1.175 494 350 822 287 507 968 736 537 222 197 6 Converted to 32 Bit Single Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 1.175 494 350 822 287 507 968 736 537 222 197 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
1.175 494 350 822 287 507 968 736 537 222 197 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. First, convert to binary (in base 2) the integer part: 1.
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;
  • 1 ÷ 2 = 0 + 1;

2. 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.

1(10) =

1(2)


3. Convert to binary (base 2) the fractional part: 0.175 494 350 822 287 507 968 736 537 222 197 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) 0.175 494 350 822 287 507 968 736 537 222 197 6 × 2 = 0 + 0.350 988 701 644 575 015 937 473 074 444 395 2;
  • 2) 0.350 988 701 644 575 015 937 473 074 444 395 2 × 2 = 0 + 0.701 977 403 289 150 031 874 946 148 888 790 4;
  • 3) 0.701 977 403 289 150 031 874 946 148 888 790 4 × 2 = 1 + 0.403 954 806 578 300 063 749 892 297 777 580 8;
  • 4) 0.403 954 806 578 300 063 749 892 297 777 580 8 × 2 = 0 + 0.807 909 613 156 600 127 499 784 595 555 161 6;
  • 5) 0.807 909 613 156 600 127 499 784 595 555 161 6 × 2 = 1 + 0.615 819 226 313 200 254 999 569 191 110 323 2;
  • 6) 0.615 819 226 313 200 254 999 569 191 110 323 2 × 2 = 1 + 0.231 638 452 626 400 509 999 138 382 220 646 4;
  • 7) 0.231 638 452 626 400 509 999 138 382 220 646 4 × 2 = 0 + 0.463 276 905 252 801 019 998 276 764 441 292 8;
  • 8) 0.463 276 905 252 801 019 998 276 764 441 292 8 × 2 = 0 + 0.926 553 810 505 602 039 996 553 528 882 585 6;
  • 9) 0.926 553 810 505 602 039 996 553 528 882 585 6 × 2 = 1 + 0.853 107 621 011 204 079 993 107 057 765 171 2;
  • 10) 0.853 107 621 011 204 079 993 107 057 765 171 2 × 2 = 1 + 0.706 215 242 022 408 159 986 214 115 530 342 4;
  • 11) 0.706 215 242 022 408 159 986 214 115 530 342 4 × 2 = 1 + 0.412 430 484 044 816 319 972 428 231 060 684 8;
  • 12) 0.412 430 484 044 816 319 972 428 231 060 684 8 × 2 = 0 + 0.824 860 968 089 632 639 944 856 462 121 369 6;
  • 13) 0.824 860 968 089 632 639 944 856 462 121 369 6 × 2 = 1 + 0.649 721 936 179 265 279 889 712 924 242 739 2;
  • 14) 0.649 721 936 179 265 279 889 712 924 242 739 2 × 2 = 1 + 0.299 443 872 358 530 559 779 425 848 485 478 4;
  • 15) 0.299 443 872 358 530 559 779 425 848 485 478 4 × 2 = 0 + 0.598 887 744 717 061 119 558 851 696 970 956 8;
  • 16) 0.598 887 744 717 061 119 558 851 696 970 956 8 × 2 = 1 + 0.197 775 489 434 122 239 117 703 393 941 913 6;
  • 17) 0.197 775 489 434 122 239 117 703 393 941 913 6 × 2 = 0 + 0.395 550 978 868 244 478 235 406 787 883 827 2;
  • 18) 0.395 550 978 868 244 478 235 406 787 883 827 2 × 2 = 0 + 0.791 101 957 736 488 956 470 813 575 767 654 4;
  • 19) 0.791 101 957 736 488 956 470 813 575 767 654 4 × 2 = 1 + 0.582 203 915 472 977 912 941 627 151 535 308 8;
  • 20) 0.582 203 915 472 977 912 941 627 151 535 308 8 × 2 = 1 + 0.164 407 830 945 955 825 883 254 303 070 617 6;
  • 21) 0.164 407 830 945 955 825 883 254 303 070 617 6 × 2 = 0 + 0.328 815 661 891 911 651 766 508 606 141 235 2;
  • 22) 0.328 815 661 891 911 651 766 508 606 141 235 2 × 2 = 0 + 0.657 631 323 783 823 303 533 017 212 282 470 4;
  • 23) 0.657 631 323 783 823 303 533 017 212 282 470 4 × 2 = 1 + 0.315 262 647 567 646 607 066 034 424 564 940 8;
  • 24) 0.315 262 647 567 646 607 066 034 424 564 940 8 × 2 = 0 + 0.630 525 295 135 293 214 132 068 849 129 881 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).


4. 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:


0.175 494 350 822 287 507 968 736 537 222 197 6(10) =

0.0010 1100 1110 1101 0011 0010(2)

5. Positive number before normalization:

1.175 494 350 822 287 507 968 736 537 222 197 6(10) =

1.0010 1100 1110 1101 0011 0010(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 0 positions to the left, so that only one non zero digit remains to the left of it:


1.175 494 350 822 287 507 968 736 537 222 197 6(10) =

1.0010 1100 1110 1101 0011 0010(2) =

1.0010 1100 1110 1101 0011 0010(2) × 20


7. 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 0 (a positive number)

Exponent (unadjusted): 0

Mantissa (not normalized):
1.0010 1100 1110 1101 0011 0010


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

0 + 2(8-1) - 1 =

(0 + 127)(10) =

127(10)


9. 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;
  • 127 ÷ 2 = 63 + 1;
  • 63 ÷ 2 = 31 + 1;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

127(10) =

0111 1111(2)


11. 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, by removing the excess bits, from the right (if any of the excess bits is set on 1, we are losing precision...).


Mantissa (normalized) =

1. 001 0110 0111 0110 1001 1001 0 =

001 0110 0111 0110 1001 1001


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0111 1111

Mantissa (23 bits) =
001 0110 0111 0110 1001 1001


Decimal number 1.175 494 350 822 287 507 968 736 537 222 197 6 converted to 32 bit single precision IEEE 754 binary floating point representation:

0 - 0111 1111 - 001 0110 0111 0110 1001 1001

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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