 484 300 950 732 8;
  • 22) 0.001 556 396 484 300 950 732 8 × 2 = 0 + 0.003 112 792 968 601 901 465 6;
  • 23) 0.003 112 792 968 601 901 465 6 × 2 = 0 + 0.006 225 585 937 203 802 931 2;
  • 24) 0.006 225 585 937 203 802 931 2 × 2 = 0 + 0.012 451 171 874 407 605 862 4;
  • 25) 0.012 451 171 874 407 605 862 4 × 2 = 0 + 0.024 902 343 748 815 211 724 8;
  • 26) 0.024 902 343 748 815 211 724 8 × 2 = 0 + 0.049 804 687 497 630 423 449 6;
  • 27) 0.049 804 687 497 630 423 449 6 × 2 = 0 + 0.099 609 374 995 260 846 899 2;
  • 28) 0.099 609 374 995 260 846 899 2 × 2 = 0 + 0.199 218 749 990 521 693 798 4;
  • 29) 0.199 218 749 990 521 693 798 4 × 2 = 0 + 0.398 437 499 981 043 387 596 8;
  • 30) 0.398 437 499 981 043 387 596 8 × 2 = 0 + 0.796 874 999 962 086 775 193 6;
  • 31) 0.796 874 999 962 086 775 193 6 × 2 = 1 + 0.593 749 999 924 173 550 387 2;
  • 32) 0.593 749 999 924 173 550 387 2 × 2 = 1 + 0.187 499 999 848 347 100 774 4;
  • 33) 0.187 499 999 848 347 100 774 4 × 2 = 0 + 0.374 999 999 696 694 201 548 8;
  • 34) 0.374 999 999 696 694 201 548 8 × 2 = 0 + 0.749 999 999 393 388 403 097 6;
  • 35) 0.749 999 999 393 388 403 097 6 × 2 = 1 + 0.499 999 998 786 776 806 195 2;
  • 36) 0.499 999 998 786 776 806 195 2 × 2 = 0 + 0.999 999 997 573 553 612 390 4;
  • 37) 0.999 999 997 573 553 612 390 4 × 2 = 1 + 0.999 999 995 147 107 224 780 8;
  • 38) 0.999 999 995 147 107 224 780 8 × 2 = 1 + 0.999 999 990 294 214 449 561 6;
  • 39) 0.999 999 990 294 214 449 561 6 × 2 = 1 + 0.999 999 980 588 428 899 123 2;
  • 40) 0.999 999 980 588 428 899 123 2 × 2 = 1 + 0.999 999 961 176 857 798 246 4;
  • 41) 0.999 999 961 176 857 798 246 4 × 2 = 1 + 0.999 999 922 353 715 596 492 8;
  • 42) 0.999 999 922 353 715 596 492 8 × 2 = 1 + 0.999 999 844 707 431 192 985 6;
  • 43) 0.999 999 844 707 431 192 985 6 × 2 = 1 + 0.999 999 689 414 862 385 971 2;
  • 44) 0.999 999 689 414 862 385 971 2 × 2 = 1 + 0.999 999 378 829 724 771 942 4;
  • 45) 0.999 999 378 829 724 771 942 4 × 2 = 1 + 0.999 998 757 659 449 543 884 8;
  • 46) 0.999 998 757 659 449 543 884 8 × 2 = 1 + 0.999 997 515 318 899 087 769 6;
  • 47) 0.999 997 515 318 899 087 769 6 × 2 = 1 + 0.999 995 030 637 798 175 539 2;
  • 48) 0.999 995 030 637 798 175 539 2 × 2 = 1 + 0.999 990 061 275 596 351 078 4;
  • 49) 0.999 990 061 275 596 351 078 4 × 2 = 1 + 0.999 980 122 551 192 702 156 8;
  • 50) 0.999 980 122 551 192 702 156 8 × 2 = 1 + 0.999 960 245 102 385 404 313 6;
  • 51) 0.999 960 245 102 385 404 313 6 × 2 = 1 + 0.999 920 490 204 770 808 627 2;
  • 52) 0.999 920 490 204 770 808 627 2 × 2 = 1 + 0.999 840 980 409 541 617 254 4;
  • 53) 0.999 840 980 409 541 617 254 4 × 2 = 1 + 0.999 681 960 819 083 234 508 8;
  • 54) 0.999 681 960 819 083 234 508 8 × 2 = 1 + 0.999 363 921 638 166 469 017 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:


0.000 000 000 742 147 676 611 4(10) =

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

6. Positive number before normalization:

0.000 000 000 742 147 676 611 4(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:


0.000 000 000 742 147 676 611 4(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 -0.000 000 000 742 147 676 611 4 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